The availability of step-by-step ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(j) can make challenging problems more manageable.

## S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(j)

Solve the following systems of equations by matrix method.

Question 1.
2x – 3y = 1
3x – 2y = 4
Solution:
The give system of eqn’s is equivalent to AX = B
where A = $$\left[\begin{array}{ll} 2 & -3 \\ 3 & -2 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \end{array}\right]$$
& B = $$\left[\begin{array}{l} 1 \\ 4 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{ll} 2 & -3 \\ 3 & -2 \end{array}\right|$$ = – 4 + 9 = 5 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
Also given system of equations has unique solution.
The cofactors of R1 are ; – 2 ; – 3
The cofactors of R2 are ; 3 ; 2
∴ adj A = $$\left[\begin{array}{cc} -2 & -3 \\ 3 & 2 \end{array}\right]^{\prime}=\left[\begin{array}{ll} -2 & 3 \\ -3 & 2 \end{array}\right]$$
Thus, A-1 = $$\frac{1}{5}\left[\begin{array}{ll} -2 & 3 \\ -3 & 2 \end{array}\right]$$
Since AX = B ⇒ X = A-1B
⇒ $$\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{5}\left[\begin{array}{ll} -2 & 3 \\ -3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 4 \end{array}\right]$$
⇒ $$\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{5}\left[\begin{array}{c} -2+12 \\ -3+8 \end{array}\right]$$
= $$\frac{1}{5}\left[\begin{array}{c} 10 \\ 5 \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \end{array}\right]$$
Thus, x = 2 ; y = 1

Question 2.
2x + 3y = 23
3x + 4y = 32
Solution:
The give system of eqn’s is equivalent to AX = B
where A = $$\left[\begin{array}{ll} 2 & 3 \\ 3 & 4 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \end{array}\right]$$
& B = $$\left[\begin{array}{l} 23 \\ 32 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{ll} 2 & 3 \\ 3 & 4 \end{array}\right|$$ = 8 – 9 = 1 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
Also given system of equations has unique solution.
The cofactors of R1 are ; 4 ; – 3
The cofactors of R2 are ; – 3 ; 2
∴ adj A = $$\left[\begin{array}{cc} 4 & -3 \\ -3 & 2 \end{array}\right]^{\prime}=\left[\begin{array}{cc} 4 & -3 \\ -3 & 2 \end{array}\right]$$
∴ A-1 = $$\frac{1}{-1}\left[\begin{array}{cc} 4 & -3 \\ -3 & 2 \end{array}\right]=\left[\begin{array}{cc} -4 & 3 \\ 3 & -2 \end{array}\right]$$
Since AX = B ⇒ X = A-1B
⇒ X = $$\left[\begin{array}{cc} -4 & 3 \\ 3 & -2 \end{array}\right]\left[\begin{array}{l} 23 \\ 32 \end{array}\right]$$
⇒ $$\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -92+96 \\ 69-64 \end{array}\right]=\left[\begin{array}{l} 4 \\ 5 \end{array}\right]$$
Thus, x = 4 ; y = 5 be the sequared solution.

Question 3.
3x + y + z = 3
2x – y – z = 2
– x – y + z – 1
Solution:
The give system of eqn’s is equivalent to AX = B
where A = $$\left[\begin{array}{ccc} 3 & 1 & 1 \\ 2 & -1 & -1 \\ -1 & -1 & 1 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$
& B = $$\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{ccc} 3 & 1 & 1 \\ 2 & -1 & -1 \\ -1 & -1 & 1 \end{array}\right|$$
Expanding along R1,
= 3(- 1 – 1) – 2(2 – 1) + 1(- 2 – 1)
= – 6 – 1 – 3 = – 10 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A …(1)
Thus, given system of eqn’s has unique solution.
The cofactors of R1 are ;
$$\left|\begin{array}{cc} -1 & -1 \\ -1 & 1 \end{array}\right| ;-\left|\begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array}\right| ;\left|\begin{array}{cc} 2 & -1 \\ -1 & -1 \end{array}\right|$$
i.e. – 2 ; – 1 ; – 3
The cofactors of R2 are ;
–$$\left|\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right| ;\left|\begin{array}{cc} 3 & 1 \\ -1 & 1 \end{array}\right| ;\left|\begin{array}{cc} 3 & 1 \\ -1 & -1 \end{array}\right|$$
i.e. – 2 ; 4 ; 2
The cofactors of R3 are ;

i.e. x = 1 & y = – 1 & z = 1
Hence the required solution of given system of eqns. be, x = 1 & y = – 1 & z = 1

Question 4.
x – y – z = 2
2x – y = 0
2y – z = 1
Solution:
The give system of eqn’s is equivalent to AX = B
where A = $$\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & -1 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$
& B = $$\left[\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & -1 \end{array}\right|$$
= 1(1 – 0) + 1(- 2 – 0) + 1(4 – 0)
= 1 – 2 + 4 = 3 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
Thus, given system has unique solution.
The cofactors of R1are ;
$$\left|\begin{array}{cc} -1 & 0 \\ 2 & -1 \end{array}\right| ;-\left|\begin{array}{cc} 2 & 0 \\ 0 & -1 \end{array}\right| ;\left|\begin{array}{cc} 2 & -1 \\ 0 & 2 \end{array}\right|$$
i.e. 1 ; 2 ; 4
The cofactors of R2 are ;
$$-\left|\begin{array}{cc} -1 & 1 \\ 2 & -1 \end{array}\right| ;\left|\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right| ;-\left|\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right|$$
i.e. 1 ; – 1 ; – 2
The cofactors of R3 are ;

Question 5.
$$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}$$ = 4
$$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}$$ = 1
$$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}$$ = 2
Solution:
Putting $$\frac { 1 }{ x }$$ = u; $$\frac { 1 }{ y }$$ = v & $$\frac { 1 }{ z }$$ = w in
2u + 3v + 10w = 4
4u + 6v + 5w = 1
6u + 9v – 20w = 2
The given system of eqn’s is eqrualant to AX = B
where A = $$\left[\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right]$$; X = $$\left[\begin{array}{l} u \\ v \\ w \end{array}\right]$$
& B = $$\left[\begin{array}{l} 4 \\ 1 \\ 2 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right|$$
Taking 2, 3 & 5 common from R1; R2 & R3 respectively
= 2 x 3 x 5$$\left|\begin{array}{ccc} 1 & 1 & 2 \\ 2 & -2 & 1 \\ 3 & 3 & -4 \end{array}\right|$$;
Expanding along R1,
= 30[1(8 – 3) – 1(- 8 – 3) + 2(6 + 6)]
= 30[5 + 11 + 24] = 30 x 40
= 1200 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
Thus, given system has unique solution.
The cofactors of R1 are ;
$$\left|\begin{array}{cc} -6 & 5 \\ 9 & -20 \end{array}\right| ;-\left|\begin{array}{cc} 4 & 5 \\ 6 & -20 \end{array}\right| ;\left|\begin{array}{cc} 4 & -6 \\ 6 & 9 \end{array}\right|$$
i.e. 75 ; 110 ; 72
The cofactors of R2 are ;
$$-\left|\begin{array}{cc} 3 & 10 \\ 9 & -20 \end{array}\right| ;\left|\begin{array}{cc} 2 & 10 \\ 6 & -20 \end{array}\right| ;-\left|\begin{array}{ll} 2 & 3 \\ 6 & 9 \end{array}\right|$$
i.e. 150 ; – 100 ; 0
The cofactors of R3 are ;
$$\left|\begin{array}{cc} 3 & 10 \\ -6 & 5 \end{array}\right| ;-\left|\begin{array}{cc} 2 & 10 \\ 4 & 5 \end{array}\right| ;\left|\begin{array}{cc} 2 & 3 \\ 4 & -6 \end{array}\right|$$
i.e. 75 ; 30 ; – 24

Question 6.
x + y = 5
z + y = 7
z + x = 6
Solution:
The give system of eqn’s is equivalent to AX = B
where A = $$\left[\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$
& B = $$\left[\begin{array}{l} 5 \\ 7 \\ 6 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right|$$; Expanding along R1
= 1(1 – 0) – 1(0 – 1) + 0
= 1 + 1 = 2 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
Also, given system of eqn’s has unique solution.

Thus, x = 2 ; y = 3 & z = 4
Hence the reqd. sol of given system be x = 2 ; y = 3 & z = 4

Question 7.
5x – y = – 7
2x + 3z = 1
3y – z = 5
Solution:
The give system of eqn’s is equivalent to AX = B
where A = $$\left[\begin{array}{ccc} 5 & -1 & 0 \\ 2 & 0 & 3 \\ 0 & 3 & -1 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$
& B = $$\left[\begin{array}{c} -7 \\ 1 \\ 5 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{ccc} 5 & -1 & 0 \\ 2 & 0 & 3 \\ 0 & 3 & -1 \end{array}\right|$$; Expanding along R1
= 5(0 – 9) + 1(- 2 – 1)
= – 45 – 2 = – 47 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
Also, given system has unique solution.
The cofactors of R1 are ;

Question 8.
(i) Find the inverse of the matrix $$\left(\begin{array}{rr} 0.8 & -0.6 \\ 0.6 & 0.8 \end{array}\right)$$ and use it in solving the equations
0.8x1 – 0.6x2 – 10, 0.6x1 + 0.8x2 = 20.
(ii) Find the inverse of the matrix $$\left(\begin{array}{ll} 6 & 7 \\ 4 & 5 \end{array}\right)$$, and use it to solve the simultaneous equations 6x + 7y = 2, 4x + 5y = 3.
Solution:
(i) Let A = $$\left[\begin{array}{rr} 0.8 & -0.6 \\ 0.6 & 0.8 \end{array}\right]$$;
and |A| = $$\left|\begin{array}{rr} 0.8 & -0.6 \\ 0.6 & 0.8 \end{array}\right|$$ = 0.64 + 0.36 = 1.0
∴ |A| = 1 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
The cofactors of R1 are ; 0.8 ; -0.6
The cofactors of R2 are ; 0.6 ; 0.8
∴ adj A = $$\left[\begin{array}{rr} 0.8 & -0.6 \\ 0.6 & 0.8 \end{array}\right]^{\prime}=\left[\begin{array}{rr} 0.8 & 0.6 \\ -0.6 & 0.8 \end{array}\right]$$
Thus, A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A = $$\left[\begin{array}{rr} 0.8 & 0.6 \\ -0.6 & 0.8 \end{array}\right]$$
Given system of eqn’s is equivalent to AX = B
where A = $$\left[\begin{array}{rr} 0.8 & -0.6 \\ 0.6 & 0.8 \end{array}\right]$$; X = $$\left[\begin{array}{l} x_1 \\ x_2 \end{array}\right]$$
& B = $$\left[\begin{array}{l} 10 \\ 20 \end{array}\right]$$
Since AX = B ⇒ X = A-1B
⇒ \begin{aligned} {\left[\begin{array}{l} x_1 \\ x_2 \end{array}\right]=\left[\begin{array}{rr} 0.8 & 0.6 \\ -0.6 & 0.8 \end{array}\right]\left[\begin{array}{l} 10 \\ 20 \end{array}\right] } & =\left[\begin{array}{c} 8+12 \\ -6+16 \end{array}\right]=\left[\begin{array}{l} 20 \\ 10 \end{array}\right] \end{aligned}
Thus, x1 = 20 & ; x2 = 10 be the required solution.

(ii) Let A = $$\left[\begin{array}{ll} 6 & 7 \\ 4 & 5 \end{array}\right]$$;
Here |A| = $$\left|\begin{array}{ll} 6 & 7 \\ 4 & 5 \end{array}\right|$$
= 30 – 28 = 2 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
The cofactors of R1 are ; 5 ; – 4
The cofactors of R2 are ; – 7 ; 6
∴ adj A = $$\left[\begin{array}{rr} 5 & -4 \\ -7 & 6 \end{array}\right]^{\prime}=\left[\begin{array}{rr} 5 & -7 \\ -4 & 6 \end{array}\right]$$
Thus, A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A = $$\frac{1}{2}\left[\begin{array}{rr} 5 & -7 \\ -4 & 6 \end{array}\right]$$
Given system of eqn’s is equivalent to AX = B
where A = $$\left[\begin{array}{ll} 6 & 7 \\ 4 & 5 \end{array}\right] ; \mathrm{X}=\left[\begin{array}{l} x \\ y \end{array}\right] \& \mathrm{~B}=\left[\begin{array}{l} 2 \\ 3 \end{array}\right]$$
⇒ X = A-1B
⇒ X = $$\begin{array}{r} {\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{2}\left[\begin{array}{rr} 5 & -7 \\ -4 & 6 \end{array}\right]\left[\begin{array}{l} 2 \\ 3 \end{array}\right]} \\ =\frac{1}{2}\left[\begin{array}{l} 10-21 \\ -8+18 \end{array}\right] \end{array}$$
⇒ $$\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -11 / 2 \\ 5 \end{array}\right]$$
∴ x = $$\frac { -11 }{ 2 }$$; y = 5

Question 9.
(i) If A = $$\left[\begin{array}{rrr} 4 & -5 & -11 \\ 1 & -3 & 1 \\ 2 & 3 & -7 \end{array}\right]$$, find A-1, the system of equations
4x – 5y – 11z = 12, x – 3y + z = 1, 2x + 3y – 7z = 2.
(ii) If A = $$\left[\begin{array}{rrr} 8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6 \end{array}\right]$$, find A-1 solve the following system of linear equations : 8x – 4y + z = 5, 10x + 6z = 4, 8x + y + 6z = $$\frac { 5 }{ 2 }$$.
Solution:
(i) Given A = $$\left[\begin{array}{rrr} 4 & -5 & -11 \\ 1 & -3 & 1 \\ 2 & 3 & -7 \end{array}\right]$$;
Here |A| = $$\left|\begin{array}{rrr} 4 & -5 & -11 \\ 1 & -3 & 1 \\ 2 & 3 & -7 \end{array}\right|$$;
Expanding along R1.
= 4(+21 – 3) + 5 (-7 – 2) – 11(3 + 6)
= +72 – 45 – 99 = – 72 ≠ 0
∴ A-1 exists & A-1= $$\frac{1}{|\mathrm{~A}|}$$ adj A … (1)
The cofactors of R1 are ;

Given system of eqn’s is equivalent to AX = B
where X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & B = $$\left[\begin{array}{c} 12 \\ 1 \\ 2 \end{array}\right]$$
Since |A| ≠ 0
∴ given system of eqns has unique soln
⇒ X = A-1B

Expanding along R1.
= 8(0 – 6) + 4(60 – 48) + 1(10 – 0)
= – 48 + 48 + 10 = 10 ≠ 0
∴ A-1 exists & A-1= $$\frac{1}{|\mathrm{~A}|}$$ adj A … (1)
The cofactors of R1 are ;

The Given system of eqn’s is equivalent to AX = B
where X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & B = $$\left[\begin{array}{c} 5 \\ 4 \\ 5 / 2 \end{array}\right]$$
Since |A| ≠ 0 ∴ given system of eqns has unique solution and is given by X = A-1B

Question 10.
Find the product of two matrices A and B, where A = $$\left[\begin{array}{rrr} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{array}\right]$$ and B = $$\left[\begin{array}{lll} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right]$$ and use it to solve the system of equations x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2.
Solution:
Here

The given system of eqn’s be
x + y + 2z = 1,
3x + 2y + z = 7,
2x + y + 3z = 2.
and this system is equivalent to BX = C
where X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & C = $$\left[\begin{array}{l} 1 \\ 7 \\ 2 \end{array}\right]$$
Here |B| = $$\left|\begin{array}{lll} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right|$$; Expanding along R1.
= 1(6 – 1) – 1(9 – 2) + 2(3 – 4)
= 5 + 7 – 2 = – 4 ≠ 0 ∴ B-1 exists
and given system of eqn’s has unique solution
Since BX = C ⇒ X = B-1C

Use matrix method to examine the following systems of equations of consistency or inconsistency :

Question 11.
3x – 2y = 5
6x – 4y = 9
Solution:
Given system of equations is equivalent to AX = B
Where A = $$\left[\begin{array}{ll} 3 & -2 \\ 6 & -4 \end{array}\right] \mathrm{X}=\left[\begin{array}{l} x \\ y \end{array}\right] \text { \& } \mathrm{B}\left[\begin{array}{l} 5 \\ 9 \end{array}\right]$$;
Here |A| = $$\left|\begin{array}{ll} 3 & -2 \\ 6 & -4 \end{array}\right|$$ = – 12 + 12 = 0
∴ A be singular matrix.
The cofactors of R1 are ; – 4 ; – 6
The cofactors of R2 are ; 2 ; 3
∴ adj A = $$\left[\begin{array}{rr} -4 & -6 \\ 2 & 3 \end{array}\right]^{\prime}=\left[\begin{array}{ll} -4 & 2 \\ -6 & 3 \end{array}\right]$$
Now (adj A)B = $$\left[\begin{array}{ll} -4 & 2 \\ -6 & 3 \end{array}\right]\left[\begin{array}{l} 5 \\ 9 \end{array}\right]$$
= $$\left[\begin{array}{l} -20+18 \\ -30+27 \end{array}\right]=\left[\begin{array}{l} -2 \\ -3 \end{array}\right]$$ ≠ 0
∴ given system of eqn’s has no solution. Thus the given system is inconsistent.

Question 12.
4x – 2y = 3
6x – 3y = 5
Solution:
The Given system of equations is equivalent to AX = B
Where A = $$\left[\begin{array}{ll} 4 & -2 \\ 6 & -3 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ & B = $$\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$$
Here A = $$\left|\begin{array}{ll} 4 & -2 \\ 6 & -3 \end{array}\right|$$ = – 12 + 12 = 0
∴ A be singular matrix.
The cofactors of R1 are ; – 3 ; – 6
The cofactors of R2 are ; 2 ; 4
∴ adj A = $$\left[\begin{array}{rr} -3 & -6 \\ 2 & 4 \end{array}\right]^{\prime}=\left[\begin{array}{ll} -3 & 2 \\ -6 & 4 \end{array}\right]$$
Now (adj A)B = $$\left[\begin{array}{ll} -3 & 2 \\ -6 & 4 \end{array}\right]\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$$
= $$\left[\begin{array}{c} -9+10 \\ -18+20 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \end{array}\right]$$ ≠ 0
The given system has no solution and given system is inconsistent.

Question 13.
x + 5y = 3
2x + 10y = 6
Solution:
The Given system of equations is equivalent to AX = B
Where A = $$\left[\begin{array}{cc} 1 & 5 \\ 2 & 10 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ & B = $$\left[\begin{array}{l} 3 \\ 6 \end{array}\right]$$
Here A = $$\left|\begin{array}{cc} 1 & 5 \\ 2 & 10 \end{array}\right|$$ = 10 – 10 = 0
∴ A be singular.
The cofactors of R1 are ; 10 ; – 2
The cofactors of R2 are ; -5 ; 1
∴ adj A = $$\left[\begin{array}{rr} 10 & -2 \\ -5 & 1 \end{array}\right]^{\prime}=\left[\begin{array}{cc} 10 & -5 \\ -2 & 1 \end{array}\right]$$
Here (adj A)B = $$\left[\begin{array}{cc} 10 & -5 \\ -2 & 1 \end{array}\right]\left[\begin{array}{l} 3 \\ 6 \end{array}\right]$$
= $$\left[\begin{array}{c} 30-30 \\ -6+6 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \end{array}\right]$$ = 0
The given system has infinitly many solutions and is consistent. Let y = k and putting y = k in first eqn.
x + 5k = 3 ⇒ x = 3 – 5 k
putting x = 3 – 5k and y = k in second eqn., we have
2(3 – 5k) + 10k = 6
⇒ 6 – 10k + 10k = 6
⇒ 6 = 6, which is true.
Thus, given system of equations is consistent and has infinitly many solutions given by x = 3 – 5x; y = k ∈ R

Question 14.
3x – y + 2z = 3
2x + y + 3z = 5
x – 2y – z = 1
Solution:
The Given system of equations is equivalent to AX = B
where A = $$\left[\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & -2 & -1 \end{array}\right]$$ & X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & B = $$\left[\begin{array}{l} 3 \\ 5 \\ 1 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & -2 & -1 \end{array}\right|$$
Expanding along R1.
= 3(- 1 + 6) + 1(- 2 – 3) + 2(- 4 – 1)
= 15 – 5 – 10 = 0
Thus, the given system of eqns has either no solution or infinitly many solution.
The cofactors of R1 are ;
$$\left|\begin{array}{cc} 1 & 3 \\ -2 & -1 \end{array}\right| ;-\left|\begin{array}{cc} 2 & 3 \\ 1 & -1 \end{array}\right| ;\left|\begin{array}{cc} 2 & 1 \\ 1 & -2 \end{array}\right|$$
i.e. 5 ; 5 ; -5
The cofactors of R2 are ;
$$-\left|\begin{array}{cc} -1 & 2 \\ -2 & -1 \end{array}\right| ;\left|\begin{array}{cc} 3 & 2 \\ 1 & -1 \end{array}\right| ;-\left|\begin{array}{cc} 3 & -1 \\ 1 & -2 \end{array}\right|$$
i.e. -5 ; -5 ; 5
The cofactors of R3 are ;

The given system of eqn’s has no solution and system is inconsistent.

Question 15.
x + y + z = 6
x + 2y + 3z = 14
x + 4y + Iz = 30
Solution:
The Given system of equations is equivalent to AX = B
where X = $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right]$$ & X = $$=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & B = $$\left[\begin{array}{c} 6 \\ 14 \\ 30 \end{array}\right]$$
Here A = $$\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right|$$;
Expanding along R1.
= 1(14 – 12) -1(7 – 3) + 1(4 – 2)
= 2 – 4 + 2 = 0
∴ A be singular matrix.
Thus, the given system of eqn’s has either no solution or infinitly many solution.
The cofactors of R1 are ;

The given system of eqn’s is consistent and has infinitly many solution.
Putting z = k in first two eqn. of given system, we get
x + y = 6 – k …(1)
x + 2y = 14 – 3k …(2)
on solving (1) & (2); we have
y = 14 – 3k – 6 + k = 8 – 2k & x = k – 2 and the values of x, y and z satisfies the 3rd eqn. of given system of eqns.
Thus, x = k – 2 ;y = 8 – 2k ; z = k,
where K ∈ R be the required infinitely many solution.

Question 16.
2x – y + 3z = 1
x + 2y – z = 2
5y – 5z = 3
Solution:
The Given system of equations is equivalent to AX = B
where X = $$\left[\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 2 & -1 \\ 0 & 5 & -5 \end{array}\right]$$ & X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & B = $$\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$$
Here A = $$\left|\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 2 & -1 \\ 0 & 5 & -5 \end{array}\right|$$;
Expanding along R1.
= 1(-10 + 5) + 1(-5 – 0) + 3(5 – 0)
= – 10 – 5 + 15 = 0
A be singular matrix.
Thus, the given system of eqn’s has either no solution or infinitly many solution.
The cofactors of R1 are ;

Thus given system of eqn’s is consistent and have infinitly may solution.
Putting z = k in first two eqn. of given system, we have
2x – y = 1 – 3k …(1)
x + 2y = 2 + k …(2)
Multiplying eqn. (1) by 2 + eqn. (2); we get
5x = 2 – 6k + 2 + k ⇒ x = $$\frac{4-5 k}{5}$$
∴ from (1) ; we have
y = $$\frac{8-10 k}{5}-1+3 k=\frac{3+5 k}{5}$$
Putting x = $$\frac{4-5 k}{5}$$ , y = $$\frac{3+5 k}{5}$$, z = k in third eqn. 5y – 5z =3
i.e. $$\left(\frac{3+5 k}{5}\right)$$ – 5k = 3 ⇒ 3 = 3, which is true.
Hence the given system of eqn’s has infinite many solutions and is given by
x = $$\frac{4-5 k}{5}$$ , y = $$\frac{3+5 k}{5}$$ & z = k
where k ∈ R.

Question 17.
Show that the following system of equations is consistent x-2y + z = 0, y-z = 3, 2x-3z = 10 Also, find the solution using matrix method.
Solution:
The given system of eqns is equivalent to AX = B where A = $$\left[\begin{array}{rrr} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 2 & 0 & -3 \end{array}\right]$$, X = $$\left[\begin{array}{l} x \\ y \\ y \end{array}\right]$$ & B = $$\left[\begin{array}{c} 0 \\ 3 \\ 10 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 2 & 0 & -3 \end{array}\right|$$;
Expanding along R1
= 1 (- 3 – 0) + 2(0 + 2) + 1(0 – 2)
= – 3 + 4 – 2 = – 1 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adjA
Thus given system of eqn’s has unique solution.
The cofactors of R1 are :

Question 18.
Find k so that the equations
3x – 2y + 2z = 1, 2x + y + 3z = – 1, x – 3y + kz = 0 may have a unique solution.
Solution:
Given system of equations is equivalent to AX = B where A = $$\left[\begin{array}{rrr} 3 & -2 & 2 \\ 2 & 1 & 3 \\ 1 & -3 & k \end{array}\right]$$ & X = $$\left[\begin{array}{l} x \\ y \\ y \end{array}\right]$$ & B = $$\left[\begin{array}{c} 1 \\ -1 \\ 0 \end{array}\right]$$
Since given system of eqn’s has unique solution.
∴ |A| ≠ 0 ⇒ $$\left|\begin{array}{rrr} 3 & -2 & 2 \\ 2 & 1 & 3 \\ 1 & -3 & k \end{array}\right|$$ ≠ 0;
Expanding along R1.
⇒ 3(k + 9) + 2(2k – 3) + 2(- 6 – 1) ≠ 0
⇒ 3k + 27 + 4k – 6 – 14 ≠ 0
⇒ 7k + 7 ≠ 0
⇒ k ≠ – 1

Question 19.
For what value off k, do the equations
2x – 3y + 2z = a
5x + 4y – 2z = – 3
x – 13y + kz = 9
not have a unique solution.
Solution:
The given system of equations is equivalent to AX = B where A = $$\left[\begin{array}{rrr} 2 & -3 & 2 \\ 5 & 4 & -2 \\ 1 & -13 & k \end{array}\right]$$ & X = $$\left[\begin{array}{l} x \\ y \\ y \end{array}\right]$$ & B = $$\left[\begin{array}{c} a \\ -3 \\ 9 \end{array}\right]$$
Here, |A| = $$\left|\begin{array}{rrr} 2 & -3 & 2 \\ 5 & 4 & -2 \\ 1 & -13 & k \end{array}\right|$$
Expanding along R1
= 2(4k – 26) + 3(5k + 2) + 2(- 65 – 4)
= 8k – 52 + 15k + 6 – 138
= 23k – 184
Since the given system do not have unique solution
∴ |A| = 0
⇒ 23k = 184
⇒ k = 8

Question 20.
Suppose the demand curve for automobiles over some time period can be written x1 = 15000 – 0.2x2 where x1 is the price of an automobile and x2 is the correponding quamtity. Suppose the supply curve is x1 = 600 + 0.4 x2. Use matrix theory to obtain x1.
Solution:
Given system of equations be,
x1 = 15000 – 0.2x2
⇒ x1 + 0.2x2 = 15000
x1 = 600 + 0.4 x2
⇒ x1 – 0.4 x2 = 600
The given system of equations is equivalent to AX = B where
A = $$\left[\begin{array}{cc} 1 & 0.2 \\ 1 & -0.4 \end{array}\right] ; \mathrm{X}=\left[\begin{array}{l} x_1 \\ x_2 \end{array}\right] \& \mathrm{~B}=\left[\begin{array}{c} 15000 \\ 600 \end{array}\right]$$
Here, |A| = $$\left|\begin{array}{cc} 1 & 0.2 \\ 1 & -0.4 \end{array}\right|$$ = – 0.4 – 0.2 = – 0.6 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
and the given system of eqn’s has unique solution.
The cofactors of R1 are : -0.4; -1
The cofactors of R2 are : -0.2; 1

Question 21.
Gaurav purchases 3 pens, 2 bags and 1 instrument box and pays ₹ 41. From the same shop, Dheeraj purchases 2 pens, 1 bag and 2 instrument boxes and pays ₹ 29, while Ankur purchases 2 pens, 2 bags and 2 instrument boxes and pays 44. Translate the problem into a system of equations. Solve the system of equation by matrix method and hence find the cost of one pen one bag and one instrument box.
Solution:
Let the cost of one pen be Rs x and of one bag be Rs y and of one instrument box be Rs z.
Thus the mathematical modiling of given problem be given as under :
3x + 2y + z = 41
2x + y + 2z = 29
2x + 2y + 2z = 44
Given system of eqns is equivalent to
AX = B where A = $$\left[\begin{array}{lll} 3 & 2 & 1 \\ 2 & 1 & 2 \\ 2 & 2 & 2 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & B = $$\left[\begin{array}{l} 41 \\ 29 \\ 44 \end{array}\right]$$
Expanding along R1
= 3(2 – 4) – 2(4 – 4) + 1(4 – 2)
= – 6 – 0 + 2 = – 4 ≠ 0
∴ A-1 exists & A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adjA
Also, given system of eqn’s has unique solution.
The cofactors of R1 are ;

Hence the required cost of each pen be Rs 2, each bag be Rs 15 and cost of each instrument be Rs. 5

Question 22.
A school wants to award its students for the value of Honesty, Regularity and Hard work with a total cash award of ₹ 6000. Three times the award money for Hardwork added to that given for Honesty amounts to ₹ 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically, and find the award money for each value, using matrix method.
Solution:
Let the award money for honesty. Regurity and hardwork be ₹ x, ₹ y and ₹ z respectively.
The mathematical modelling of given problem is given as under :
x + y + z = 6000
x + 3z = 11,000
x + z = 2y
Given system of eqns is equivalent to
AX = B where A = $$\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & B = $$\left[\begin{array}{c} 6000 \\ 11000 \\ 0 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{array}\right|$$;
Expanding along R1
= 1(0 + 6) – 1(1 – 3) + 1(- 2 – 0)
= 6 + 2 – 2 = 6 ≠ 0
∴ A-1 exists and given system of eqn’s has unique solution.
The cofactors of R1 are :

Thus, x = 500 ; y = 2000 ; z = 3500
Hence the award money given by honesty, regularity and hard work be ₹ 500, ₹ 2000 & ₹ 3500 respectively.

Examples

Question 1.
Using matrices, solve the following system of equations :
x1 – 2x2 + 3x3 = 4
2x1 + x2 – 3x3 = 5
– x1 + x2 + 2x3 = 3
Solution:
Given system of eqn’s is equivalent to AX =B
where A = $$\left[\begin{array}{rrr} 1 & -2 & 3 \\ 2 & 1 & -3 \\ -1 & 1 & 2 \end{array}\right]$$; X = $$\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]$$; B = $$\left[\begin{array}{l} 4 \\ 5 \\ 3 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & -2 & 3 \\ 2 & 1 & -3 \\ -1 & 1 & 2 \end{array}\right|$$;
Expanding along R1
= 1(2 + 3) + 2(4 – 3) + 3(2 + 1)
= 5 + 2 + 9 = 16 ≠ 0
∴ A-1 exists
and system of eqns has unique solution given by X = A-1 B
The cofactors of R1 are :
$$\left|\begin{array}{rr} 1 & -3 \\ 1 & 2 \end{array}\right| ;-\left|\begin{array}{rr} 2 & -3 \\ -1 & 2 \end{array}\right| ;\left|\begin{array}{rr} 2 & 1 \\ -1 & 1 \end{array}\right|$$
i.e. 5; – 1 ; 3
The cofactors of R2 are ;
$$-\left|\begin{array}{rr} -2 & 3 \\ 1 & 2 \end{array}\right| ;\left|\begin{array}{rr} 1 & 3 \\ -1 & 2 \end{array}\right| ;-\left|\begin{array}{rr} 1 & -2 \\ -1 & 1 \end{array}\right|$$
i.e. 7; 5; +1
The cofactors of R3 are :
$$\left|\begin{array}{rr} -2 & 3 \\ 1 & -3 \end{array}\right| ;-\left|\begin{array}{rr} 1 & 3 \\ 2 & -3 \end{array}\right| ;\left|\begin{array}{rr} 1 & -2 \\ 2 & 1 \end{array}\right|$$
i.e. 3; 9; 5

Question 2.
Using matrix method, solve the following system of equations :
x – 2y + 3z = 6
x + 4y + z = 12
x – 3y + 2z = 1
Solution:
The given system of eqn’s is equivalent
where A = $$\left[\begin{array}{rrr} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \\ y \end{array}\right]$$; B = $$\left[\begin{array}{c} 6 \\ 12 \\ 1 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{array}\right|$$;
Expanding along R1
= 1(8 + 3) + 2(2 – 1) + 3(-3 – 4)
= 11 + 2 – 21 = – 8 ≠ 0
∴ A-1 exists
Thus given system of eqn’s has unique solution given by X = A-1 B
The cofactors of R1 are :

Thus, x = 1, y = 2 ; z = 3

Question 3.
Find the inverse of the matrix A = $$\left[\begin{array}{ll} 3 & 1 \\ 4 & 2 \end{array}\right]$$.
Solution:
Given A = $$\left[\begin{array}{ll} 3 & 1 \\ 4 & 2 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{ll} 3 & 1 \\ 4 & 2 \end{array}\right|$$ = 6 – 4 = 2 ≠ 0
∴ A-1 exists
The cofactors of R1 are : 2, – 4
The cofactors of R2 are : – 1, 3
∴ adj A = $$\left[\begin{array}{cc} 2 & -4 \\ -1 & 3 \end{array}\right]^{\prime}=\left[\begin{array}{cc} 2 & -1 \\ -4 & 3 \end{array}\right]$$
Thus, A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A = $$\frac{1}{2}\left[\begin{array}{cc} 2 & -1 \\ -4 & 3 \end{array}\right]$$

Question 4.
If $$\left[\begin{array}{lll} x^2 & 3 & 4 \\ 1 & 9 & 8 \end{array}\right]+\left[\begin{array}{crr} -3 x & 1 & -5 \\ -3 & -2 & -6 \end{array}\right]$$ = $$\left[\begin{array}{rrr} 4 & 4 & -1 \\ -2 & 7 & 2 \end{array}\right]$$, find the value sof x.
Solution:
Given $$\left[\begin{array}{lll} x^2 & 3 & 4 \\ 1 & 9 & 8 \end{array}\right]+\left[\begin{array}{crr} -3 x & 1 & -5 \\ -3 & -2 & -6 \end{array}\right]$$ = $$\left[\begin{array}{rrr} 4 & 4 & -1 \\ -2 & 7 & 2 \end{array}\right]$$
⇒ $$\left[\begin{array}{ccc} x^2-3 x & 3+1 & 4-5 \\ 1-3 & 9-2 & 8-6 \end{array}\right]=\left[\begin{array}{ccc} 4 & 4 & -1 \\ -2 & 7 & 2 \end{array}\right]$$
⇒ $$\left[\begin{array}{ccc} x^2-3 x & 4 & -1 \\ -2 & 7 & +2 \end{array}\right]=\left[\begin{array}{ccc} 4 & 4 & -1 \\ -2 & 7 & 2 \end{array}\right]$$
Thus their corresponding entries are equal.
∴ x² – 3x = 4
⇒ x² – 3x – 4 = 0
⇒ (x + 1)(x – 4) = 0
⇒ x = – 1, 4

Question 5.
Using matrices, solve the following system of equations :
x + 2y = 5, y + 2z = 8, 2x + z = 5
Solution:
The given system of eqn’s is equivalent to AX = B
where A = $$\left[\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & B = $$\left[\begin{array}{l} 5 \\ 8 \\ 5 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1 \end{array}\right|$$;
Expanding along R1
= 1(1 – 0) – 2(0 – 4) + 0
∴ |A| = 1 + 8 = 9 ≠ 0
∴ A-1 exists given system of eqn’s has unique solution given by X = A-1 B
The cofactors of R1 are :

Thus, x = 1 ; y = 2 ; z = 3

Question 6.
If the matrix A = $$\left[\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right]$$, show that A-1 = $$\frac { 1 }{ 19 }$$A.
Solution:
Given A = $$\left[\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right|$$ = – 4 – 15 = – 19 ≠ 0
Thus A-1 exists and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A … (1)
The cofactors of R1 are : -2 ; -5
The cofactors of R2 are : -3 ; 2
∴ adj A = $$\left[\begin{array}{cc} -2 & -5 \\ -3 & 2 \end{array}\right]^{\prime}=\left[\begin{array}{cc} -2 & -3 \\ -5 & 2 \end{array}\right]$$
∴ from (1); A-1 = – $$\frac{1}{19}\left[\begin{array}{cc} -2 & -3 \\ -5 & 2 \end{array}\right]$$
= $$\frac{1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]=\frac{1}{19}$$A

Question 7.
Given that A = $$\left[\begin{array}{rr} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right]$$ and A (adj A) = k$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$, find the value of k.
Solution:

Question 8.
Given the matrix A = $$\left[\begin{array}{rrr} 1 & 0 & 2 \\ -2 & 1 & 0 \\ 0 & -1 & 2 \end{array}\right]$$, compute A-1.
Solution:
Given A = $$\left[\begin{array}{rrr} 1 & 0 & 2 \\ -2 & 1 & 0 \\ 0 & -1 & 2 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & 0 & 2 \\ -2 & 1 & 0 \\ 0 & -1 & 2 \end{array}\right|$$;
Expanding along R1
= 1(2 – 0) + 0 + 2(2 – 0)
= 2 + 4 ≠ 0
∴ A-1 exists and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A … (1)
The cofactors of R1 are :

Question 9.
Find the adjoint of the matrix A = $$\left[\begin{array}{rrr} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & -6 & -7 \end{array}\right]$$ and hrence find the matrix A-1.
Solution:
Given A = $$\left[\begin{array}{rrr} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & -6 & -7 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & -6 & -7 \end{array}\right|$$;
Expanding along R1
= 1(- 28 + 30) + 0 – 1(- 18 – 0)
= 2 + 18 = 20 ≠ 0
∴ A-1 exists and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A … (1)

Question 10.
If A = $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]$$, show that A² – 3I = 2A.
Solution:
Given A = $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]$$
L.H.S = A² – 3I
= $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ll} 1+4 & 2+2 \\ 2+2 & 4+1 \end{array}\right]-\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]$$
= $$\left[\begin{array}{ll} 5-3 & 4-0 \\ 4-0 & 5-3 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 4 & 2 \end{array}\right]$$
= 2$$\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]$$ = 2A
= R.H.S
Thus, A² – 3I = 2A

Question 11.
Solve the following system of equations using matrices :
x + y + z = 6, x – y + z = 2,
Solution:
The given system of eqns is equivalent to AX = B where A = $$\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{array}\right]$$; X = $$\left[\begin{array}{l} x \\ y \\ y \end{array}\right]$$; B = $$\left[\begin{array}{l} 6 \\ 2 \\ 1 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{array}\right|$$;
Expanding along R1
|A| = 1(1 – 1) – 1(- 1 – 2) + 1(1 + 2)
= 0 + 3 + 3 ≠ 0
∴ A-1 exists and given system of eqn’s has unique solution given by X = A-1 B
The cofactors of R1 are :

Question 12.
Find x and y, if x + y = $$\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]$$ and x – y = $$\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]$$.
Solution:

Question 13.
If A = $$\left[\begin{array}{rrr} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{array}\right]$$, find A-1 and hence solve the following system of linear equations:
x + 2y – 3z = – 4, 2x + 3y + 2z = 2, 3x – 3y – 4z = 11.
Solution:
Given A = $$\left[\begin{array}{rrr} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{array}\right|$$;
Expanding along R1
= 1(- 12 + 6) – 2(- 8 – 6) – 3(- 6 – 9)
= – 6 + 28 + 45 = 67 ≠ 0
∴ A-1 exists and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A … (1)
The cofactors of R1 are :

Thus, x = 3 ; y = – 2 & z = 1 be the required solution.

Question 14.
If M (θ) = $$\left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$$, show that M(x) M(y) = M(x + y)
Solution:

Question 15.
Solve the following linear equations using the matrix method :
x + y + z = 9,
2x + 5y + 7z = 52,
2x + y – z = 0.
Solution:
The given system of eqns is equivalent to AX = B where A = $$\left[\begin{array}{rrr} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{array}\right]$$ & X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ ; B = $$\left[\begin{array}{c} 9 \\ 52 \\ 0 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{array}\right|$$;
Expanding along R1
|A| = 1(- 5 – 7) – 1(- 2 – 14) + 1(2 – 10)
= – 12 + 16 – 8 = – 4 ≠ 0
∴ A-1 exists and given system of eqn’s has unique solution given by X = A-1 B
The cofactors of R1 are :

Thus, x = 1; y = 3 ; z = 5 be the required solution

Question 16.
If the matrix A = $$\left[\begin{array}{rrr} 6 & x & 2 \\ 3 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right]$$ is a singular matrix, find the value of x.
Solution:
Given A = $$\left[\begin{array}{rrr} 6 & x & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right]$$
Since A be a singular matrix.
∴ |A| = 0
i.e., $$\left|\begin{array}{rrr} 6 & x & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|$$ = 0
Expanding along R1;
we have 6(- 2 – 10) – x(4 + 20) + 2(10 – 10) = 0 ⇒ – 72 – 24x = 0 ⇒ x = – 3

Question 17.
Using matrix method, solve the following equations:
5x + 3y + z = 16, 2x + y + 3z = 19, x + 2y + 4z = 25.
Solution:
The given system of eqns is equivalent to AX = B where A = $$\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right]$$ & X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ ; B = $$\left[\begin{array}{l} 16 \\ 19 \\ 25 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right|$$; Expanding along R1
= 5(4 – 6) – 3(8 – 3) + 1(4 – 1)
= – 10 – 15 + 3 = – 22 ≠ 0
∴ A-1 exists and given system of eqn’s has unique solution given by X = A-1 B … (1)
The cofactors of R1 are :

Thus, x = 1 ; y = 2 ; z = 5 be the required solution

Question 18.
If A = $$\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]$$, find x such that A² = xA – 2I. Hence, find A-1.
Solution:
Given A = $$\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right|$$ = – 6 + 8 = 2 ≠ 0
∴ A-1 exists
Since A² = xA – 2I … (1)
$$\left[\begin{array}{ll} 3 & –2 \\ 4 & -2 \end{array}\right]\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]=x\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
⇒ $$\left[\begin{array}{cc} 9-8 & -6+4 \\ 12-8 & -8+4 \end{array}\right]=\left[\begin{array}{cc} 3 x & -2 x \\ 4 x & -2 x \end{array}\right]-\left[\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right]$$
$$\left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 x-2 & -2 x \\ 4 x & -2 x-2 \end{array}\right]$$
Thus corresponding elements
∴ 1 = 3x – 2 ⇒ 3x = 3 ⇒ x = 1
-2x = – 2 ⇒ x = 1
Also, 4x = 4 ⇒ x = 1
and – 4 = – 2x – 2 ⇒ x = 1
Hence, x = 1
∴ from (1) ; A² = A – 2I
2I = A – A² … (2)
pre multiplying both sides of eqn (2) by A-1 ; we have

Question 19.
Solve the following system of equations using matrix method :
$$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4, \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1,$$, $$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$$
Solution:
Putting $$\frac { 1 }{ x }$$ = u; $$\frac { 1 }{ y }$$ = v & $$\frac { 1 }{ z }$$ = w in
given system of eqn’s ; we have
2u + 3v + 10w = 4
4u + 6v + 5w = 1
6u + 9v – 20w = 2
The given system of eqn’s is eqrualant to AX = B
where A = $$\left[\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right]$$ & X = $$\left[\begin{array}{l} u \\ v \\ w \end{array}\right]$$ ; B = $$\left[\begin{array}{l} 4 \\ 1 \\ 2 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right|$$
Taking 2, 3 & 5 common from R1; R2 & R3 respectively
= 2 x 3 x 5$$\left|\begin{array}{ccc} 1 & 1 & 2 \\ 2 & -2 & 1 \\ 3 & 3 & -4 \end{array}\right|$$ ;
Expanding along R1
= 30[1(8 – 3) – 1(- 8 – 3) +2(6 + 6)]
= 30[5 + 11 + 24]
= 30 x 40
= 1200 ≠ 0
∴ A-1 exists and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A … (1)
Thus, given system has unique solution.
The cofactors of R1 are ;

Question 20.
Solve for x and y if
$$\left[\begin{array}{l} x^2 \\ y^2 \end{array}\right]+2\left[\begin{array}{l} 2 x \\ 3 y \end{array}\right]=3\left[\begin{array}{c} 7 \\ -3 \end{array}\right]$$.
Solution:
Given $$\left[\begin{array}{l} x^2 \\ y^2 \end{array}\right]+2\left[\begin{array}{l} 2 x \\ 3 y \end{array}\right]=3\left[\begin{array}{c} 7 \\ -3 \end{array}\right]$$
⇒ $$\left[\begin{array}{l} x^2 \\ y^2 \end{array}\right]+\left[\begin{array}{l} 4 x \\ 6 y \end{array}\right]=\left[\begin{array}{c} 21 \\ -9 \end{array}\right]$$
⇒ $$\left[\begin{array}{l} x^2+4 x \\ y^2+6 y \end{array}\right]=\left[\begin{array}{l} 21 \\ -9 \end{array}\right]$$
Thus corresponding entries are equal.
∴ x² + 4x – 21 = 0
⇒ (x – 3) (x + 7) = 0
⇒ x = 3, – 7
& y² + 6y = – 9
⇒ y² + 6y + 9 = 0
⇒ (y + 3)² = 0
⇒ y = – 3

Question 21.
Find the product of the matrices A and B, where
A = $$\left[\begin{array}{rrr} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{array}\right]$$, B = $$\left[\begin{array}{lll} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right]$$.
Hence, solve the following equations by matrix method :
x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2.
Solution:

The given system of eqn’s be
x + y + 2z = 1,
3x + 2y + z = 7,
2x + y + 3z = 2.
and this system is equivalent to BX = C
where X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ & C = $$\left[\begin{array}{l} 1 \\ 7 \\ 2 \end{array}\right]$$
Here |B| = $$\left|\begin{array}{lll} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right|$$; Expanding along R1
= 1(6 – 1) -1(9 – 2) +2(3- 4)
= 5 + 7 – 2 = – 4 ≠ 0
∴ B-1 exists
and given system of eqn’s has unique solution
Since BX = C ⇒ X = B-1C
⇒ $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{4}\left[\begin{array}{ccc} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{array}\right]\left[\begin{array}{l} 1 \\ 7 \\ 2 \end{array}\right]$$
= $$\frac{1}{4}\left[\begin{array}{c} -5+7+6 \\ 7+7-10 \\ 1-7+2 \end{array}\right]=\frac{1}{4}\left[\begin{array}{c} 8 \\ 4 \\ -4 \end{array}\right]$$
⇒ $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ 1 \\ -1 \end{array}\right]$$
∴ x = 2 ; y = 1 and z = – 1

Question 22.
If (A – 2I) (A – 3I) = 0, when A = $$\left(\begin{array}{rr} 4 & 2 \\ -1 & x \end{array}\right)$$ and I = $$\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)$$, find the value of x.
Solution:
Given A = $$\left(\begin{array}{rr} 4 & 2 \\ -1 & x \end{array}\right)$$ and I = $$\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)$$

Thus their corresponding entries are equal.
2x – 2 = 0
⇒ x = 1; 1 – x = 0 ⇒ x = 1
x² – 5x + 4 = 0 ⇒ x = 1, 4
Hence, the common value of x be 1

Question 23.
Find A-1, where A = $$\left[\begin{array}{rrr} 4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{array}\right]$$.
Hence, solve the following system of linear equation :
4x + 2y + 3z = 2
x + y + z = 1
3x + y – 2z = 5
Solution:
Given A = $$\left[\begin{array}{rrr} 4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{array}\right|$$;
Expanding along R1
= 4(-2 – 1) – 2(-2 – 3) + 3(1 – 3)
= -12 + 10 – 6 = – 18 ≠ 0
∴ A-1 exists and given system of eqn’s has unique solution

Question 24.
If A = $$\left[\begin{array}{ll} 3 & 1 \\ 7 & 5 \end{array}\right]$$, find the values of x and y such that A² + xI2 = yA.
Solution:
Given A = $$\left[\begin{array}{ll} 3 & 1 \\ 7 & 5 \end{array}\right]$$
Also A² + xI2 = yA
⇒ $$\left[\begin{array}{ll} 3 & 1 \\ 7 & 5 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 7 & 5 \end{array}\right]+x\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=y\left[\begin{array}{ll} 3 & 1 \\ 7 & 5 \end{array}\right]$$
⇒ $$\left[\begin{array}{cc} 9+7 & 3+5 \\ 21+35 & 7+25 \end{array}\right]+\left[\begin{array}{ll} x & 0 \\ 0 & x \end{array}\right]=\left[\begin{array}{cc} 3 y & y \\ 7 y & 5 y \end{array}\right]$$
⇒ $$\left[\begin{array}{cc} 16+x & 8 \\ 56 & 32+x \end{array}\right]=\left[\begin{array}{cc} 3 y & y \\ 7 y & 5 y \end{array}\right]$$
Thus their corresponding entries are equal
∴ 16 + x = 3y … (1) ; y = 8
∴ from (1); 16 + x = 24 ⇒ x = 8
Also 7y = 56 ⇒ y = 8
& 32 + x = 5y ⇒ 5y = 40 ⇒ y = 8
Thus x = 8 & y = 8

Question 25.
Using matrix method, solve the following system of equations :
x – 2y = 10, 2x + y + 3z = 8 and – 2y + z = 7.
Solution:
The given system of eqns is equivalent to AX = B where A = $$\left[\begin{array}{rrr} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right]$$
& X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$; B = $$\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right|$$;
Expanding along R1
= 1(1 + 6) + 2(2 – 0) + 0 = 11 ≠ 0
A-1 exists and given system of eqn’s has unique solution given by X = A-1 B … (1)
The cofactors of R1 are :
Thus, x = 4 ; y = – 3 ; z = 1 be the required solution.

Question 26.
Find the value of k if M = $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]$$ and M² – AM – I2 = 0.
Solution:
Given M = $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]$$

Thus corresponding entries are equal.
∴ 5 – k = 0 ⇒ k = 4
8 – 2k = 0 ⇒ k = 4
Also k = 4 satisfies 13 – 3k – 1 = 0
Thus required value of k = 4

Question 27.
Given two matrices A and B
A = $$\left[\begin{array}{rrr} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{array}\right]$$ and B = $$\left[\begin{array}{rrr} 11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{array}\right]$$, find AB and use this result to solve the following system of equations :
x – 2y + 3z = 6, x + 4y + z = 12, x – 3y + 2z = 1.
Solution:
Given A = $$\left[\begin{array}{rrr} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{array}\right]$$

Expanding along R1; we have
= 1(8 + 3) + 2(2 – 1) + 3(- 3 – 4)
= 11 + 2 – 21 = – 8 ≠ 0
∴ A-1 exists and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A … (1)
Since AB = – 8I3 ; pre-multiplying both sides by A-1 ; we have

Thus, x = 1; y = 2 & z = 3 be the required solution.

Question 28.
Find the matrix x for which
$$\left[\begin{array}{ll} 5 & 4 \\ 1 & 1 \end{array}\right] X=\left[\begin{array}{rr} 1 & -2 \\ 1 & 3 \end{array}\right]$$.
Solution:

Question 29.
Solve the following system of linear equations using matrix method :
3x + y + z = 1, 2x + 2z = 0, 5x + y + 2z = 2
Solution:
The given system of eqns is equivalent to AX = B; where A = $$\left[\begin{array}{rrr} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right]$$
& X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$; B = $$\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right|$$;
Expanding along R1 we have
= 3(0 – 2) – 1(4 – 10) + 1(2 – 0)
= – 6 + 6 + 2 = 2 ≠ 0
∴ A-1 exists and given system of eqn’s has unique solution
The cofactors of R1 are :

Question 30.
For a 3 x 3 matrix A = [aij] where elements are given by aij = $$\frac{|i-j|}{2}$$ then the value of the element a23 is ……………….
Solution:
Given aij = $$\frac{|i-j|}{2}$$
∴ a23 = $$\frac{|2-3|}{2}=\frac{1}{2}$$

Question 31.
(i) If a matrix has 5 elements, then all possible orders it can have are ………………
(ii) The product of m x p and p x n matrix is an ……………… matrix.
Solution:
(i) We know that, a matrix of order m x n have mn elements. Now we want to find all possible orders of matrix having 5 elements, thus we find all ordered pairs, the product of whose components is 5.
Thus, such possible ordered pairs are (1, 5), (5, 1).
Hence possible orders are 1 x 5 ; 5 x 1.

(ii) We know that, the product of two matrices is possible.
When no. of columns in first matrix = no. of rows in second matrix.
Thus, the product of m x p and p x n matrix is a m x n matrix.

Question 32.
If $$\left[\begin{array}{cc} x y & 4 \\ z+6 & x+y \end{array}\right]=\left[\begin{array}{cc} 8 & w \\ 0 & 6 \end{array}\right]$$, write the value of (x + y + z) = ………………
Solution:
Given
Thus, their corresponding elements are equal.
∴ xy = 8 …(1);
w = 4; z + 6 = 0 ⇒ z = – 6
and x + y = 6 …(2)
from (1) ; y = $$\frac { 8 }{ x }$$
∴ from (2); x + $$\frac { 8 }{ x }$$ = 6
⇒ x² – 6x + 8 = 0
⇒ (x – 2) (x – 4) = 0 ⇒ x = 2, 4
When x = 2 ∴y = $$\frac { 8 }{ 2 }$$ = 4
When x = 4 ∴ y = 2
When x = 4 ; y = 4 ; z = – 6
Then x + y + z = 1 + 2 – 6 = 0
When x = 2, y = 4 ; z = – 6
Then x + y + z = 2 + 4 – 6 = 0

Question 33.
If 2$$\left[\begin{array}{ll} 1 & 0 \\ 0 & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 0 \\ 1 & 8 \end{array}\right]$$, then x + y = ………………..
Solution:
Given, 2$$\left[\begin{array}{ll} 1 & 0 \\ 0 & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 0 \\ 1 & 8 \end{array}\right]$$
⇒ $$\left[\begin{array}{cc} 2+y & 0 \\ 1 & 2 x+2 \end{array}\right]=\left[\begin{array}{ll} 5 & 0 \\ 1 & 8 \end{array}\right]$$
∴ 2 + y = 5 ⇒ y = 3
and 2x + 2 = 8 ⇒ x = 3
∴ x + y = 3 + 3 = 6

Question 34.
If $$\left[\begin{array}{cc} a+b & 2 \\ 5 & b \end{array}\right]=\left[\begin{array}{ll} 6 & 2 \\ 5 & 2 \end{array}\right]$$, then find a = ……………
Solution:
Given $$\left[\begin{array}{cc} a+b & 2 \\ 5 & b \end{array}\right]=\left[\begin{array}{ll} 6 & 2 \\ 5 & 2 \end{array}\right]$$
Thus, their corresponding elements are equal.
∴ a + b = 6 … (1)
and b = 2
∴ from (1); a = 6 – 2 = 4

Question 35.
If [2x 4]$$\left[\begin{array}{r} x \\ -8 \end{array}\right]$$ value of x is ……………..
Solution:
Given [2x 4]
⇒ [2x² – 32] = 0 ⇒ 2x² – 32 = 0 ⇒ x = ± 4

Question 36.
A 2 x 2 matrix which is both symmetric and skew symmetric is ……………….
Solution:
Given A is symmetric matrix ∴ A’ = A
and A is skew symmetric ∴ A’ = – A
∴ A = – A ⇒ 2A = O
⇒ A = 0 = $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$

Question 37.
A square matrix A = [aij]nxn is said to be
(i) Symmetric if ………………. and
(ii) skew symmetric if ……………….
Solution:
(i) A’ = A
(ii) A’ = – A

Question 38.
If A and B are square matrices of the same order, then
(i) (AB)’ = ……………….
(ii) (A’)’ = ……………….
Solution:
(i) (AB)’ = B’A’
(ii) (A’)’ = A

Question 39.
If A and B are invertible matrices and AB = BA = I, then (AB)-1 = ……………….
Solution:
Given AB = BA = I
∴ (AB)-1 = I-1 = I

Question 40.
The adjoint of a square matrix is the transpose of the matrix of ……………….
Solution:
cofactors

Question 41.
If A is a square matrix, then A is not invertible if ……………….
Solution:
Since A-1 = $$\frac{1}{|A|}$$ adj A
∴ A-1 exists if |A| ≠ 0
Thus A is not invertible if | A | = 0

Question 42.
If A is a matrix of order 3 x 3, then | kA | is equal to …………..
Solution:
Then | kA | = kn|A|

Question 43.
If A = $$\left[\begin{array}{ll} 5 & 2 \\ 7 & 3 \end{array}\right]$$, then (adj A) = ………………
Solution:
Given A = $$\left[\begin{array}{ll} 5 & 2 \\ 7 & 3 \end{array}\right]$$
The cofactors of R1 are ; 3 ; – 7
The cofactors of R2 are ; – 2 ; 5
∴ adj A = $$\left[\begin{array}{rr} 3 & -7 \\ -2 & 5 \end{array}\right]^{\prime}=\left[\begin{array}{rr} 3 & -2 \\ -7 & 5 \end{array}\right]$$

Question 44.
If $$\left[\begin{array}{c} x-y-z \\ -y+z \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 5 \\ 3 \end{array}\right]$$, then the values of x, y and z are respectively.
(a) 5, 2, 2
(b) 1, – 2, 3
(c) 0, – 3, 3
(d) 11, 8, 3
Solution:
Given $$\left[\begin{array}{c} x-y-z \\ -y+z \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 5 \\ 3 \end{array}\right]$$
∴ x – y – z = 0; – y + z = 5 and z = 3
⇒ – y + 3 = 5
⇒ y = – 2
∴ x + 2 – 3 = 0
⇒ x = 1

Question 45.
If A = $$\left[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}\right]$$ and A² is the identity matrix, then x is equal to
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
Given A = $$\left[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}\right]$$ and A² = I
⇒ $$\left[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
⇒ $$\left[\begin{array}{cc} x^2+1 & x \\ x & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
∴ x² + 1 = 1
⇒ x² = 0 ⇒ x = 0

Question 46.
If A = $$\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]$$ and I is the unit matrix of order 2, then A² equals
(a) 4A – 31
(b) 3A – 4I
(c) A – I
(d) A + I
Solution:
A² = A . A
= $$\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{rr} 5 & -4 \\ -4 & 5 \end{array}\right]$$
Now 4A – 3I = 4$$\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rr} 8 & -4 \\ -4 & 8 \end{array}\right]-\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]=\left[\begin{array}{rr} 5 & -4 \\ -4 & 5 \end{array}\right]$$

Question 47.
If x$$\left[\begin{array}{r} -3 \\ 4 \end{array}\right]+y\left[\begin{array}{l} 4 \\ 3 \end{array}\right]=\left[\begin{array}{r} 10 \\ -5 \end{array}\right]$$, then
(a) x = -2, y = 1
(b) x = – 9, y = 10
(c) x = 22, y = 1
(d) x = 2, y = – 1
Solution:
x$$\left[\begin{array}{r} -3 \\ 4 \end{array}\right]+y\left[\begin{array}{l} 4 \\ 3 \end{array}\right]=\left[\begin{array}{r} 10 \\ -5 \end{array}\right]$$
⇒ $$\left[\begin{array}{r} -3 x+4 y \\ 4 x+3 y \end{array}\right]=\left[\begin{array}{r} 10 \\ -5 \end{array}\right]$$
∴ – 3x + 4y = 10 …(1)
and 4x + 3y = – 5 …(2)
3 x eqn. (1) – 4 x eqn. (2); we have
– 25x = 50 ⇒ x = – 2
∴ from (1); 6 + 4y = 10 ⇒ y = 1

Question 48.
If A = $$\left[\begin{array}{cc} 3 & x-1 \\ 2 x+3 & x+2 \end{array}\right]$$ is a symmetric matrix, then the value of x is
(a) 4
(b) 3
(c) – 4
(d) – 3
Solution:
Given A = $$\left[\begin{array}{cc} 3 & x-1 \\ 2 x+3 & x+2 \end{array}\right]$$ is symmetric matrix.
∴ A’ = A
⇒ $$\left[\begin{array}{cc} 3 & 2 x+3 \\ x-1 & x+2 \end{array}\right]=\left[\begin{array}{cc} 3 & x-1 \\ 2 x+3 & x+2 \end{array}\right]$$
∴ 2x + 3 = x – 1 ⇒ x = – 4

Question 49.
If A is a square matrix, then
(a) A + AT is symmetric
(b) AAT is skew symmetric
(c) AT + A is skew symmetric
(d) ATA is skew symmetric
Solution:
Let B = A + AT
∴ BT = (A + AT)T = AT + (AT)T
= AT + A = A + AT = B
∴ B is symmetric.

Question 50.
For what value of x is the matrix A = $$\left[\begin{array}{rrr} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{array}\right]$$ is a skew matrix?
(a) – 2
(b) 0
(c) 2
(d) 3
Solution:
Given A = $$\left[\begin{array}{rrr} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{array}\right]$$
and A is skew-symmetric matrix
∴ A’ = – A
⇒ $$\left[\begin{array}{rrr} 0 & -1 & x \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{array}\right]=-\left[\begin{array}{rrr} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{array}\right]$$
⇒ $$\left[\begin{array}{rrr} 0 & -1 & x \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{array}\right]=\left[\begin{array}{rrr} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -x & 3 & 0 \end{array}\right]$$
∴ x = 2

Question 51.
If A = $$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{array}\right]$$, then A² is equal to
(a) a null matrix
(b) a unit matrix
(c) – A
(d) A
Solution:
A² = A.A = $$\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{array}\right]\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{array}\right]$$
= $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ = I3
which is a unit matrix.

Question 52.
If A + I = $$\left[\begin{array}{rr} 3 & -2 \\ 4 & 1 \end{array}\right]$$, then (A + I) (A – I) is equal to
(a) $$\left[\begin{array}{rr} -5 & -4 \\ 8 & -9 \end{array}\right]$$
(b) $$\left[\begin{array}{ll} -5 & 4 \\ -8 & 9 \end{array}\right]$$
(c) $$\left[\begin{array}{ll} 5 & 4 \\ 8 & 9 \end{array}\right]$$
(d) $$\left[\begin{array}{ll} -5 & -4 \\ -8 & -9 \end{array}\right]$$
Solution:
A + I = $$\left[\begin{array}{rr} 3 & -2 \\ 4 & 1 \end{array}\right]$$
A – I = A + I – 2 – I
= $$\left[\begin{array}{rr} 3 & -2 \\ 4 & 1 \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & -2 \\ 4 & -1 \end{array}\right]$$
∴ (A + I) (A – I) = $$\left[\begin{array}{rr} 3 & -2 \\ 4 & 1 \end{array}\right]\left[\begin{array}{rr} 1 & -2 \\ 4 & -1 \end{array}\right]$$
= $$\left[\begin{array}{rr} -5 & -4 \\ 8 & -9 \end{array}\right]$$

Question 53.
If A = $$\left[\begin{array}{rr} 1 & -2 \\ 4 & 5 \end{array}\right]$$ and f(t) = t² – 3t + 7, f(A) + $$\left[\begin{array}{rr} 3 & 6 \\ -12 & -9 \end{array}\right]$$ is equal to
(a) $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
(b) $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$
(c) $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$
(d) $$\left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$$
Solution:

Question 54.
Let A = $$\left[\begin{array}{ll} 5 & 0 \\ 1 & 0 \end{array}\right]$$ and B = $$\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right]$$. If 4A + 5B – C = 0, then C is
(a) $$\left[\begin{array}{rr} 5 & 25 \\ -1 & 0 \end{array}\right]$$
(b) $$\left[\begin{array}{rr} 20 & 5 \\ -1 & 0 \end{array}\right]$$
(c) $$\left[\begin{array}{cc} 5 & -1 \\ 0 & 25 \end{array}\right]$$
(d) $$\left[\begin{array}{rr} 5 & 25 \\ -1 & 5 \end{array}\right]$$
Solution:
Given A = $$\left[\begin{array}{ll} 5 & 0 \\ 1 & 0 \end{array}\right]$$ and B = $$\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right]$$
and 4A + 5B – C = 0
⇒ C = 4A + 5B
⇒ C = 4$$\left[\begin{array}{ll} 5 & 0 \\ 1 & 0 \end{array}\right]+5\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right]$$
= $$\left[\begin{array}{cc} 20 & 5 \\ -1 & 0 \end{array}\right]$$

Question 55.
If $$\left[\begin{array}{ll} x+1 & x-1 \\ x-3 & x+2 \end{array}\right]=\left[\begin{array}{ll} 4 & 1 \\ 1 & 3 \end{array}\right]$$, then write the value of x.
(a) 4
(b) – 3
(c) 2
(d) – 1
Solution:
Given $$\left[\begin{array}{ll} x+1 & x-1 \\ x-3 & x+2 \end{array}\right]=\left[\begin{array}{ll} 4 & 1 \\ 1 & 3 \end{array}\right]$$
Thus their corresponding entries are equal.
x + 2 = 4 ⇒ x = 2; x – 1 = 1 ⇒ x = 2
x – 3 = – 1 ⇒ x = 2 ; x + 1 = 3 ⇒ x = 2

Question 56.
If A = $$\left[\begin{array}{rr} 1 & 2 \\ 3 & -1 \end{array}\right]$$ and B = $$\left[\begin{array}{rr} 1 & 3 \\ -1 & 1 \end{array}\right]$$ write the value of |AB|.
|AB| = |A| |B| = $$\left|\begin{array}{rr} 1 & 2 \\ 3 & -1 \end{array}\right|\left|\begin{array}{rr} 1 & 3 \\ -1 & 1 \end{array}\right|$$
= (- 1 – 6) (1 + 3) = – 7 x 4 = – 28

Question 57.
If for any 2 x 2 square matrix A, A (adj A) = $$\left|\begin{array}{ll} 8 & 0 \\ 0 & 8 \end{array}\right|$$, then write the value of | A |.
(a) 8
(b) 16
(c) 0
(d) 64
Solution:
Given A adj A = $$\left[\begin{array}{ll} 8 & 0 \\ 0 & 8 \end{array}\right]$$
= 8 $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$ = 8 I2
We know that,
A adjA = |A | I2
∴ | A | = 8

Question 58.
If A is a square non-singular matrix of order 3 such that | adj A | = 64, find | A |.
(a) 32
(b) 4
(c) ±8
(d) None of these
Solution:
We know that,
If A be a square matrix of order n.
Then | adj A | = | A |n-1
Given A be a non-singular matrix of order 3
Then | adj A | = | A |3-1 = | A |²
also, | adj A | = 64
∴ | A |² = 64
∴ | A | = ± 8

Question 59.
For what value of x, is the given matrix singular ?
(a) 1
(b) – 1
(c) 2
(d) None of these
Solution:
We know that a matrix A is singular
∴ | A | = 0
⇒ $$\left|\begin{array}{cc} 3-2 x & x+1 \\ 2 & 4 \end{array}\right|$$
⇒ 4 (3 – 2x) – 2 (x + 1) = 0
⇒ 10 – 10x = 0
⇒ x = 1

Question 60.
Write A-1 for A = $$\left[\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right]$$.
Solution:
(a) $$\left[\begin{array}{rr} 2 & 5 \\ 3 & -2 \end{array}\right]$$
(b) $$\left[\begin{array}{rr} -2 & -3 \\ -5 & 2 \end{array}\right]$$
(c) $$\frac{1}{19}\left[\begin{array}{rr} -2 & -5 \\ -3 & 2 \end{array}\right]$$
(d) $$\frac{1}{19}\left[\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right]$$
Solution:

Question 61.
Which of the following is correct?
(a) Determinant is a square matrix.
(b) Determinant is a number associated to a matrix.
(c) Determinant is a number associated to a square matrix.
(d) All of the above.
Solution:
(c) Determinant is a number associated to a square matrix.

Question 62.
If A = $$\left[\begin{array}{rr} \log x & -1 \\ -\log x & 2 \end{array}\right]$$ and if det (A) = 2, then the value of x is equal to
(a) 2
(b) e²
(c) – 2
(d) e
(e) log 2
Solution:
Given A = $$\left[\begin{array}{rr} \log x & -1 \\ -\log x & 2 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rr} \log x & -1 \\ -\log x & 2 \end{array}\right|$$
= 2 log x – log x = log x
and det (A) = 2
∴ log x = 2 ⇒ x = e²

Question 63.
If $$\left[\begin{array}{cc} x+y & x-y \\ 2 x+z & x+z \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 1 & 1 \end{array}\right]$$ then the value of x, y, z are respectively
(a) 0, 0, 1
(b) 1, 1, 0
(c) 0, 0, 0
(d) 1, 1, 1
Solution:
Given $$\left[\begin{array}{cc} x+y & x-y \\ 2 x+z & x+z \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 1 & 1 \end{array}\right]$$
∴ x + y = 0 …(1)
x – y = 0 …(2)
2x + z = 1 …(3)
x + z = 1 …(4)
On solving eqns. (1) and (2); we have x = y = 0
∴ from (3); z = 1

Question 64.
If [x 1] $$\left[\begin{array}{rr} 1 & 0 \\ -2 & 0 \end{array}\right]$$ = 0, then x equals
(a) 0
(b) – 2
(c) – 2
(d) 2
Solution:
[x 1]1×2 $$\left[\begin{array}{rr} 1 & 0 \\ -2 & 0 \end{array}\right]$$2×2 = 0
⇒ $$\left[\begin{array}{c} x-2 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \end{array}\right]$$
∴ x – 2 = 0
⇒ x = 2

Question 65.
$$\left[\begin{array}{lll} 7 & 1 & 5 \\ 8 & 0 & 0 \end{array}\right]\left[\begin{array}{l} 2 \\ 3 \\ 1 \end{array}\right]+5\left[\begin{array}{l} 1 \\ 0 \end{array}\right]$$ is equal to
(a) $$\left[\begin{array}{l} 16 \\ 27 \end{array}\right]$$
(b) $$\left[\begin{array}{l} 27 \\ 16 \end{array}\right]$$
(c) $$\left[\begin{array}{c} 15 \\ 16 \end{array}\right]$$
(d) $$\left[\begin{array}{l} 16 \\ 15 \end{array}\right]$$
(e) $$\left[\begin{array}{l} 16 \\ 16 \end{array}\right]$$
Solution:
$$\left[\begin{array}{lll} 7 & 1 & 5 \\ 8 & 0 & 0 \end{array}\right]\left[\begin{array}{l} 2 \\ 3 \\ 1 \end{array}\right]+5\left[\begin{array}{l} 1 \\ 0 \end{array}\right]$$
= $$\left[\begin{array}{l} 14+3+5 \\ 16+0+0 \end{array}\right]+\left[\begin{array}{l} 5 \\ 0 \end{array}\right]=\left[\begin{array}{l} 22 \\ 16 \end{array}\right]+\left[\begin{array}{l} 5 \\ 0 \end{array}\right]=\left[\begin{array}{l} 27 \\ 16 \end{array}\right]$$

Question 66.
If the square of the matrix $$\left[\begin{array}{rr} a & b \\ a & -a \end{array}\right]$$
is the unit matrix, then b is equal to
(a) $$\frac{a}{1+a^2}$$
(b) $$\frac{1-a^2}{a}$$
(c) $$\frac{1+a^2}{a}$$
(d) $$\frac{a}{1-a^2}$$
Solution:
Let A = $$\left[\begin{array}{rr} a & b \\ a & -a \end{array}\right]$$ s.t A² = I
⇒ $$\frac { 1 }{ 2 }$$
⇒ $$\frac { 1 }{ 2 }$$
∴ a² + ab = 1 ⇒ b = $$\frac { 1 }{ 2 }$$

Question 67.
(i) If A = $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$, then A² is equal to
(a) $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$
(b) $$\left[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right]$$
(c) $$\left[\begin{array}{ll} 0 & 1 \\ 0 & 1 \end{array}\right]$$
(d) $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
(ii) Alternative Form. If matrix A = [aij]2×2, where aij= 1, if i ≠ j and = 0 if i = j, then A² is equal to (a) I
(b) A
(c) O
(d) None of these
Solution:
(i) A² = A . A
= $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$ = I

(ii) A = $$\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$
∴ A² = $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= I

Question 68.
If U = [2 – 1 4], X = [0 2 3], V = $$\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right]$$ and X = $$\left[\begin{array}{l} 2 \\ 2 \\ 4 \end{array}\right]$$, then UV + XY is equal to
(a) [24]
(b) 20
(c) [- 20]
(d) – 20
Solution:
UV + XY = $$\left[\begin{array}{lll} 2 & -1 & 4 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right]+\left[\begin{array}{lll} 0 & 2 & 3 \end{array}\right]$$
$$\left[\begin{array}{l} 2 \\ 2 \\ 4 \end{array}\right]$$
= [6 – 2 + 4] + [0 + 4 + 12] = [8] + [16]
= [24]

Question 69.
If P = $$\left[\begin{array}{lll} 1 & 2 & 1 \\ 1 & 3 & 1 \end{array}\right]$$, Q = PPT, then the value of the determinant of Q is equal to
(a) 2
(b) – 2
(c) 1
(d) 0
Solution:

Question 70.
If A = $$\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]$$,
(a) a27
(b) a9
(c) a6
(d) a²
Solution:
A = $$\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right|$$
Expanding along R1
= a$$\left|\begin{array}{ll} a & 0 \\ 0 & a \end{array}\right|$$ = a x a² = a³
If A be a matrix of order n.
Then | adj A | = | A |n-1
Here A be a matrix of order 3.
Then | adj A | = | A |3-1 = (a³)² = a6

Question 71.
If A is a 3 x 3 matrix such that | A | = 8, then | 3A | is equal to
(a) 8
(b) 24
(c) 72
(d) 216
Solution:
(d) 216
We know that if A be a square matrix of order n. Then det (kA) = kn det A
∴ | 3A | = 3³ | A | = 27 x 8 = 216

Question 72.
If A is a square matrix of order 3 such that A (adj A) = 10 I, then | adj A | is equal to
(a) 1
(b) 10
(c) 100
(d) 101
Solution:
Given A be a square matrix of order 3
s.t A (adj A) = 101
also, A (adj A) = | A |I
∴ | A | = 10
Thus | adj A | = | A |3-1 = | A |² = 10² = 100

Question 73.
If A = $$\left[\begin{array}{cc} 2-k & 2 \\ 1 & 3-k \end{array}\right]$$ is a singular matrix, then the value of 5k – k³ is
(a) 0
(b) 6
(c) – 6
(d) 4
Solution:
Given A = $$\left[\begin{array}{cc} 2-k & 2 \\ 1 & 3-k \end{array}\right]$$ is a singular matrix
∴ | A | = 0 ⇒ $$\left|\begin{array}{cc} 2-k & 2 \\ 1 & 3-k \end{array}\right|$$ = 0
⇒ (2 – k)(3 – k) – 2 = 0 ⇒ k² – 5k + 4 = 0
⇒ (k – 1) (k – 4) = 0 ⇒ k = 1, 4
When k = 1 ; ∴ 5k – k³ = 5 x 1 – 1³ = 4
When k = 4 ; ∴ 5k – k³ = 20 – 64 = – 44

Question 74.
If A-1 = $$\left[\begin{array}{rr} 5 & -2 \\ -7 & 3 \end{array}\right]$$ and B-1 = $$\frac{1}{2}\left[\begin{array}{rr} 9 & -7 \\ -8 & 6 \end{array}\right]$$ then (AB)-1 = ?
(a) $$\left[\begin{array}{cc} 94 & -39 \\ -82 & 34 \end{array}\right]$$
(b) $$\left[\begin{array}{rr} 94 & -82 \\ -39 & 34 \end{array}\right]$$
(c) $$\left[\begin{array}{rr} -47 & 46 \\ -\frac{39}{2} & -17 \end{array}\right]$$
(d) $$\left[\begin{array}{rr} 47 & -\frac{39}{2} \\ -41 & 17 \end{array}\right]$$
Solution:
Given A-1 = $$\left[\begin{array}{rr} 5 & -2 \\ -7 & 3 \end{array}\right]$$ and B-1 = $$\frac{1}{2}\left[\begin{array}{rr} 9 & -7 \\ -8 & 6 \end{array}\right]$$
∴ (AB)-1 = B-1A-1 = $$\frac{1}{2}\left[\begin{array}{rr} 9 & -7 \\ -8 & 6 \end{array}\right]\left[\begin{array}{rr} 5 & -2 \\ -7 & 3 \end{array}\right]=\frac{1}{2}\left[\begin{array}{rr} 94 & -39 \\ -82 & 34 \end{array}\right]=\left[\begin{array}{rr} 47 & -\frac{39}{2} \\ -41 & 17 \end{array}\right]$$

Question 75.
If A = $$\left[\begin{array}{ll} 2 & 1 \\ 0 & x \end{array}\right]$$ and A-1 = $$\left[\begin{array}{cc} 1 / 2 & 1 / 6 \\ 0 & 1 / x \end{array}\right]$$, then the value of x is equal to
(a) – 3
(b) 3
(c) – 2
(d) 6
(e) – 6
Solution:
We know that AA-1 = I
⇒ $$\left[\begin{array}{cc} 2 & 1 \\ 0 & x \end{array}\right]\left[\begin{array}{cc} 1 / 2 & 1 / 6 \\ 0 & 1 / x \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \quad \Rightarrow\left[\begin{array}{cc} 1 & \frac{1}{3}+\frac{1}{x} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
∴ $$\frac { 1 }{ 3 }$$ + $$\frac { 1 }{ x }$$ = 0 ⇒ x = – 3

Question 76.
If A = $$\left[\begin{array}{rr} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$$, then AA-1 =
(a) A
(b) zero matrix
(c) A’
(d) I
Solution:
AA’ = $$\left[\begin{array}{rr} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{rr} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$$
= $$\left[\begin{array}{cc} \cos ^2 \alpha+\sin ^2 \alpha & -\cos \alpha \sin \alpha+\sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha+\cos \alpha \sin \alpha & \sin ^2 \alpha+\cos ^2 \alpha \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$ = I

Question 77.
Matrix A = $$\left[\begin{array}{ccr} 0 & 2 b & -2 \\ 3 & 1 & 3 \\ 3 a & 3 & -1 \end{array}\right]$$ is given to be symmetric, then the values of a and b respectively are
(a) $$\frac{3}{2},-\frac{2}{3}$$
(b) $$-\frac{2}{3}, \frac{3}{2}$$
(c) 2, – 3
(d) 3, – 2
Solution:
Given A = $$\left[\begin{array}{ccr} 0 & 2 b & -2 \\ 3 & 1 & 3 \\ 3 a & 3 & -1 \end{array}\right]$$ is given to be symmetric
∴ A’ = A
⇒ $$\left[\begin{array}{rrr} 0 & 3 & 3 a \\ 2 b & 1 & 3 \\ -2 & 3 & -1 \end{array}\right]=\left[\begin{array}{ccr} 0 & 2 b & -2 \\ 3 & 1 & 3 \\ 3 a & 3 & -1 \end{array}\right]$$
Thus their corresponding entries are equal.
∴ 3 = 2b ⇒ b = $$\frac { 3 }{ 2 }$$
and 3a = – 2 ⇒ a = – $$\frac { 2 }{ 3 }$$

Question 78.
In the matrix A = $$\left[\begin{array}{ccc} a & 1 & x \\ 2 & \sqrt{3} & x^2-y \\ 0 & 5 & -\frac{2}{5} \end{array}\right]$$,
write (i) the order of the matrix A (ii) the number of the element (iii) elements a23, a31 and a12
Solution:
Given A = $$\left[\begin{array}{ccc} a & 1 & x \\ 2 & \sqrt{3} & x^2-y \\ 0 & 5 & -\frac{2}{5} \end{array}\right]$$
(i) matrix A has 3 rows and 3 columns and it is of order 3×3
(ii) matrix A have 3 x 3 i.e. 9 elements.
(iii) a23 = x² – y ; a31 = 0 ; a12 = 1

Question 79.
(i) If a matrix has 5 elements, write all possible orders it can have.
(ii) If A is matrix of order 3×4 and B is a matrix of order 4×3, then the order of matrix (AB) is ………….
Solution:
(i) We know that, if a matrix of order m x n then it has mn elements. Thus to find possible orders of matrix containing 5 elements we have to find all ordered pairs (a, b) such that a and b are factors of 5. Hence such possible ordered pairs are (1, 5) and (5, 1).
∴ possible orders are 1 x 5 and 5×1.

(ii) Given A is a matrix of order 3×4 and B be a matrix of 4 x 3.
Thus, the order of matrix AB is 3 x 3.

Question 80.
If matrix A = [1 2 3], then find AA’ where A’ is the transpose of matrix A.
Solution:
Given A = [1 2 3] and A’ = $$\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$$
∴ AA’ = $$\left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right]_{1 \times 3}\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]_{3 \times 1}$$
= [1 + 4 +9] = [14]

Question 81.
(i) Write the number of all possible matrices of order 2×2 with each entry 1, 2 or 3.
(ii) Similar Question. The number of 3 x 3 matrices with entries – 1 or + 1 is
(a) 24
(b) 25
(c) 26
(d) 29
Solution:
(i) Let A = $$\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]_{2 \times 2}$$
since each entry is 1, 2 or 3. i.e. each entry is filled with 1, 2 or 3.
Thus, each aij can be filled by 3 ways and there are 4 elements.
Hence the total number of 2 x 2 matrices with each entry is either 1, 2 or 3 = 34 = 81.

(ii) Now the matrices of order 3×3 contains 9 entries and each entry can be 1 or – 1 i.e. there are two ways to fill each entry of given matrix.
∴ Total no. of such matrices = 29

Question 82.
If A is a skew matrix of order 3, then prove that det A = 0. [Remember this result]
Solution:
Given A be a skew symmetric matrix of order 3 x 3
∴ A’ = – A ⇒ | A’| = |- A | = (- 1)³ | A | [If A be a matrix of order n x n Then | AA | = kn | A |]
⇒ | A | + | A | = 0 [∵ | A’ | = | A | ]
⇒ 2 | A | = 0 ⇒ | A | = 0

Question 83.
If A is a square matrix such that A² = I, then find the simplified value of (A – I)³ + (A + I)³ – 7 A.
Solution:
(A – I)³ = (A-I)(A-I)² = (A-I)(A-I) (A-I) = (A-I) (A²- AI – IA + I²)
= (A – I)(I – A – A + I)
[∵ A² = I and IA = A]
= (A – I) (2I – 2A) =- 2 (A²-AI-IA + I²) = – 2 (I – A – A + I)
= – 2 (2I – 2A) = + 4 (A – I)
(A + I)³ = (A + I) (A + I) (A + I)
= (A + I) (A² + AI + IA + I²)
= (A + I) (I + A + A + I)
= 4 (A + I)
∴ (A – I)³ + (A + I)³ – 7A = 4 (A – I) + 4 (A + I) – 7A = A

Question 84.
Verify that A² = I, when
A = $$\left[\begin{array}{rrr} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]$$
Solution:
A² = $$\left[\begin{array}{rrr} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]$$
= $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ = I

Question 85.
If A and B are matrices of order 3 and | A | = 5, | B | = 3, then find the value of | 3AB |.
Solution:
Given A and B be matrices of order 3 then AB be also a matrix of order 3.
Then | A | = 5 ; | B | = 3
∴ | 3AB | = 3³ | A | | B | = 27 x 5 x 3 = 405
[if A be a matrix of order n. Then | kA | = kn|A| ]

Question 86.
If A is a square matrix of order 3, with | A | = 9, then write the value of | 2 adj A|.
Solution:
If A be a matrix of order 3. Then adj A is also a matrix of order 3.
Then | 2 adj A | = 2³ | adj A |
= 8 | A |3-1 = 8 x | A |²
= 8 x 9² = 8 x 81 = 648

Question 87.
If A and B are square matrices of the same order 3, such that | A | = 2 and AB = 2 1, write the value of | B |.
Solution:
Given | A | = 2 and AB = 2I
⇒ | AB | = | 2I | ⇒ | A||B | = 2³ |I| ⇒ 2 x |B| = 8 x 1 ⇒ | B | = 4

Question 88.
If A is a non-singular matrix of order 3 and | adj A | = | A |k, then write the value of k.
Solution:
If A be a non-singular matrix of order n.
Then | Adj A | = | A |n-1
also given | adj A | = | A |k
Here n = 3 ; | adj A | = | A |3-1 = | A |²
∴ k = 2

Question 89.
(i) If A is an invertible matrix of order 2 and det (A)=4, then write the value of det (A-1 ).
(ii) If A is a 3 x 3 invertible matrix then, what will be the k if det (A-1) = (det A)k .
Solution:
(i) Given A be an invertible matrix of order 2
∴ | A | ≠ 0 and | A | = 4
det (A-1) = | A-1 | = | A|-1 = 4-1 = $$\frac{1}{4}$$

(ii) If A be an invertible matrix of order 3 s.t | A | ≠ 0
Given det (A-1) = (det A)k
Since, det (A-1) = (det A)k
∴ k = – 1

Question 90.
Given A = $$\left[\begin{array}{rrr} 4 & 2 & 5 \\ 2 & 0 & 3 \\ -1 & 1 & 0 \end{array}\right]$$, value of det. (2 AA-1)
Solution:
Here |A| = $$\left|\begin{array}{rrr} 4 & 2 & 5 \\ 2 & 0 & 3 \\ -1 & 1 & 0 \end{array}\right|$$;
Expanding along R1 = 4 (0 – 3) – 2 (0 + 3) + 5(2 – 0) = – 12 – 6 + 10 = – 8 ≠ 0
∴ A-1 exists
We know that AA-1 = I = A-1A
∴ det (2 A A-1) = det (2I)
= 2³ det (I) = 8 x 1=8
[if A be a matrix of order n. Then | kA | = kn|A|]