ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.5

Effective ML Aggarwal Class 12 Solutions Chapter 4 Determinants Ex 4.5 can help bridge the gap between theory and application.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.5

Question 1.
Use matrix method to solve the following system of eqautions :
(i) 5x + 2y = 4
7x + 3y = 5

(ii) 4x – 3y = 3
3x – 5y = 7.
Solution:
(i) The given system of eqn’s is equivalent to AX = B,
where A = \(\left[\begin{array}{ll}
5 & 2 \\
7 & 3
\end{array}\right]\);
X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and
B = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ll}
5 & 2 \\
7 & 3
\end{array}\right|\)
= 15 – 14
= 1 ≠ 0
∴ A^-1 exists
and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
Thus given system of eqn’s has inique solution given by
X = A^-1B …………..(1)
Here, A₁₁ = 3 ;
A₁₂ = – 7 ;
A₂₁ = – 2 ;
A₂₂ = 5
Thus adj A = \(\left[\begin{array}{cc}
3 & -2 \\
-7 & 5
\end{array}\right]\)
∴ A^-1 = \(\left[\begin{array}{cc}
3 & -2 \\
-7 & 5
\end{array}\right]\)
∴ from (1) ; we have
\(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{cc}
3 & -2 \\
-7 & 5
\end{array}\right]\left[\begin{array}{l}
4 \\
5
\end{array}\right]=\left[\begin{array}{c}
2 \\
-3
\end{array}\right]\)
Thus x = 2 and y = 3.

(ii) Given system of eqns. is equivalent to AX = B
where A = \(\left[\begin{array}{ll}
4 & -3 \\
3 & -5
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) ;
B = \(\left[\begin{array}{l}
3 \\
7
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ll}
4 & -3 \\
3 & -5
\end{array}\right|\)
= – 20 + 9
= – 11 ≠ 0
∴ A^-1 exists
and Given system of eqautions has a unique solution
i.e., X = A^-1B
Here,
A₁₁ = – 5 ;
A₁₂ = – 3 ;
A₂₁ = + 3 ;
A₂₂ = 4
∴ adj A = \(\left[\begin{array}{rr}
-5 & -3 \\
3 & 4
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{ll}
-5 & 3 \\
-3 & 4
\end{array}\right]\)
Thus A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
= \(-\frac{1}{11}\left[\begin{array}{ll}
-5 & 3 \\
-3 & 4
\end{array}\right]\)
from (1) ;
X = \(-\frac{1}{11}\left[\begin{array}{ll}
-5 & 3 \\
-3 & 4
\end{array}\right]\left[\begin{array}{l}
3 \\
7
\end{array}\right]\)
= \(-\frac{1}{11}\left[\begin{array}{c}
6 \\
19
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
-6 / 11 \\
-19 / 11
\end{array}\right]\)
∴ x = \(\frac{-6}{11}\) and y = \(\frac{-19}{11}\)
Thus, the required solution of given system of equations be given by
x = \(\frac{-6}{11}\) and
y = \(\frac{-19}{11}\)

Question 2.
Using matrix method, determine whether the following system of eqautions is consistent or inconsistent:
(i) x + 2y = 2
2x + 3y = 3 (NCERT)

(ii) x + 2y = 9
2x + 4y = 7

(iii) 3x – y + 2z = 3
2x + y + 3z = 5
x – 2y – z = 1 (NCERT)

(iv) 5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = – 1
Solution:
(i) The given system of eqns. is
x + 2y = 2 …………(1)
and 2x + 3y = 3 ………..(2)
∴ Given system of eqns. has unique solution and is, consistent.

(ii) The given system of equations is equivalent to
AX = B
where A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) ;
B = \(\left[\begin{array}{l}
9 \\
7
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right|\)
= 4 – 4 = 0
Thus, given system may or may not be consistent.
Here, A₁₁ = 4 ;
A₁₂ = – 2 ;
A₂₁ = – 2 ;
A₂₂ = 1
∴ adj A = \(\left[\begin{array}{rr}
4 & -2 \\
-2 & 1
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{rr}
4 & -2 \\
-2 & 1
\end{array}\right]\)
Now (adj A) B = \(\left[\begin{array}{rr}
4 & -2 \\
-2 & 1
\end{array}\right]\left[\begin{array}{l}
9 \\
7
\end{array}\right]\)
= \(\left[\begin{array}{r}
22 \\
-11
\end{array}\right]\) ≠ O
Thus given system of eqn’s is inconsistent.

(iii) Given system of eqns. is equivalent to AX = B
where A = \(\left[\begin{array}{rrr}
3 & -1 & 2 \\
2 & 1 & 3 \\
1 & -2 & -1
\end{array}\right]\);
X = \(]\) ;
B = \(\left[\begin{array}{l}
3 \\
5 \\
1
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
3 & -1 & 2 \\
2 & 1 & 3 \\
1 & -2 & -1
\end{array}\right|\) ;
expnading along R₁
= 3 (- 1 + 6) + 1 (- 2 – 3) + 2 (- 4 – 1)
= 15 – 5 – 10 = 0
Thus given system may or may not be consistent.

Thus given system of eqns. are inconsistent and have no solution.

(iv) The given system of eqns. is equivalent to
AX = B ………..(1)
where A = \(\) ;
X = \(\)
and B = \(\)
Here |A| = 5 (28) + 1 (- 13) + 4 (- 19)
= 140 – 13 – 76
= 51 ≠ 0
∴ The given system of eqns. is consistent and has unique solution.

Question 3.
Sol;ve the following system of equations by using matrix method :
x + y = 1
7x + 7y = 7.
Solution:
Given system of eqn’s is equivalent to
AX = B
where A = \(\left[\begin{array}{ll}
1 & 1 \\
7 & 7
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) ;
B = \(\left[\begin{array}{l}
1 \\
7
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ll}
1 & 1 \\
7 & 7
\end{array}\right|\)
= 7 – 7 = 0
Thus given system may or may not be consistent.
Here A₁₁ = 7 ;
A₁₂ = – 7 ;
A₂₁ = – 1 ;
A₂₂ = 1
∴ adj A = \(\left[\begin{array}{rr}
7 & -7 \\
-1 & 1
\end{array}\right]^{\mathrm{T}}\)
= \(\left[\begin{array}{rr}
7 & -1 \\
-7 & +1
\end{array}\right]\)
Here (adj A) B = \(\left[\begin{array}{rr}
7 & -1 \\
-7 & 1
\end{array}\right]\left[\begin{array}{l}
1 \\
7
\end{array}\right]\)
= \(\left[\begin{array}{l}
0 \\
0
\end{array}\right]\)
= O
∴ given system is consistent and has infinitely many solutions.

For solution:
We put y = k in first eqn. of given system, we have
x = 1 – k
Clearly, x = 1 – k and y = k
satisfies the second equation of given system.
Hence the required solution of given system be x = 1 – k ; y = k, where k be any number.

Question 4.
Using matrix method, solve the following system of equations:
(i) x – 2y + 3z = 6
x + 4y + z = 12
x – 3y + 2z = 1 (ISC 2001)

(ii) x + y + z = 6
x – y + z = 2
2x + y – z = 1 (ISC 2007)

(iii) x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0 (ISC 2009)

(iv) x – 2y = 10
2x – y – z = 8
– 2y + z = 7

(v) 3x + y + z = 1
2x + 2z = 0
5x + y + 2z = 2 (ISC 2016)

(vi) 2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
Solution:
(i) The given system of equations is equivalent to
AX = B
where A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
1 & 4 & 1 \\
1 & -3 & 2
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
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
Thus A^-1 exists and given system has a unique solution.
The cofactors of R₁ are ;

∴ x = 1 ;
y = 2 and
z = 3
Hence the requires solution of given system be
x = 1 ;
y = 2 and
z = 3.

(ii) The given system of equations is equivalent to
AX = B ………….(1)
where A = \(\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B = \(\left[\begin{array}{l}
6 \\
2 \\
1
\end{array}\right]\)
Here, |A| = 1 (0) – 1 (- 3) + 1 (3)
= 6 ≠ 0
∴ A^-1 exists
and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A ………..(2)
∴ given system of eqns. has unique solution.
Now A₁₁ = 0 ;
A₁₂ = 3 ;
A₁₃ = 3 ;
A₂₁ = 2 ;
A₂₂ = – 3 ;
A₂₃ = 1 ;
A₃₁ = 2 ;
A₃₂ = 0 ;
A₃₃ = – 2 ;
∴ adj A = \(\left[\begin{array}{rrr}
0 & 2 & 2 \\
3 & -3 & 0 \\
3 & 1 & -2
\end{array}\right]\)
∴ From (2) ;
A^-1 = \(\frac{1}{6}\left[\begin{array}{rrr}
0 & 2 & 2 \\
3 & -3 & 0 \\
3 & 1 & -2
\end{array}\right]\)
From (1) ;
AX = B
⇒ X = A^-1B
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{6}\left[\begin{array}{rrr}
0 & 2 & 2 \\
3 & -3 & 0 \\
3 & 1 & -2
\end{array}\right]\left[\begin{array}{l}
6 \\
2 \\
1
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{6}\left[\begin{array}{r}
6 \\
12 \\
12
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
2
\end{array}\right]\)
⇒ x = 1, y = 2, z = 2.

(iii) Given system of equations is equivalent to AX = B
where A = \(\left[\begin{array}{rrr}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B = \(\left[\begin{array}{r}
9 \\
52 \\
0
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{rrr}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right|\) ;
Expanding along R₁
= 1 (- 5 – 7) – 1 (- 2 – 14) + 1 (2 – 10)
= – 12 + 16 – 8
= – 4 ≠ 0
Thus A^-1 exists and given system has unique solution is given by
X = A^-1B ………………(1)
The cofactors of R₁ are ;

∴ x = 1 ;
y = 3 and
z = 5
Thhus, required solution of given system of eqautions be,
x = 1 ;
y = 3 and
z = 5.

(iv) The given system of equations is equivalent to AX = B
where A = \(\left[\begin{array}{rrr}
1 & -2 & 0 \\
2 & -1 & -1 \\
0 & -2 & 1
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
B = \(\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{rrr}
1 & -2 & 0 \\
2 & -1 & -1 \\
0 & -2 & 1
\end{array}\right|\) ;
Expanding along R₁
= 1 (- 1 – 2) + 2 (2 – 0)
= – 3 + 4
= 1 ≠ 0
∴ A^-1 exists
and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
Thus, given system of equations has unique solution.

Since AX = B
⇒ A^-1 (AX) = A^-1 B
⇒ (A^-1 A) X = A^-1 B
⇒ IX = A^-1 B
⇒ X = A^-1 B

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{lll}
-3 & 2 & 2 \\
-2 & 1 & 1 \\
-4 & 2 & 3
\end{array}\right]\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right]\)
= \(\left[\begin{array}{r}
-30+16+14 \\
-20+8+7 \\
-40+16+21
\end{array}\right]\)
= \(\left[\begin{array}{r}
0 \\
-5 \\
-3
\end{array}\right]\)
∴ x = 0 ;
y = – 5
and z = – 3

(v) Given system of equations is equivalent to AX = B
where A = \(\left[\begin{array}{lll}
3 & 1 & 1 \\
2 & 0 & 2 \\
5 & 1 & 2
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B = \(\left[\begin{array}{l}
1 \\
0 \\
2
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{lll}
3 & 1 & 1 \\
2 & 0 & 2 \\
5 & 1 & 2
\end{array}\right|\) ;
expanding along R₁
= 3 (0 – 2) – 1 (4 – 10) + 1 (2 – 0)
= – 6 + 6 + 2
= 2 ≠ 0
Thus A^-1 exists and
given system of eqautions has unique solution and is given by
X = A^-1B ……………(1)
The cofactors of R₁ are ;

∴ x = 1 ; y = – 1 and z = – 1
Hence the required solution of given system of eqautions be
x = 1 ; y = – 1 and z = – 1.

(vi) The given system of equations is equivalent to AX = B
where A = \(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
B = \(\left[\begin{array}{c}
11 \\
-5 \\
3
\end{array}\right]\)
Here, |A| = \(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\) ;
expanding along R₁
= 2 (- 4 + 4) + 3 (- 6 + 4) + 5 (3 – 2)
= – 6 + 5
= – 1 ≠ 0
∴ A^-1 exists.
A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj. A
and given system of eqautions has unique solution.
The cofactors of R₁ are :
\(\left|\begin{array}{ll}
2 & -4 \\
1 & -2
\end{array}\right| ;-\left|\begin{array}{ll}
3 & -4 \\
1 & -2
\end{array}\right| ;\left|\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right|\)
i.e., 0 ; 2 ; 1

The cofactors of R₂ are :
\(\left|\begin{array}{cc}
-3 & 5 \\
1 & -2
\end{array}\right| ;\left|\begin{array}{cc}
2 & 5 \\
1 & -2
\end{array}\right| ;-\left|\begin{array}{cc}
2 & -3 \\
1 & 1
\end{array}\right|\)
i.e., – 1 ; – 9 ; – 5

The cofactors of R₃ are :
\(\left|\begin{array}{cc}
-3 & 5 \\
2 & -4
\end{array}\right| ;-\left|\begin{array}{cc}
2 & 5 \\
3 & -4
\end{array}\right| ;\left|\begin{array}{cc}
2 & -3 \\
3 & 2
\end{array}\right|\)
i.e., 2 ; 23 ; 13

∴ adj A = \(\left[\begin{array}{ccc}
0 & -1 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{array}\right]\)
∴ A^-1 = – \(\left[\begin{array}{ccc}
0 & -1 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{array}\right]\)
Since AX = B
⇒ X = A^-1B
= \(\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\left[\begin{array}{c}
11 \\
-5 \\
-3
\end{array}\right]\)

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
-5+6 \\
-22-45+69 \\
-11-25+39
\end{array}\right]\)
= \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
∴ x = 1 ;
y = 2 ;
z = 3.

Question 4 (old).
(iv) x + 2y = 5
y + 2z = 8
2x + z = 5 (ISC 2003)
(vi) 2x – y + z = 3
– x + 2y – z = – 4
x – y + 2z = 1
Solution:
(iv) The given solution of eqautions 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
= 1 + 8
= 9 ≠ 0
∴ A^-1 exists
and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj. A …………(1)
Also, given system of eqn’s has unique solution and is given by
X = A^-1B ……………(2)

Thus x = 1, y = 2, z = 3
Hence the required solution of given system of eqn’s be
x = 1 ;
y = 2 and
z = 3.

(vi) Given system of eqautions is equivalent to
AX = B
where A = \(\left[\begin{array}{rrr}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
B = \(\left[\begin{array}{r}
3 \\
-4 \\
1
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{rrr}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right|\) ;
expanding along R₁
= 2 (4 – 1) + 1 (- 2 + 1) + 1 (1 – 2)
= 6 – 1 – 1
= 4 ≠ 0
Thus A^-1 exists and given system of eqautions has unique solution and is given by
X = A^-1B ………….(1)
The cofactors of R₁ are ;

∴ x = 1 ;
y = – 2 ;
z = – 1
Thus, required solution of given system be x = 1, y = – 2 and z = – 1.

Question 5.
Solve the following system of linear equations using matrix method:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) = 9,
\(\frac{2}{x}+\frac{5}{y}+\frac{7}{z}\) = 52,
\(\frac{2}{x}+\frac{1}{y}-\frac{1}{z}\) = 0.
Solution:
Putting \(\frac{1}{x}\) = u; \(\frac{1}{y}\) = v and \(\frac{1}{z}\) = w
in given system of equations we have,
u + v + w = 9
2u + 5v + 7w = 52
2u + v – w = 0
The given system of eqautions is equivalent to
AU = B
where A = \(\left[\begin{array}{rrr}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right]\) ;
U = \(\left[\begin{array}{l}
u \\
v \\
w
\end{array}\right]\) ;
B = \(\left[\begin{array}{r}
9 \\
52 \\
0
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right|\)
expanding along R₁
= 1 (- 5 – 7) – 1 (- 2 – 14) + 1 (2 – 10)
= – 12 + 16 – 8
= – 4 ≠ 0
Thus, system of eqautions have unique solution.
Further A^-1 exists and
A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactors of R₁ are

∴ u = 1
⇒ \(\frac{1}{x}\) = 1
⇒ x = 1

v = 3
⇒ \(\frac{1}{y}\) = 3
⇒ y = \(\frac{1}{3}\)

and w = 5
⇒ \(\frac{1}{z}\) = 5
⇒ z = \(\frac{1}{5}\)

Question 6.
State the condition under which the following equations have a unique solution:
x + 2y – 2z + 5 = 0
– x + 3y + 4 = 0
– 2y + z – 4 = 0
Using Martin’s rule, find the unique solution of the above system of linear eqautions.
Solution;
Given system of equations can be written as :
x + 2y – 2z + 5 = 0
– x + 3y + 4 = 0
– 2y + z – 4 = 0
The given system is equivalent to
AX = B
where A = \(\left[\begin{array}{rrr}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
B = \(\left[\begin{array}{r}
-5 \\
-4 \\
4
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{rrr}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right|\)
= 1 (3 – 0) – 2 (- 1 – 0) – 2 (2 – 0)
= 3 + 2 – 4
= 1 ≠ 0
∴ A^-1 exists and given system of eqautions has unique solution and is given by
X = A^-1 B ……………(1)
The cofactors of R₁ are ;

⇒ X = \(\left[\begin{array}{r}
-15-8+24 \\
-5-4+8 \\
-10-8+20
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
1 \\
-1 \\
2
\end{array}\right]\)
∴ x = 1 ; y = – 1 and z = 2
Thus, the required solution of given system be
x = 1 ; y = – 1 and z = 2.

Question 7.
The sum of three numbers is 6. If we multiply third number by 3 and add second to it we get 11. By adding first and third numbers, we get double of the
second number. Represent it algebraically and find the numbers using matrix method. (NCERT)
Solution:
Let the three numbers be x, y and z.
According to given question, we have
x + y + z = 6 …………….(1)
0.x + y + 3z = 11 …………….(2)
and x – 2y + z = 0
The given system of eqns. is equivalent to
AX = B …………….(4)
where A = \(\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & -2 & 1
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B = \(\left[\begin{array}{r}
6 \\
11 \\
0
\end{array}\right]\)
Here, |A| = 1 (7) – 1 (- 3) + 1 (- 1)
= 9 ≠ 0
∴ Given system of eqns. has unique solution.
Further A^-1 exists
and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A ………….(5)
Cofactors of 1st row are ; 7 ; 3 ; – 1
Cofactors of 2nd row are ; – 3 ; 0 ; 3
Cofactors of 3rd row are ; 2 ; – 3 ; 1
∴ adj A = \(\frac{1}{9}\left[\begin{array}{rrr}
7 & -3 & 2 \\
3 & 0 & -3 \\
-1 & 3 & 1
\end{array}\right]\)
∴ From (5) ;
A^-1 = \(\frac{1}{9}\left[\begin{array}{rrr}
7 & -3 & 2 \\
3 & 0 & -3 \\
-1 & 3 & 1
\end{array}\right]\)
∴ From (4) ;
X = A^-1B
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{9}\left[\begin{array}{rrr}
7 & -3 & 2 \\
3 & 0 & -3 \\
-1 & 3 & 1
\end{array}\right]\left[\begin{array}{r}
6 \\
11 \\
0
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{9}\left[\begin{array}{r}
9 \\
18 \\
27
\end{array}\right]\)
= \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
∴ x = 1 ;
y = 2
and z = 3.
Thus the required three numbers are 1, 2 and 3.

Question 8.
If A = \(\left[\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right]\), find A^-1. Using A^-1, solve the system of linear equations
x + 2y – 3z = – 4, 2x + 3y + 2z = 2, 3x – 3y – 4z = 11. (ISC 2008)
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right]\)
Here |A| = 1 (- 6) – 2 (- 14) – 3 (- 15)
= – 6 + 28 + 45
= 67 ≠ 0
∴ A is non-singular and A^-1 exists
∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A ………….(1)
Cofactors of 1st row are ;
– 6 ; 14 ; – 15
Cofactors of 2nd row are ;
17 ; 5 ; 9
Cofactors of 3rd row are ;
13 ; – 8 ; – 1
∴ adj A = \(\left[\begin{array}{rrr}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right]\)
∴ From (1) ;
A^-1 = \(\frac{1}{67}\left[\begin{array}{rrr}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right]\)
The given system of equations is equivalent to AX = B
where X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
B = \(\left[\begin{array}{r}
-4 \\
2 \\
11
\end{array}\right]\)
Since AX = B
⇒ X = A^-1B
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{67}\left[\begin{array}{rrr}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right]\left[\begin{array}{r}
-4 \\
2 \\
11
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{67}\left[\begin{array}{r}
201 \\
-134 \\
67
\end{array}\right]\)
= \(\left[\begin{array}{r}
3 \\
-2 \\
1
\end{array}\right]\)
∴ x = 3 ; y = – 2 and z = 1.

Question 9.
If A = \(\left[\begin{array}{rrr}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\), find A^-1. Using A^-1, solve the following system of equations:
2x – 3y + 5z = 11, 3x + 2y – 4z = – 5, x + y – 2z = 3.
Solution:
Given A = \(\left[\begin{array}{rrr}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\)
Here |A| = 2 (- 4 + 4) + 3 (- 6 + 4) + 5 (3 – 2)
= 0 – 6 + 5
= – 1 ≠ 0
∴ A^-1 exists
and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A ………….(1)

⇒ x = 1 ; y = 2 and z = 3
Hence x = 1 ; y = 2 and z = 3 be the required solution of the given equations.

Question 10.
If A = \(\left[\begin{array}{rr}
3 & 5 \\
-2 & 3
\end{array}\right]\), find A^-1 and use it to solve the system of eqautions:
3x – 2y = 7, 5x + 3y = 1.
Solution:
Given A = \(\left[\begin{array}{rr}
3 & 5 \\
-2 & 3
\end{array}\right]\)
∴ |A| = 9 + 10
= 19 ≠ 0
∴ A^-1 exists and
∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A …………..(1)
Here, A₁₁ = 3
A₁₂ = 2
A₂₁ = – 5
A₂₂ = 3
∴ adj A = \(=\left[\begin{array}{rr}
3 & 2 \\
-5 & 3
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{rr}
3 & -5 \\
2 & 3
\end{array}\right]\)
∴ from (1) ;
A^-1 = \(\frac{1}{19}\left[\begin{array}{rr}
3 & -5 \\
2 & 3
\end{array}\right]\) ……………..(2)
The given system of eqn’s be
3x – 2y = 7
5x + 3y = 1
and it is equivalent to A’X = B
where A’ = \(\left[\begin{array}{rr}
3 & -2 \\
5 & 3
\end{array}\right]\) ;
X = \(=\left[\begin{array}{l}
x \\
y
\end{array}\right]\) ;
B = \(\left[\begin{array}{l}
7 \\
1
\end{array}\right]\)
Now |A’| = |A|
= 19 ≠ 0
∴ A^-1 exists
and given system has unique solution.
∴ X = (A’)^-1 B
= (A^-1 )’B
[∵ (A’)^-1 = (A^-1)’]
⇒ X = \(\frac{1}{19}\left[\begin{array}{rr}
3 & -5 \\
2 & 3
\end{array}\right]^{\prime}\left[\begin{array}{l}
7 \\
1
\end{array}\right]\)
⇒ X = \(\frac{1}{19}\left[\begin{array}{rr}
3 & 2 \\
-5 & 3
\end{array}\right]\left[\begin{array}{l}
7 \\
1
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\frac{1}{19}\left[\begin{array}{r}
23 \\
-32
\end{array}\right]\)
= \(\left[\begin{array}{r}
\frac{23}{19} \\
\frac{-32}{19}
\end{array}\right]\)
∴ x = \(\frac{23}{19}\)
and y = \(\frac{32}{19}\)
Thus, the required solution of given system be
x = \(\frac{23}{19}\)
and y = \(\frac{32}{19}\).

Question 11.
If A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\), find a^-1 and use it to solve the system of eqautions:
x + 2y + z = 4, – x + y + z = 0, x – 3y + z = 2.
Soution:
Given,
A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\)
Here |A| = 1 (4) + 1 95) + 1 (1)
= 10 ≠ 0
∴ A^-1 exists and
A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A …………..(1)
Cofactors of 1st row are,
\(\left|\begin{array}{cc}
1 & -3 \\
1 & 1
\end{array}\right| ;-\left|\begin{array}{cc}
2 & -3 \\
1 & 1
\end{array}\right| ;\left|\begin{array}{cc}
2 & 1 \\
1 & 1
\end{array}\right|\)
i.e., 4, – 5, 1

Cofactors of 2nd row are,
\(-\left|\begin{array}{cc}
-1 & 1 \\
1 & 1
\end{array}\right| ;\left|\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right| ;-\left|\begin{array}{cc}
1 & -1 \\
1 & 1
\end{array}\right|\)
i.e., 2, 0 , – 2

Cofactors of 3rd row are,
\(\left|\begin{array}{cc}
-1 & 1 \\
1 & -3
\end{array}\right| ;-\left|\begin{array}{cc}
1 & 1 \\
2 & -3
\end{array}\right| ;\left|\begin{array}{cc}
1 & -1 \\
2 & 1
\end{array}\right|\)
i.e., 2, 5, 3

∴ adj A = \(\left[\begin{array}{rrr}
4 & +2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right]\)
∴ From (1) ;
A^-1 = \(\frac{1}{10}\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right]\)
The given system of eqns. is equivalent to
AX = B
where X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B = \(\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right]\)
Since AX = B
⇒ X = A^-1B
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{10}\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{10}\left[\begin{array}{r}
20 \\
-10 \\
10
\end{array}\right]\)
= \(\left[\begin{array}{r}
2 \\
-1 \\
1
\end{array}\right]\)
∴ x = 2 ; y = – 1 and z = 1.

Question 12.
Given two matrices A and B where 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. (ISC 2015)
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
1 & 4 & 1 \\
1 & -3 & 2
\end{array}\right]\)
and B = \(\left[\begin{array}{rrr}
11 & -5 & -14 \\
-1 & -1 & 2 \\
-7 & 1 & 6
\end{array}\right]\)
∴ AB = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
1 & 4 & 1 \\
1 & -3 & 2
\end{array}\right]\left[\begin{array}{rrr}
11 & -5 & -14 \\
-1 & -1 & 2 \\
-7 & 1 & 6
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-8 & 0 & 0 \\
0 & -8 & 0 \\
0 & 0 & -8
\end{array}\right]\)

⇒ AB = – 8 I₃
⇒ A (- \(\frac{1}{8}\) B) = I₃
⇒ A^-1 = – \(\frac{1}{8}\) B …………..(1)
[by def. of inverse]
Also |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
The given system of equations is equivalent to AX = C
where A = \(\left[\begin{array}{lll}
1 & -2 & 3 \\
1 & +4 & 1 \\
1 & -3 & 2
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
C = \(\left[\begin{array}{r}
6 \\
12 \\
1
\end{array}\right]\)
Since |A| ≠ 0
∴ A^-1 exists
and hence given system has unique solution
i.e., X = A^-1C
⇒ X = – \(\frac{1}{8}\) BC [using (1)]
⇒ X = \(-\frac{1}{8}\left[\begin{array}{rrr}
11 & -5 & -14 \\
-1 & -1 & 2 \\
-7 & 1 & 6
\end{array}\right]\left[\begin{array}{r}
6 \\
12 \\
1
\end{array}\right]\)
⇒ X = \(-\frac{1}{8}\left[\begin{array}{r}
66-60-14 \\
-6-12+2 \\
-42+12+6
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=-\frac{1}{8}\left[\begin{array}{r}
-8 \\
-16 \\
-24
\end{array}\right]\)
= \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
Thus, x = 1 ; y = 2 and z = 3
Hence the required solution ofgiven system of eqns. be
x = 1 ; y = 2 and z = 3.

Question 13.
Using matrices, solve the following systems of homogeneous equations :
(i) 5x + 5y + 2z = 0
2x + 5y + 4z = 0
4x + 5y + 2z = 0

(ii) 3x + y – 2z = 0
x + y + z = 0
x – 2y + z = 0
Solution:
(i) The given system of eqn’s is equivalent to AX = O
where A = \(\left[\begin{array}{lll}
5 & 5 & 2 \\
2 & 5 & 4 \\
4 & 5 & 2
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{lll}
5 & 5 & 2 \\
2 & 5 & 4 \\
4 & 5 & 2
\end{array}\right|\) ;
expanding along R₁
= 5 (10 – 20) – 5 (4 – 16) + 2 (10 – 20)
= – 50 + 60 – 20
= – 10 ≠ 0
Thus the given system has tivial or zero solution.
i.e., x = y = z = 0

(ii) Given system of eqns is equivalent to
AX = O
where A = \(\left[\begin{array}{rrr}
3 & 1 & -2 \\
1 & 1 & 1 \\
1 & -2 & 1
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{rrr}
3 & 1 & -2 \\
1 & 1 & 1 \\
1 & -2 & 1
\end{array}\right|\) ;
expanding along R₁
= 3 (3) – 1 (1 – 1) – 2 (- 2 – 1)
= 9 + 6
= 15 ≠ 0
∴ given system of homogeneous eqn’s have trivial solution given by x = y = z = 0.

Question 14.
Using matrices, solve the following systems of homogeneous equations:
(i) 2x – 3y – z = 0
x + 3y – 2z = 0
x – 3y = 0

(ii) x + y – z = 0
x – 2y + z = 0
3x + 6y – 5z = 0.
Solution:
(i) The given system of eqn’s is equivalent to AX = O
where A = \(\left[\begin{array}{rrr}
2 & -3 & -1 \\
1 & 3 & -2 \\
1 & -3 & 0
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
2 & -3 & -1 \\
1 & 3 & -2 \\
1 & -3 & 0
\end{array}\right|\) ;
expanding along R₁
= 2 (0 – 6) + 3 (0 + 2) – 1 (- 3 – 3)
= – 12 + 6 + 6 = 0
Thus the given system has infinitely many solution.

For solution:
putting z = k in first two equations of given system;
we have
2x – 3y = k
x + 3y = 2k, where k be any number
∴ The given system is equivalent to CX = B
where C = \(\left[\begin{array}{rr}
2 & -3 \\
1 & 3
\end{array}\right]\) ;
X = \(=\left[\begin{array}{l}
x \\
y
\end{array}\right]\) ;
B = \(\left[\begin{array}{r}
k \\
2 k
\end{array}\right]\)
Here, |C| = \(\left|\begin{array}{rr}
2 & -3 \\
1 & 3
\end{array}\right|\)
= 6 + 3
= 9 ≠ 0
∴ C^-1 exists and given system has unique solution and is given by
X = C^-1B …………..(1)
Here, C₁₁ = 3
C₁₂ = – 1
C₂₁ = 3
C₂₂ = 2
∴ adj A = \(\left[\begin{array}{rr}
3 & -1 \\
3 & 2
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{rr}
3 & 3 \\
-1 & 2
\end{array}\right]\)
Thus C^-1 = \(\frac{1}{|\mathrm{C}|}\) adj C
= \(\frac{1}{9}\left[\begin{array}{rr}
3 & 3 \\
-1 & 2
\end{array}\right]\)
from (1) ;
X = \(\frac{1}{9}\left[\begin{array}{rr}
3 & 3 \\
-1 & 2
\end{array}\right]\left[\begin{array}{r}
k \\
2 k
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
x \\
y \\
z
\end{array}\right]=\frac{1}{9}\left[\begin{array}{c}
9 k \\
3 k
\end{array}\right]\)
= \(\left[\begin{array}{r}
k \\
k / 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
k \\
k / 3
\end{array}\right]\)
∴ x = k ; y = k/3.
Also, x = k, y = k/3 and z = k satisfies the third eqn. of given system.
Hence, the required solution of given system of eqn’s be
x = k, y k/3 and z = k ; where k be any number.

(ii) The given eqns are,
x + y – z = 0 …………..(1)
x – 2y + z = 0 ……………(2)
and 3x + 6y – 5z …………(3)
∴ given system of eqns is equivalent to
AX = O
Here |A| = \(\left|\begin{array}{ccc}
1 & 1 & -1 \\
1 & -2 & 1 \\
3 & 6 & -5
\end{array}\right|\)
= 1 (4) – 1 (- 8) – 1 (12)
= 4 + 8 – 12 = 0
Thus, given system of eqns has infinitely many solution.
Let z = k;
from (1);
x + y = k …………..(4)
from (2);
x – 2y = – k …………..(5)
Solving (4) and (5);
3y = 2k
⇒ y = \(\frac{2 k}{3}\) ;
x = \(\frac{k}{3}\)
Putting in eqn. (3); we have
L.H.S. = 3 (k/3) + 6 (2k/3) – 5k = 0
= R.H.S.
∴ eqn. (3) is satisfied by
x = \(\frac{k}{3}\) ;
y = \(\frac{k}{3}\);
z = k ∀ k ∈ R.
Thus system of eqns lias infinitely many solution, is given by
x = \(\frac{k}{3}\) ;
y = \(\frac{k}{3}\);
and z = k ∀ k ∈ R.