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## ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Chapter Test

Question 1.
(i) Construct a 2 × 3 matrix whose elements in the ith row and the jth column are given by aij = $$\frac{i+3 j}{2}$$
(ii) Construct a 2 × 2 matrix A = [aij] who: elements are given by
aij = $$\begin{cases}i-j & \text { if } \quad i \geq j \\ i+j & \text { if } \quad i<j\end{cases}$$
Solution:
(i) Let A = [aij]2 × 3
where 1 ≤ i ≤ 2 ; 1 ≤ j ≤ 3
when i = 1 = j ;
a11 = $$\frac{1+3}{2}$$ = 2
when i = 1, j = 2
a12 = $$\frac{1+6}{2}=\frac{7}{2}$$
when i = 1, j = 3
a12 = $$\frac{1+9}{2}$$ = 5
when i = 2, j = 1
a12 = $$\frac{2+3}{2}=\frac{5}{2}$$
when i = 2, j = 2
a12 = $$\frac{2+6}{2}$$ = 4
when i = 2, j = 3
a12 = $$\frac{2+9}{2}=\frac{11}{2}$$
∴ A = $$\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{array}\right]$$
= $$\left[\begin{array}{ccc} 2 & 7 / 2 & 5 \\ 5 / 2 & 4 & 11 / 2 \end{array}\right]_{2 \times 3}$$.

(ii) A = [aij]2 × 2 where 1 ≤ i, j ≤ 2
Here aij = $$\left\{\begin{array}{lll} i-j & \text { if } & i \geq j \\ i+j & \text { if } & i<j \end{array}\right.$$
∴ a11 = 1 – 1 = 0 [when i = 1, j = 1]
when i = 1, j = 2 i.e., i < j
∴ a12 = 1 + 2 = 3
when i = 2, j = 1 i.e., i > j
∴ a21 = 2 – 1 = 1
when i = j = 2;
a22 = 2 – 2 = 0
∴ A = $$\left[\begin{array}{lll} a_{11} & & a_{12} \\ a_{21} & & a_{22} \end{array}\right]$$
= $$\left[\begin{array}{ll} 0 & 3 \\ 1 & 0 \end{array}\right]$$.

Question 2.
Construct a 2 × 3 matrix A = [aij] whose elements aij are given by aij = $$\left[\frac{i}{j}\right]$$, where [x] represents greatest integer function.
Solution:
Given A = [aij]2 × 3
where 1 ≤ i ≤ 2 ; 1 ≤ j ≤ 3
where aij = $$\left[\frac{i}{j}\right]$$
when i = 1 = j;
a11 = $$\left[\frac{1}{1}\right]$$
= [1] = 1
When i = 1, j = 2;
a12 = $$\left[\frac{1}{2}\right]$$
= [0.5] = 0
When i = 1, j = 3 ;
a13 = $$\left[\frac{1}{3}\right]$$ = 0
When i = 2, j = 1 ;
a21 = $$\left[\frac{2}{1}\right]$$ = 2
When i = 2 = j = 2 ;
a22 = $$\left[\frac{2}{2}\right]$$ = 1
When i = 2, j = 3 ;
a23 = $$\left[\frac{2}{3}\right]$$ = 0
∴ A = $$\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{array}\right]$$
= $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \end{array}\right]_{2 \times 3}$$

Question 3.
Determine the matrices A and B when A + 2B = $$\left[\begin{array}{rrr} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{array}\right]$$ and 2A – B = $$\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]$$.
Solution:
Given,
A + 2B = $$\left[\begin{array}{rrr} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{array}\right]$$ ………..(1)
2A – B = $$\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]$$ ……………(2)
Multiplying eqn. (2) by 2 and adding to eqn. (1) ; we have
5A = $$2\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]+\left[\begin{array}{rrr} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{array}\right]$$
⇒ 5A = $$\left[\begin{array}{rrr} 4 & -2 & 10 \\ 4 & -2 & 12 \\ 0 & 2 & 4 \end{array}\right]+\left[\begin{array}{rrr} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 5 & 0 & 10 \\ 10 & -5 & 15 \\ -5 & 5 & 15 \end{array}\right]$$
⇒ A = $$\frac{1}{5}\left[\begin{array}{rrr} 5 & 0 & 10 \\ 10 & -5 & 15 \\ -5 & 5 & 5 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{array}\right]$$

∴ from eqn. (2) ; we have
B = 2A – $$\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]$$
= $$2\left[\begin{array}{rrr} 1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{array}\right]-\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]$$

B = $$\left[\begin{array}{rrr} 2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2 \end{array}\right]-\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{array}\right]$$.

Question 4.
If A = $$\left[\begin{array}{rrr} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{array}\right]$$ and B = $$\left[\begin{array}{rrr} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{array}\right]$$, show that AB + BA = A + B.
Solution:
Given A = $$\left[\begin{array}{rrr} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{array}\right]$$
and B = $$\left[\begin{array}{rrr} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{array}\right]$$

∴ AB = $$\left[\begin{array}{rrr} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{array}\right]\left[\begin{array}{rrr} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 4+3-5 & -4-9+10 & -8-12+15 \\ -2-4+5 & 2+12-10 & 4+16-15 \\ 2+3-4 & -2-9+8 & -4-12+12 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{array}\right]$$
= A

BA = $$\left[\begin{array}{rrr} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{array}\right]\left[\begin{array}{rrr} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 4+2-4 & -6-8+12 & -10-10+16 \\ -2-3+4 & 3+12-12 & 5+15-16 \\ 2+2-3 & -3-8+9 & -5-10+12 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{array}\right]$$
= B
Thus AB + BA = A + B [Hence proved].

Question 5.
If A = $$\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]$$, find a, b so that A2 = aA – bI.
Solution:
Given A = $$\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]$$
and A2 = aA – bI
⇒ $$\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=a\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]-b\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
⇒ $$\left[\begin{array}{rr} 7 & 12 \\ 4 & 7 \end{array}\right]=\left[\begin{array}{rr} 2 a & 3 a \\ a & 2 a \end{array}\right]-\left[\begin{array}{ll} b & 0 \\ 0 & b \end{array}\right]$$
⇒ $$\left[\begin{array}{rr} 7 & 12 \\ 4 & 7 \end{array}\right]=\left[\begin{array}{cc} 2 a-b & 3 a \\ a & 2 a-b \end{array}\right]$$
∴ 2a – b = 7
3a = 12
⇒ a = 4
∴ from (1) ;
8 – b = 7
⇒ b = 1.

Question 6.
If A = $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$ and B = $$\left[\begin{array}{rr} 0 & -x \\ x & 0 \end{array}\right]$$, verify that (A + B) = A + B. (NCERT Exampler)
Solution:
Given B = $$\left[\begin{array}{rr} 0 & -x \\ x & 0 \end{array}\right]$$ ;
A = $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$
∴ A + B = $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]+\left[\begin{array}{rr} 0 & -x \\ x & 0 \end{array}\right]$$
= $$\left[\begin{array}{cc} 0 & 1-x \\ x+\mathrm{i} & 0 \end{array}\right]$$

∴ L.H.S. = (A + B)2
= (A + B) . (A – B)
= $$\left[\begin{array}{cc} 0 & 1-x \\ x+1 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & 1-x \\ x+1 & 0 \end{array}\right]$$
= $$\left[\begin{array}{cc} 1-x^2 & 0 \\ 0 & 1-x^2 \end{array}\right]$$

and R.H.S. = A2 + B2
= $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$
= $$\left[\begin{array}{rr} 0 & -x \\ x & 0 \end{array}\right]\left[\begin{array}{rr} 0 & -x \\ x & 0 \end{array}\right]$$
= $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{cc} -x^2 & 0 \\ 0 & -x^2 \end{array}\right]$$
= $$\left[\begin{array}{cc} 1-x^2 & 0 \\ 0 & 1-x^2 \end{array}\right]$$

∴ L.H.S. = R.H.S.
⇒ (A + B)2 = A2 + B2

Question 7.
Find a and b if $$\left\{3\left[\begin{array}{rrr} 2 & 1 & -3 \\ 1 & 4 & 2 \end{array}\right]-2\left[\begin{array}{lll} 1 & -2 & 0 \\ 2 & -1 & 3 \end{array}\right]\right\}$$ $$\left[\begin{array}{r} 2 \\ 0 \\ -1 \end{array}\right]$$ = $$\left[\begin{array}{l} a \\ b \end{array}\right]$$.
Solution:
Given, $$\left\{3\left[\begin{array}{rrr} 2 & 1 & -3 \\ 1 & 4 & 2 \end{array}\right]-2\left[\begin{array}{lll} 1 & -2 & 0 \\ 2 & -1 & 3 \end{array}\right]\right\}$$ $$\left[\begin{array}{r} 2 \\ 0 \\ -1 \end{array}\right]$$ = $$\left[\begin{array}{l} a \\ b \end{array}\right]$$

⇒ $$\left\{\left[\begin{array}{rrr} 6 & 3 & -9 \\ 3 & 12 & 6 \end{array}\right]-\left[\begin{array}{rrr} 2 & -4 & 0 \\ 4 & -2 & 6 \end{array}\right]\right\}\left[\begin{array}{r} 2 \\ 0 \\ -1 \end{array}\right]=\left[\begin{array}{l} a \\ b \end{array}\right]$$

⇒ $$\left[\begin{array}{rrr} 4 & 7 & -9 \\ -1 & 14 & 0 \end{array}\right]\left[\begin{array}{r} 2 \\ 0 \\ -1 \end{array}\right]=\left[\begin{array}{l} a \\ b \end{array}\right]$$

⇒ $$\left[\begin{array}{c} 4 \times 2+7 \times 0-9 \times(-1) \\ -1 \times 2+14 \times 0+0 \times(-1) \end{array}\right]=\left[\begin{array}{l} a \\ b \end{array}\right]$$

⇒ $$\left[\begin{array}{c} 17 \\ -2 \end{array}\right]=\left[\begin{array}{l} a \\ b \end{array}\right]$$
Thus their corresponding elements are equal.
∴ a = 17
and b = – 2.

Question 8.
Given A = $$\left[\begin{array}{rr} 3 & -1 \\ 1 & 2 \end{array}\right]$$, B = $$\left[\begin{array}{l} 3 \\ 1 \end{array}\right]$$ and C = $$\left[\begin{array}{r} 1 \\ -2 \end{array}\right]$$, find the matrix X such that AX = 3B + 2C.
Solution:
Given A = $$\left[\begin{array}{rr} 3 & -1 \\ 1 & 2 \end{array}\right]_{2 \times 2}$$ ;
B = $$\left[\begin{array}{l} 3 \\ 1 \end{array}\right]_{2 \times 1}$$
and C = $$\left[\begin{array}{r} 1 \\ -2 \end{array}\right]_{2 \times 1}$$
and AX = 3B + 2C ……………….(1)
Since B and C are matrices of order 2 × 1.
∴ 3B + 2C must be a matrix of order 2 × 1.
Now No. of rows in X = No. of columns in A = 2
and No. of columns in X = No. of columns in (3B + 2C)
∴ X be a matrix of order 2 × 1
Let X = $$\left[\begin{array}{l} u \\ v \end{array}\right]$$
∴ from (1) ;
We have
$$\left[\begin{array}{rr} 3 & -1 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} a \\ b \end{array}\right]=3\left[\begin{array}{l} 3 \\ 1 \end{array}\right]+2\left[\begin{array}{r} 1 \\ -2 \end{array}\right]$$
⇒ $$\left[\begin{array}{l} 3 a-b \\ a+2 b \end{array}\right]=\left[\begin{array}{l} 9 \\ 3 \end{array}\right]+\left[\begin{array}{r} 2 \\ -4 \end{array}\right]$$
$$\left[\begin{array}{c} 11 \\ -1 \end{array}\right]$$

∴ 3a – b = 11 ……………….(1)
a + 2b = – 1 ………………(2)
Multiply eqn. (1) by (2) + eqn. (2) we have
7a = 22 – 1 = 21
⇒ a = 3
∴ from (1) ;
b = 9 – 11 = – 2
Thus X = $$\left[\begin{array}{l} a \\ b \end{array}\right]=\left[\begin{array}{r} 3 \\ -2 \end{array}\right]$$

Question 9.
Given $$\left[\begin{array}{rr} 2 & 1 \\ -3 & 2 \end{array}\right] A+\left[\begin{array}{rr} -5 & 0 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{rr} 3 & -9 \\ 7 & 1 \end{array}\right]$$, find matrix A.
Solution:
Given $$\left[\begin{array}{rr} 2 & 1 \\ -3 & 2 \end{array}\right] A+\left[\begin{array}{rr} -5 & 0 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{rr} 3 & -9 \\ 7 & 1 \end{array}\right]$$ ……………(1)
Since all matrices in eqn. (1) be of same order i.e., 2 × 2
∴ A must be a matrix of order 2 × 2.
Let A = $$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$$
∴ eqn. (1) becomes;
$$\left[\begin{array}{rr} 2 & 1 \\ -3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]+\left[\begin{array}{rr} -5 & 0 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{rr} 3 & -9 \\ 7 & 1 \end{array}\right]$$

⇒ $$\left[\begin{array}{rr} 2 a+c & 2 b+d \\ -3 a+2 c & -3 b+2 d \end{array}\right]=\left[\begin{array}{rr} 3 & -9 \\ 7 & 1 \end{array}-\left[\begin{array}{rr} -5 & 0 \\ 2 & 4 \end{array}\right]\right.$$

⇒ $$\left[\begin{array}{rr} 2 a+c & 2 b+d \\ -3 a+2 c & -3 b+2 d \end{array}\right]=\left[\begin{array}{rr} 8 & -9 \\ 5 & -3 \end{array}\right]$$

∴ 2a + c = 8 ………….(1)
2b + d = – 9 ………….(2)
– 3a + 2c = 5 ………….(3)
– 3b + 2d = – 3 ………….(4)
Multiplying eqn. (1) by 3 + eqn. (3) by 2 ;
we have
7c = 24 + 10
⇒ c = $$\frac{34}{7}$$
∴ from (1);
2a = 8 – $$\frac{34}{7}$$
= $$\frac{22}{7}$$
⇒ a = $$\frac{11}{7}$$
Multiplying eqn. (2) by 3 + eqn. (4) by 2 ;
we have
7d = – 27 – 6
⇒ d = $$-\frac{33}{7}$$
∴ from (2) ;
2b = – 9 + $$\frac{33}{7}$$
= – $$\frac{30}{7}$$
⇒ b = $$-\frac{15}{7}$$
Thus A = $$\left[\begin{array}{ll} 11 / 7 & -15 / 7 \\ 34 / 7 & -33 / 7 \end{array}\right]$$

Question 10.
If A = $$\left[\begin{array}{rrr} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5 \end{array}\right]$$ and B = $$\left[\begin{array}{rrr} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{array}\right]$$, compute A2B2.
Solution:
Given A = $$\left[\begin{array}{rrr} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5 \end{array}\right]$$
and B = $$\left[\begin{array}{rrr} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{array}\right]$$

∴ A2 = A . A
= $$\left[\begin{array}{rrr} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5 \end{array}\right]\left[\begin{array}{rrr} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5 \end{array}\right]$$
= $$\left[\begin{array}{rrr} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5 \end{array}\right]$$
= A

B2 = B . B
= $$\left[\begin{array}{rrr} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{array}\right]\left[\begin{array}{rrr} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{array}\right]$$
= $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
= I
∴ A2B2 = A . I
= A = $$\left[\begin{array}{rrr} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5 \end{array}\right]$$

Question 11.
If A = $$\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]$$, show that A2 – 5A + 7I = O. use this result to find A4.
Solution:
Here,
A2 – 5A + 7I = $$\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]-5\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]+7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rr} 8 & 5 \\ -5 & 3 \end{array}\right]-\left[\begin{array}{rr} 15 & 5 \\ -5 & 10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]$$
= $$\left[\begin{array}{cc} 8-15+7 & -5+5+0 \\ -5+5+0 & 3-10+7 \end{array}\right]$$
= $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$
= O
Thus, A2 = 5A + 7I ; pre-multiply bothsides by A
⇒ A3 = 5A2 – 71A
⇒ A3 = 5 (5A – 71) A [∵ IA = A = AI]
⇒ A3 = 25A – 35I – 7A
= 18A – 35I
⇒ A4 = 18A2 – 351A
⇒ A4 = 18 (5A – 71) – 35A
= 90A – 126I – 35A
= 55A – 126I
= $$55\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]-126\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rr} 165 & 55 \\ -55 & 110 \end{array}\right]-\left[\begin{array}{rr} 126 & 0 \\ 0 & 126 \end{array}\right]$$
= $$\left[\begin{array}{rr} 39 & 55 \\ -55 & -16 \end{array}\right]$$

Question 12.
If A = $$\left[\begin{array}{rrr} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{array}\right]$$, then show that A3 – 23A – 40I = O.
Solution:
Given A = $$\left[\begin{array}{rrr} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{array}\right]$$

Question 13.
If A = $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]$$, prove that An = $$\left[\begin{array}{ccc} 1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{array}\right]$$ for all positive integers n.
Solution:
We shall prove the given result by principle of mathematical induction on n.
A’ = A = $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ccc} 1 & 1 & \frac{1(1+1)}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]$$

∴ result is true for n = 1.
Let us assume that result is true for n = m.
i.e., Am = $$\left[\begin{array}{ccc} 1 & m & m(m+1) \\ 0 & 1 & m \\ 0 & 0 & 1 \end{array}\right]$$
We shall prove the given result for n = m + 1 ;
i.e., Am+1 = Am . A
= $$\left[\begin{array}{ccc} 1 & m & m(m+1) / 2 \\ 0 & 1 & m \\ 0 & 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ccc} 1 & 1+m & 1+m+\frac{m(m+1)}{2} \\ 1 & 1 & 1+m \\ 0 & 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ccc} 1 & m+1 & \frac{(m+1)(m+1+1)}{2} \\ 1 & 1 & m+1 \\ 0 & 0 & 1 \end{array}\right]$$

Question 14.
If A = $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$, prove that An = $$\left[\begin{array}{lll} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{array}\right]$$ for all positive integers n. (NCERT)
Solution:
We shall prove the given result by using mathematical induction on n ∈ N.
For n = 1 ;
A’ = A
= $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$
= $$\left[\begin{array}{lll} 3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1} \end{array}\right]$$
∴ result is true for n = 1
Let us assume that result is true for n = m

∴ result is true for n = m + 1
Hence by mathematical induction result is true for all n ∈ N.

Question 15.
If A = $$\left[\begin{array}{rrr} 2 & 4 & -1 \\ -1 & 0 & 2 \end{array}\right]$$ and B = $$\left[\begin{array}{rr} 3 & -4 \\ -1 & 2 \\ 2 & 1 \end{array}\right]$$, find (AB)’.
Solution:
Given A = $$\left[\begin{array}{rrr} 2 & 4 & -1 \\ -1 & 0 & 2 \end{array}\right]$$
and B = $$\left[\begin{array}{rr} 3 & -4 \\ -1 & 2 \\ 2 & 1 \end{array}\right]$$
AB = $$\left[\begin{array}{rrr} 2 & 4 & -1 \\ -1 & 0 & 2 \end{array}\right]\left[\begin{array}{rr} 3 & -4 \\ -1 & 2 \\ 2 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rr} 6-4-2 & -8+8-1 \\ -3+0+4 & 4+0+2 \end{array}\right]$$
= $$\left[\begin{array}{rr} 0 & -1 \\ 1 & 6 \end{array}\right]$$
∴ (AB)T = $$\left[\begin{array}{rr} 0 & 1 \\ -1 & 6 \end{array}\right]$$

Question 16.
For the matrix A = $$\left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right]$$, verify that
(i) A + A’ is a symmetric matrix
(ii) A – A’ is a skew-symmetric matrix
Solution:
(i) Given A = $$\left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right]$$
⇒ A’ = $$\left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right]$$
∴ A + A’ = $$\left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right]+\left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right]$$
= $$\left[\begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array}\right]$$
Let P = A + A’
⇒ P’ = $$\left[\begin{array}{rr} 2 & 11 \\ 11 & 14 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} 2 & 11 \\ 11 & 14 \end{array}\right]$$ = P
Thus P i.e., A + A’ is symmetric.

(ii) Since A = $$\left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right]$$
∴ A’ = $$\left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right]$$
∴ A – A’ = $$\left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right]-\left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right]$$
= $$\left[\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right]$$
Let Q = A – A’
Q’ = $$\left[\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right]^1$$
= $$\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right]$$
= $$-\left[\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right]$$
= – Q
Thus Q is skew-symmetric matrix.

Question 17.
If A and B are skcw-symmctrk matrices of the same order, prove that
(i) AB + BA is a symmetric matrix
(ii) AB – BA is a skew-symmetric matrix.
Solution:
Given A and B are skew symmetric matrices of same order.
∴ A’ = – A
and B’ = – B
(i) (AB + BA)’ = (AB)’ + (BA)’
[∵ (A + B)’ =A’ + B’]
= B’A’ + A’B’ [using reversal ‚aw]
= (- B) (- A) – (- A) (- B) [using eqn. (1)]
= BA + AB
= AB + BA [∵ A + B = B + A]
Thus, AB + BA is symmetric.

(ii) (AB – BA)’ = (AB)’ – (BA)’
[∵ (A – B)’ = A’ – B’]
= B’A’ – A’B’ [using reversal law]
= (- B) (- A) – (- A) (- B) [using eqn. (1)]
= BA – AB
= – (AB – BA)
Thus, AB – BA is skew-symmetric matrix.

Question 18.
Find the inverse of the following matrices, ¡f they exist, by using elementary operations :
(i) $$\left[\begin{array}{rr} 2 & -3 \\ -1 & 2 \end{array}\right]$$
(ii) $$\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]$$
(iii) $$\left[\begin{array}{rrr} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right]$$
Solution:
(i) We know that A = I2A
$$\left[\begin{array}{rr} 2 & -3 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$ A ;

R1 → R1 + R2

$$\left[\begin{array}{rr} 1 & -1 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]$$ A ;

R2 → R2 + R1

$$\left[\begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right]$$ A ;

R → R + R

$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]$$ A

Hence A-1 = $$\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]$$

(ii) Let A = $$\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]$$
Then A = I3A
⇒  A;
operate R2 → R2 – R1;
R3 → R3 – R1

(iii) (By elementary row transformations)
We know that A = I3A
$$\left[\begin{array}{rrr} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ A;

R2 ↔ R2 + 3R1 ;
R3 → R3 – 2R1

$$\left[\begin{array}{rrr} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{array}\right]$$ A;

R2 R3

$$\left[\begin{array}{rrr} 1 & 3 & -2 \\ 0 & -1 & 4 \\ 0 & 9 & -11 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 0 & 1 \\ 3 & 1 & 0 \end{array}\right]$$ A;

R3 → R3 + 9R2

(By elementary column transformations)
We know that
A = I3A

∴ A-1 = $$\left[\begin{array}{rrr} 1 & -2 / 5 & -3 / 5 \\ -2 / 5 & 4 / 25 & 11 / 25 \\ -3 / 5 & 1 / 25 & 9 / 25 \end{array}\right]$$ [by def. of inverse]

Question 19.
If M = {$$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]$$, x ≠ 0, x ∈ R} and * be an operation on Mdefined by A * B = AB, for all A, B ∈ M.
(i) Prove that * is a binary operation on M.
(ii) Prove that the operation * is commutative as well as associative.
(iii) Find the identity element of the operation *.
Solution:
Given M = {$$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]$$, x ≠ 0, x ∈ R}
and * be an operation on M defined by A * B = AB ∀ A, B ∈ M.

(i) ∀ A, B ∈ M,
Let A = $$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]$$
and B = $$\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]$$
where x, y ∈ R, x ≠ 0, y ≠ 0
∴ A * B = AB
= $$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]$$
= $$\left[\begin{array}{ll} x y+x y & x y+x y \\ x y+x y & x y+x y \end{array}\right]$$
= $$\left[\begin{array}{ll} 2 x y & 2 x y \\ 2 x y & 2 x y \end{array}\right]$$ ∈ M
[∵ x ≠ 0, y ≠ 0
⇒ xy ≠ 0
⇒ 2xy ≠ 0
also x, y ∈ R
⇒ xy ∈ R]

(ii) ∀ A, B ∈ M.
Let A = $$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]$$
and B = $$\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]$$
where x, y ∈ R and x ≠ 0 and y ≠ 0
Commutativity:
A * B = AB
= $$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]$$
= $$\left[\begin{array}{ll} 2 x y & 2 x y \\ 2 x y & 2 x y \end{array}\right]$$

and B * A = BA
= $$\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]$$
= $$\left[\begin{array}{ll} 2 y x & 2 y x \\ 2 y x & 2 y x \end{array}\right]$$

Since xy = yx ∀ x, y ∀ R
Thus, A * B = B * A
Hence * be commutativity on M.

Associativity:
∀ A, B, C ∈ M
Let A = $$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]$$ ;
B = $$\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]$$
and C = $$\left[\begin{array}{ll} z & z \\ z & z \end{array}\right]$$
where x, y, z ∈ R, x ≠ 0, y ≠ 0, z ≠ 0
Here (A * B) * C = (AB) * C = (AB) C
= $$\left(\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]\right)\left[\begin{array}{ll} z & z \\ z & z \end{array}\right]$$
= $$\left[\begin{array}{ll} 2 x y & 2 x y \\ 2 x y & 2 x y \end{array}\right]\left[\begin{array}{ll} z & z \\ z & z \end{array}\right]$$
= $$\left[\begin{array}{ll} 2 x y z & 2 x y z \\ 2 x y z & 2 x y z \end{array}\right]$$

and A * (B * C) = A * (BC) = A (BC)
= $$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]\left(\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]\left[\begin{array}{ll} z & z \\ z & z \end{array}\right]\right)$$
= $$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]\left[\begin{array}{ll} 2 y z & 2 y z \\ 2 y z & 2 y z \end{array}\right]$$
= $$\left[\begin{array}{ll} 2 x y z & 2 x y z \\ 2 x y z & 2 x y z \end{array}\right]$$
[∵ x, y, z ∈ R and associative law holds under multiplication]

Thus (A * B) * C = A * (B * C)
Hence * be associative on M.

(iii) Let E ∈ M be the identity element of operation * s.t.
E * A = A = A * E ∀ A ∈ M
Let E = $$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]$$
and A = $$\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]$$
where x ≠ 0, y ≠ 0 x, y ∈ R
∴ $$\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]=\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]$$
= $$\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]\left[\begin{array}{ll} x & x \\ x & x \end{array}\right]$$

⇒ $$\left[\begin{array}{ll} 2 x y & 2 x y \\ 2 x y & 2 x y \end{array}\right]=\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]$$
and $$\left[\begin{array}{ll} y & y \\ y & y \end{array}\right]=\left[\begin{array}{ll} 2 y x & 2 y x \\ 2 y x & 2 y x \end{array}\right]$$

⇒ 2xy = y
and 2yx = y
⇒ (2x – 1) y = 0
and y (2x – 1) = 0
⇒ x = $$\frac{1}{2}$$ or y = 0
since y ≠ 0
∴ E = $$\left[\begin{array}{ll} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{array}\right]$$ be the required element in operation * on M.