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

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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 a_ij = \(\frac{i+3 j}{2}\)
(ii) Construct a 2 × 2 matrix A = [a_ij] who: elements are given by
a_ij = \(\begin{cases}i-j & \text { if } \quad i \geq j \\ i+j & \text { if } \quad i<j\end{cases}\)
Solution:
(i) Let A = [a_ij]_{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 = [a_ij]_{2 × 2} where 1 ≤ i, j ≤ 2
Here a_ij = \(\left\{\begin{array}{lll}
i-j & \text { if } & i \geq j \\
i+j & \text { if } & i<j
\end{array}\right.\)
∴ 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 = [a_ij] whose elements a_ij are given by a_ij = \(\left[\frac{i}{j}\right]\), where [x] represents greatest integer function.
Solution:
Given A = [a_ij]_{2 × 3}
where 1 ≤ i ≤ 2 ; 1 ≤ j ≤ 3
where a_ij = \(\left[\frac{i}{j}\right]\)
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., A^m = \(\left[\begin{array}{ccc}
1 & m & m(m+1) \\
0 & 1 & m \\
0 & 0 & 1
\end{array}\right]\)
We shall prove the given result for n = m + 1 ;
i.e., A^m+1 = A^m . 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.