Chapter 3: Matrices

Q31Written answer

Show that if A and B are square matrices such that AB = BA, then

(A + B)² = A² + 2AB + B².

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Given, A and B are square matrices such that AB = BA.

So, (A + B)² = (A + B) . (A + B)

= A² + AB + BA + B²

= A² + AB + AB + B² [Since, AB = BA]

= A²+ 2AB + B²

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 59

Q32Written answer

Let and a = 4, b = –2.

Show that:

(a) A + (B + C) = (A + B) + C

(b) A (BC) = (AB) C

(c) (a + b)B = aB + bB

(d) a (C–A) = aC – aA

(e) (A^T)^T = A

(f) (bA)^T = b A^T

(g) (AB)^T = B^T A^T

(h) (A –B)C = AC – BC

(i) (A – B)^T = A^T – B^T

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NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 63

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 65

Q33Written answer

If \( A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \), then show that \( A^2 = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix} \).

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Solution coming soon.

Q34Written answer

If and x² = –1, then show that (A + B)² = A² + B²

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NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 67

Q35Written answer

Verify that A² = I when A =

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Q36Written answer

Prove by Mathematical Induction that (A¢)ⁿ = (Aⁿ) ¢, where n ∈ N for any square matrix A.

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Let P(n): (A¢)ⁿ = (Aⁿ) ¢

So, P(1): (A¢) = (A) ¢

A¢ = A¢

Hence, P(1) is true.

Now, let P(k) = (A¢)^k = (A^k) ¢, where k ∈ N

And,

P(k + 1): (A¢)^k+1 = (A¢)^kA¢

= (A^k) ¢A¢

= (AA^k) ¢

= (A^k+1) ¢

Hence, P(1) is true and whenever P(k) is true P(k + 1) is true.

Therefore, P(n) is true for all n ∈ N.

Q37Written answer

Find the inverse, by elementary row operations (if possible), of the following matrices

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 69

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As we obtain all the zeroes in a row of the matrix A on the L.H.S., A^-1 does not exist.

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 71

Q38Written answer

If then find values of x, y, z and w.

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In the given matrix equation,

Comparing the corresponding elements, we get

x + y = 6,

xy = 8,

z + 6 = 0 and

w = 4

From the first two equations, we have

(6 – y) . y = 8

y² – 6y + 8 = 0

(y – 2) (y – 4) = 0

y = 2 or y = 4

Hence, x = 4 and x = 2

Also, z + 6 = 0

z = -6 and w = 4

Therefore,

x = 2, y = 4 or x = 4, y = 2, z = -6 and w = 4

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 72

Q39Written answer

If find a matrix C such that 3A + 5B + 2C is a null matrix.

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Let’s consider a matrix C, such that

Comparing the terms,

2a + 48 = 0 ⇒ a = -24

20 + 2b = 0 ⇒ b = -10

56 + 2c = 0 ⇒ c = -28

And,

76 + 2d = 0 ⇒ d = -38

Therefore, the matrix C is

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 74

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 75

Q40Written answer

If , then find A² – 5A – 14I. Hence, obtain A³.

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