Class 12 Maths NCERT Solutions Chapter 3 Ex 3.2 – Operations on Matrices

(i) A+B = \begin{bmatrix} 2+1 & 4+3 \\ 3-2 & 2+5 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}

(ii) A-B = \begin{bmatrix} 2-1 & 4-3 \\ 3+2 & 2-5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}

(iii) 3A = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}\ \Rightarrow\ 3A-C = \begin{bmatrix} 6-(-2) & 12-5 \\ 9-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}

(iv) AB = \begin{bmatrix} 2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5) \end{bmatrix} = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}

(v) BA = \begin{bmatrix} 1(2)+3(3) & 1(4)+3(2) \\ -2(2)+5(3) & -2(4)+5(2) \end{bmatrix} = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}

(i) \begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}

(ii) Adding corresponding entries and grouping as perfect squares:

= \begin{bmatrix} a^2+2ab+b^2 & b^2+2bc+c^2 \\ a^2-2ac+c^2 & a^2-2ab+b^2 \end{bmatrix} = \begin{bmatrix} (a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2 \end{bmatrix}

(iii) = \begin{bmatrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix} = \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}

(iv) Using \sin^2x+\cos^2x = 1 in every entry:

(i) \begin{bmatrix} a^2+b^2 & -ab+ab \\ -ab+ab & b^2+a^2 \end{bmatrix} = \begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix}

(ii) \begin{bmatrix} 1(2) & 1(3) & 1(4) \\ 2(2) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4) \end{bmatrix} = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}

(iii) Row 1: [1(1)+(-2)(2),\ 1(2)+(-2)(3),\ 1(3)+(-2)(1)] = [-3,-4,1]; Row 2: [2(1)+3(2),\ 2(2)+3(3),\ 2(3)+3(1)] = [8,13,9]

= \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}

(iv) Multiplying row by column for each entry:

= \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}

(v) = \begin{bmatrix} 2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\ 3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \\ -1(1)+1(-1) & -1(0)+1(2) & -1(1)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix}

(vi) = \begin{bmatrix} 3(2)+(-1)(1)+3(3) & 3(-3)+(-1)(0)+3(1) \\ -1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1) \end{bmatrix} = \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}

A+B = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix}    B-C = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}

A+(B-C) = \begin{bmatrix} 1-1 & 2-2 & -3+0 \\ 5+4 & 0-1 & 2+3 \\ 1+1 & -1+2 & 1+0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}

(A+B)-C = \begin{bmatrix} 4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1+2 & 4-3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}

Multiplying every entry of A by 3 clears the denominator of 3, and multiplying every entry of B by 5 clears the denominator of 5:

3A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}    5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}

Since 3A and 5B turn out to be identical:

= \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ -\cos\theta\sin\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix}

= \begin{bmatrix} \cos^2\theta+\sin^2\theta & \cos\theta\sin\theta-\sin\theta\cos\theta \\ -\cos\theta\sin\theta+\sin\theta\cos\theta & \cos^2\theta+\sin^2\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

(i) Adding the two equations: 2X = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \Rightarrow X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}

Subtracting: 2Y = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix} \Rightarrow Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}

(ii) Multiply the first equation by 2 and the second by 3, then subtract to eliminate Y:

4X+6Y = \begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix},\quad 9X+6Y = \begin{bmatrix} 6 & -6 \\ -3 & 15 \end{bmatrix}

Subtracting: 5X = \begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix} \Rightarrow X = \begin{bmatrix} \tfrac{2}{5} & -\tfrac{12}{5} \\ -\tfrac{11}{5} & 3 \end{bmatrix}

Substituting back into 2X+3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} and solving for Y:

2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}

\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}

Comparing entries: 2+y=5 \Rightarrow y=3, and 2x+2=8 \Rightarrow x=3.

2\begin{bmatrix} x & z \\ y & t \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix} - \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 6 & 18 \\ 12 & 12 \end{bmatrix}

\begin{bmatrix} x & z \\ y & t \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ 6 & 6 \end{bmatrix}

Comparing entries gives two equations: 2x-y=10  ...(1) and 3x+y=5  ...(2).

Adding (1) and (2): 5x = 15 \Rightarrow x = 3. Substituting in (2): 9+y=5 \Rightarrow y = -4.

Comparing corresponding entries:

3x = x+4 \Rightarrow 2x = 4 \Rightarrow x = 2

3y = 6+(x+y) \Rightarrow 2y = 6+x = 6+2 = 8 \Rightarrow y = 4

3w = 2w+3 \Rightarrow w = 3

3z = -1+(z+w) \Rightarrow 2z = -1+w = -1+3 = 2 \Rightarrow z = 1

Multiplying F(x) and F(y):

F(x)F(y) = \begin{bmatrix} \cos x\cos y-\sin x\sin y & -\cos x\sin y-\sin x\cos y & 0 \\ \sin x\cos y+\cos x\sin y & -\sin x\sin y+\cos x\cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}

Using the compound angle identities \cos x\cos y-\sin x\sin y = \cos(x+y), \sin x\cos y+\cos x\sin y = \sin(x+y), and -(\cos x\sin y+\sin x\cos y) = -\sin(x+y):

F(x)F(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(x+y)

(i) \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}, while \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}.

(ii) Multiplying in the order given (first matrix × second matrix):

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}

Multiplying in the reverse order:

\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}

First compute A^2 = A \cdot A:

A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}

Now form 5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} and 6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}.

A^2-5A+6I = \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+6 \end{bmatrix}

First compute A^2 = A\cdot A = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}.

Then A^3 = A\cdot A^2 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}.

Now form 6A^2 = \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix}, 7A = \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix}, and 2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}.

A^3-6A^2+7A+2I = \begin{bmatrix} 21-30+7+2 & 0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0 & 55-78+21+2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

A^2 = A\cdot A = \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}

kA-2I = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}

Equating entries: 3k-2=1 \Rightarrow k=1; -2k=-2 \Rightarrow k=1; 4k=4 \Rightarrow k=1; -2k-2=-4 \Rightarrow k=1. All four give the same value.

Let t = \tan\tfrac{\alpha}{2}, so I+A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} and I-A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}.

Multiplying out the right-hand side:

(I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} \cos\alpha+t\sin\alpha & -\sin\alpha+t\cos\alpha \\ \sin\alpha-t\cos\alpha & t\sin\alpha+\cos\alpha \end{bmatrix}

Using the half-angle substitutions \cos\alpha = \dfrac{1-t^2}{1+t^2} and \sin\alpha = \dfrac{2t}{1+t^2}, each entry simplifies as follows:

\cos\alpha+t\sin\alpha = \dfrac{1-t^2+2t^2}{1+t^2} = \dfrac{1+t^2}{1+t^2} = 1

-\sin\alpha+t\cos\alpha = \dfrac{-2t+t(1-t^2)}{1+t^2} = \dfrac{-t(1+t^2)}{1+t^2} = -t

\sin\alpha-t\cos\alpha = \dfrac{2t-t(1-t^2)}{1+t^2} = \dfrac{t(1+t^2)}{1+t^2} = t

t\sin\alpha+\cos\alpha = \dfrac{2t^2+1-t^2}{1+t^2} = \dfrac{1+t^2}{1+t^2} = 1

So the product equals \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}, which is exactly I+A.

Let ₹x be invested in the first bond (5%) and ₹(30000 − x) in the second bond (7%). The interest earned is given by the matrix product

\begin{bmatrix} x & 30000-x \end{bmatrix}\begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix} = 0.05x + 0.07(30000-x)

(a) Total interest ₹1800: 0.05x+0.07(30000-x)=1800 \Rightarrow 2100-0.02x=1800 \Rightarrow x = 15000.

(b) Total interest ₹2000: 0.05x+0.07(30000-x)=2000 \Rightarrow 2100-0.02x=2000 \Rightarrow x = 5000.

Number of books: chemistry = 10\times12=120, physics =8\times12=96, economics =10\times12=120.

Writing quantities as a row matrix and prices as a column matrix and multiplying:

\begin{bmatrix} 120 & 96 & 120 \end{bmatrix}\begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix} = 120(80)+96(60)+120(40) = 9600+5760+4800

P has order p\times k and Y has order 3\times k. For the product PY to be defined, the number of columns of P must equal the number of rows of Y, so k=3.

With k=3, PY has order p\times3.

Also, W has order n\times3 and Y (now 3\times3) is compatible with W, so WY has order n\times3.

For PY+WY to be defined, PY and WY must have the same order, so p\times3 = n\times3 \Rightarrow p=n.

X has order 2\times n and Z has order 2\times p. For 7X-5Z to be defined, X and Z must have the same order, so n=p (which is given).

With n=p, both matrices are 2\times n, and so is their combination 7X-5Z.