Chapter 3: Matrices

Ans: Given:

$ \begin{aligned} &x\left[\begin{array}{l} 2 \\ 3 \end{array}\right]+y\left[\begin{array}{c} -1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right] \\ &{\left[\begin{array}{l} 2 x \\ 3 x \end{array}\right]+\left[\begin{array}{c} -y \\ y \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right]} \\ &{\left[\begin{array}{l} 2 x-y \\ 3 x+y \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right]} \end{aligned} $ Comparing the corresponding elements of these two matrices, we get $2 x-y=10$ and $3 x+y=5$

Adding these two equations, we have:

$ \begin{aligned} &5 \mathrm{x}=15 \\ &\Rightarrow \mathrm{x}=3 \end{aligned} $ Now, putting $x=3$ in any equation, we have: $ \begin{aligned} &\mathrm{y}=5-3 \mathrm{x} \\ &\Rightarrow \mathrm{y}=-4 \\ &\therefore \mathrm{x}=3 \text { and } \mathrm{y}=-4 \end{aligned} $

Ans: Simplifying

$\left[\begin{array}{cc}3 x & 3 y \\ 3 z & 3 w\end{array}\right]=\left[\begin{array}{cc}x+4 & 6+y+x \\ -1+z+w & 2 w+3\end{array}\right]$ Comparing the corresponding elements of these two matrices, we get: $3 x=x+4$ $\Rightarrow x=2$ $ \begin{aligned} &3 y=6+x+y \\ &\Rightarrow y=4 \end{aligned} $ $ \begin{aligned} &3 w=2 w+3 \\ &\Rightarrow w=3 \end{aligned} $ $ 3 z=-1+z+w $ $ \begin{aligned} &\Rightarrow \mathrm{z}=1 \\ &\therefore \mathrm{x}=2, \mathrm{y}=4, \mathrm{z}=1 \text { and } \mathrm{w}=3 \end{aligned} $

Ans: To show $\mathrm{F}(\mathrm{x}) \mathrm{F}(\mathrm{y})=\mathrm{F}(\mathrm{x}+\mathrm{y})$,

We first calculate $\mathrm{F}(\mathrm{x}) \mathrm{F}(\mathrm{y})$ and $\mathrm{F}(\mathrm{x}+\mathrm{y})$, and check that both are equal LHS

$F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$ and $F(y)=\left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]$ $F(x) F(y)=\left[\begin{array}{cccc}\cos x \cos y-\sin x \sin y & -\cos x \sin y-\sin x \cos y & 0 \\ \sin x \cos y+\cos x \sin y & -\sin x \sin y+\cos x \cos y & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$ $F(x) F(y)=\left[\begin{array}{ccc}\cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \operatorname{cox}(x+y) & 0 \\ 0 & 0 & 1\end{array}\right]$

$ \begin{aligned} &F(x) F(y)=\left[\begin{array}{ccc} \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{array}\right] \\ &F(x) F(y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \operatorname{cox}(x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned} $

RHS

$ F(x+y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \operatorname{cox}(x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] $

$\mathrm{LHS}=\mathrm{RHS}$, proved.

Ans: To show, we first calculate LHS and RHS

LHS

$ \begin{aligned} &{\left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 5(2)-1(3) & 5(1)-1(4) \\ 6(2)+7(3) & 6(1)+7(4) \end{array}\right]} \\ &{\left[\begin{array}{cc} 10-3 & 5-4 \\ 12+21 & 6+28 \end{array}\right]=\left[\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}\right]} \end{aligned} $ RHS $ \begin{aligned} &{\left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right]=\left[\begin{array}{ll} 2(5)+1(6) & 2(-1)+1(7) \\ 3(5)+4(6) & 3(-1)+4(7) \end{array}\right]} \\ &{\left[\begin{array}{cc} 10+6 & -2+7 \\ 15+24 & -3+28 \end{array}\right]=\left[\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array}\right]} \end{aligned} $

${\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right] }$ ${\left[\begin{array}{cc}10+6 & -2+ \\ 15+24 & -3+\end{array}\right.}$ $\therefore$ LHS $\neq$ RHS

Ans: To show, we first calculate LHS and RHS

LHS

$ \left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]=\left[\begin{array}{lll} 1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \\ 0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \\ 1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4) \end{array}\right] $

$=\left[\begin{array}{ccc} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right] $

RHS

$\begin{aligned} {\left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]}&\\=\left[\begin{array}{ccc} -1(1)+1(0)+0(1) & -1(2)+1(1)+0(1) & -1(3)+1(0)+0(0) \\ 0(1)-1(0)+1(1) & 0(2)-1(1)+1(1) & 0(3)-1(0)+1(0) \\ 2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0) \end{array}\right] \\ &\\=\left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right] \end{aligned}$

Ans: We know that, $\mathrm{A}^{2}=\mathrm{A} \times \mathrm{A}$

$ \begin{aligned} &{\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]=\left[\begin{array}{ll} 2(2)+0(2)+1(1) & 2(0)+0(1)+1(-1) & 2(1)+0(3)+1(0) \\ 2(2)+1(2)+3(1) & 2(0)+1(1)+3(-1) & 2(1)+1(3)+3(0) \\ 1(2)-1(2)+0(1) & 1(0)-1(1)+0(-1) & 1(1)-1(3)+0(0) \end{array}\right]} \\ &{\left[\begin{array}{lll} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3+0 \end{array}\right]=\left[\begin{array}{ccc} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]} \end{aligned} $

Substituting value of $\mathrm{A}^{2}, \mathrm{~A}, \mathrm{I}$ in $\mathrm{A}^{2}-5 \mathrm{~A}+6 \mathrm{I}$

$ \Rightarrow\left[\begin{array}{ccc} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-5\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]+6\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $

$\Rightarrow \begin{bmatrix} 5-10+6 &-1 &2-5 \\ 9-10&-2-5+6 & 5-15\\ -5+0 & -1+5 & -2-0+6 \end{bmatrix}\Rightarrow \begin{bmatrix} 1 & -1 &-3 \\ -1 &-1 & -10\\ -5& 4 & 4 \end{bmatrix}$

Ans: We know that, $\mathrm{A}^{2}=\mathrm{A} \times \mathrm{A}$ and $\mathrm{A}^{3}=\mathrm{A}^{2} \times \mathrm{A}$ For $\mathrm{A}^{2}$,

$ \begin{aligned} &{\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]=\left[\begin{array}{ll} 1(1)+0(0)+2(2) & 1(0)+0(2)+2(0) & 1(2)+0(1)+2(3) \\ 0(1)+2(0)+1(2) & 0(0)+2(2)+1(0) & 0(2)+2(1)+1(3) \\ 2(1)+0(0)+3(2) & 2(0)+0(2)+3(0) & 2(2)+0(1)+3(3) \end{array}\right]} \\ &{\left[\begin{array}{lll} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{array}\right]=\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]} \end{aligned} $

For $\mathrm{A}^{3}$,

Substituting value of $\mathrm{A}^{3}, \mathrm{~A}^{2}, \mathrm{~A}, \mathrm{I}$ in $\mathrm{A}^{3}-6 \mathrm{~A}^{2}+7 \mathrm{~A}+2 \mathrm{I}=0$

$ \begin{aligned} &\Rightarrow\left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-6\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]+7\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]+2\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-\left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right]+\left[\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array}\right]+\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \\ &\therefore \mathrm{A}^{3}-6 \mathrm{~A}^{2}+7 \mathrm{~A}+2 \mathrm{I}=0 \end{aligned} $

Ans: We know that, $\mathrm{A}^{2}=\mathrm{A} \times \mathrm{A}$

$ \begin{aligned} {\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] } &=\left[\begin{array}{ll} 3(3)-2(4) & 3(-2)-2(-2) \\ 4(3)-2(4) & 4(-2)-2(-2) \end{array}\right] \\ &=\left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right] \end{aligned} $ Now $\mathrm{A}^{2}=\mathrm{k} \mathrm{A}-2 \mathrm{I}$ $ \begin{aligned} &\Rightarrow\left[\begin{array}{cc} 1 & -2 \\ 4 & -4 \end{array}\right]=\mathrm{k}\left[\begin{array}{cc} 3 & -2 \\ 4 & -2 \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} 1 & -2 \\ 4 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 \mathrm{k}-2 & -2 \mathrm{k} \\ 4 \mathrm{k} & -2 \mathrm{k}-2 \end{array}\right] \end{aligned} $

Comparing the corresponding elements, we have:

Comparing the corresponding elements, we have:

$ \begin{aligned} &\Rightarrow 3 \mathrm{k}-2=1 \\ &\Rightarrow \mathrm{k}+1 \end{aligned} $

Thus, the value of $\mathrm{k}$ is 1 .

Ans: LHS

$ \mathrm{I}+\mathrm{A}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{cc} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{array}\right] $ $ =\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right] $

RHS

$ \begin{aligned} &(\mathrm{I}-\mathrm{A})\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right]\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} \cos \alpha+\sin \alpha \tan \frac{\alpha}{2} & -\sin \alpha+\cos \alpha \tan \frac{\alpha}{2} \\ -\cos \alpha \tan \frac{\alpha}{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha}{2}+\cos \alpha \end{array}\right] \end{aligned} $

$ \begin{aligned} &\Rightarrow\left[\begin{array}{cc} 1-2 \sin ^{2} \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\left(2 \cos ^{2} \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2} \\ -\left(2 \cos ^{2} \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2}+1-2 \sin ^{2} \frac{\alpha}{2} \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} 1-2 \sin ^{2} \frac{\alpha}{2}+2 \sin ^{2} \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}-\tan \frac{\alpha}{2} \\ -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin ^{2} \frac{\alpha}{2}+1-2 \sin ^{2} \frac{\alpha}{2} \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right] \end{aligned} $

We get LHS = RHS.

Ans: Let Rs $x$ be invested in the first bond. Then, the sum of money invested in the second bond pays Rs $(30,000-\mathrm{x})$

It is given that the first bond pays $5 \%$ interest per year and the second bond pay $7 \%$ interest per year.

Therefore, in order to obtain an annual total interest of Rs 1,800 , we have:

$\Rightarrow\left[\begin{array}{ll}x & 30,000-x\end{array}\right]\left[\begin{array}{c}\frac{5}{100} \\ \frac{7}{100}\end{array}\right]=1,800$

$ \begin{aligned} &\Rightarrow \frac{5 x}{100}+\frac{7(30,000-x)}{100}=1,800 \\ &\Rightarrow 5 x+2,10,000-7 x=1,80,000 \\ &\Rightarrow 2 x=30,000 \\ &\Rightarrow x=15,000 \end{aligned} $

Thus, in order to obtain an annual total interest of Rs 1,800 , the trust fund should invest Rs 15,000 in the first bond and the remaining Rs 15,000 in the second bond.

(b) Rs 2,000

Ans: Let Rs $x$ be invested in the first bond. Then, the sum of money invested in the second bond pays Rs $(30,000-\mathrm{x})$

It is given that the first bond pays $5 \%$ interest per year and the second bond pay $7 \%$ interest per year.

Therefore, in order to obtain an annual total interest of Rs 2,000 , we have:

$\Rightarrow\left[\begin{array}{ll}x & 30,000-x\end{array}\right]\left[\begin{array}{c}\frac{5}{100} \\ \frac{7}{100}\end{array}\right]=2,000$

$\begin{aligned} &\Rightarrow \frac{5 x}{100}+\frac{7(30,000-x)}{100}=2,000 \\ &\Rightarrow 5 x+2,10,000-7 x=2,00,000 \\ &\Rightarrow 2 x=10,000 \\ &\Rightarrow x=5,000 \end{aligned} $

Thus, in order to obtain an annual total interest of Rs 2,000 , the trust fund should invest Rs 5,000 in the first bond and the remaining Rs 25,000 in the second bond.

Ans: The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books. The selling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80, Rs 60 and Rs 40 . The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:

$ \begin{aligned} &\Rightarrow 12\left[\begin{array}{lll} 10 & 8 & 10 \end{array}\right]\left[\begin{array}{l} 80 \\ 60 \\ 40 \end{array}\right] \\ &\Rightarrow 12[10 \times 80+8 \times 60+10 \times 40] \\ &\Rightarrow 12(800+480+400) \\ &\Rightarrow 12(1680) \\ &\Rightarrow 20160 \end{aligned} $

Thus, the bookshop will receive Rs 20160 from the sale of all these books.