Chapter 3: Matrices

Ans: Adding the matrices

$\mathrm{A}+\mathrm{B}=\left[\begin{array}{ll} 2 & 4 \\3 & 2\end{array}\right]+\left[\begin{array}{cc} 1 & 3 \\-2 & 5\end{array}\right]=\left[\begin{array}{ll} 3 & 7 \\1 & 7 \end{array}\right] $

Ans: Subtracting the matrices

$A-B=\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]-\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right]=\left[\begin{array}{cc} 1 & 1 \\ 5 & -3 \end{array}\right] $

Ans: Subtracting the matrices

$\begin{aligned} &3 \mathrm{~A}-\mathrm{C}=3\left[\begin{array}{cc} 2 & 4 \\ 3 & 2 \end{array}\right]-\left[\begin{array}{cc} -2 & 5 \\ 3 & 4 \end{array}\right] \\ &{\left[\begin{array}{cc} 6 & 12 \\ 9 & 6 \end{array}\right]-\left[\begin{array}{cc} -2 & 5 \\3 & 4 \end{array}\right]=\left[\begin{array}{cc} 8 & 7 \\ 6 & 2 \end{array}\right]} \end{aligned}$

Ans: Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as:

$\begin{aligned} &\mathrm{AB}=\left[\begin{array}{cc} 2 & 4 \\ 3 & 2\end{array}\right]\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right] \\ &{\left[\begin{array}{ll} 2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5) \end{array}\right]=\left[\begin{array}{cc} -6 & 26 \\ 1 & 19\end{array}\right]} \end{aligned} $

Ans: Computing,

$\left[\begin{array}{cc}a & b \\ -b & a\end{array}\right]+\left[\begin{array}{cc}a & b \\ b & a\end{array}\right]=\left[\begin{array}{cc}2 a & 2 b \\ 0 & 2 a\end{array}\right]$

ii. $\left[\begin{array}{cc}a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2}\end{array}\right]+\left[\begin{array}{cc}2 a b & 2 b c \\ -2 a c & -2 a b\end{array}\right]$

Ans: Computing,

$ \begin{aligned} &{\left[\begin{array}{cc} a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2} \end{array}\right]+\left[\begin{array}{cc} 2 a b & 2 b c \\ -2 a c & -2 a b \end{array}\right]=\left[\begin{array}{cc} a^{2}+b^{2}+2 a b & b^{2}+c^{2}+2 b c \\ a^{2}+c^{2}-2 a c & a^{2}+b^{2}-2 a b \end{array}\right]} \\ &{\left[\begin{array}{ll} a^{2}+b^{2}+2 a b & b^{2}+c^{2}+2 b c \\ a^{2}+c^{2}-2 a c & a^{2}+b^{2}-2 a b \end{array}\right]=\left[\begin{array}{ll} (a+b)^{2} & (b+c)^{2} \\ (a-c)^{2} & (a-b)^{2} \end{array}\right]} \end{aligned}$

Ans: Computing,

$ \left[\begin{array}{ccc} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{array}\right]+\left[\begin{array}{ccc} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{array}\right]=\left[\begin{array}{ccc} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{array}\right] $

Ans: Computing,

$ \begin{aligned} &{\left[\begin{array}{cc} \cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x \end{array}\right]+\left[\begin{array}{cc} \sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x \end{array}\right]=\left[\begin{array}{cc} \cos ^{2} x+\sin ^{2} x & \sin ^{2} x+\cos ^{2} x \\ \sin ^{2} x+\cos ^{2} x & \cos ^{2} x+\sin ^{2} x \end{array}\right]} \\ &{\left[\begin{array}{ll} \cos ^{2} x+\sin ^{2} x & \sin ^{2} x+\cos ^{2} x \\ \sin ^{2} x+\cos ^{2} x & \cos ^{2} x+\sin ^{2} x \end{array}\right]=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right],\left(\because \sin ^{2} x+\cos ^{2} x=1\right)} \end{aligned} $

Ans: Given: $\left[\begin{array}{cc}\mathrm{a} & \mathrm{b} \\ -\mathrm{b} & \mathrm{a}\end{array}\right]\left[\begin{array}{cc}\mathrm{a} & -\mathrm{b} \\ \mathrm{b} & \mathrm{a}\end{array}\right]$

$ \begin{aligned} {\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right]\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] } &=\left[\begin{array}{cc} a(a)+b(b) & a(-b)+a(b) \\ -b(a)+b(a) & -b(-b)+a(a) \end{array}\right] \\ &=\left[\begin{array}{cc} a^{2}+b^{2} & 0 \\ 0 & a^{2}+b^{2} \end{array}\right] \end{aligned} $

Ans: Given: $\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{lll}2 & 3 & 4\end{array}\right]$

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

$ =\left[\begin{array}{ccc} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{array}\right] $

Ans: Given: $\left[\begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$

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

Ans: Given: $\left[\begin{array}{lll}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]$

$ \begin{aligned} {\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right]\left[\begin{array}{ccc} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right] } &=\left[\begin{array}{ccc} 2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\ 3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \\ 4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5) \end{array}\right] \\ &=\left[\begin{array}{ccc} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{array}\right] \end{aligned} $

Ans: Given: $\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]$

$ \begin{aligned} {\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 1 \\ -1 & 2 & 1 \end{array}\right] } &=\left[\begin{array}{ccc} 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{array}\right] \\ &=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{array}\right] \end{aligned}$

Ans: Given: $\left[\begin{array}{ccc}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]$

\[\begin{aligned} {\left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{array}\right] } &=\left[\begin{array}{cc} 3(2)-1(1)+3(3) & 3(-3)-1(0)+3(1) \\ -1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1) \end{array}\right] \\ &=\left[\begin{array}{cc} 14 & -6 \\ 4 & 5 \end{array}\right] \end{aligned}\]

Ans:

Computing $(\mathrm{A}+\mathrm{B})=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right]+\left[\begin{array}{ccc}3 & 1 & 2 \\ 4 & 3 & 5 \\ 2 & -2 & 3\end{array}\right]=\left[\begin{array}{ccc}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4\end{array}\right]$

Computing $(B-C)=\left[\begin{array}{ccc}3 & 1 & 2 \\ 4 & 3 & 5 \\ 2 & -2 & 3\end{array}\right]-\left[\begin{array}{ccc}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]=\left[\begin{array}{ccc}-1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0\end{array}\right]$

To verify $\mathrm{A}+(\mathrm{B}-\mathrm{C})=(\mathrm{A}+\mathrm{B})-\mathrm{C}$,

We calculate $\mathrm{A}+(\mathrm{B}-\mathrm{C})$ and $(\mathrm{A}+\mathrm{B})-\mathrm{C}$ first, and then check that $\mathrm{LHS}=\mathrm{RHS}$

Computing $A+(B-C)$

$ \left[\begin{array}{ccc} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right]+\left[\begin{array}{ccc} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{array}\right] \text {, and } $ Computing $(A+B)-C$ $\left[\begin{array}{ccc} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{array}\right]-\left[\begin{array}{ccc} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{array}\right]=\left[\begin{array}{ccc} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{array}\right] $

Hence, we have verified that $A+(B-C)=(A+B)-C$.

Ans: Evaluating $3 \mathrm{~A}-5 \mathrm{~B}$, ${\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]-5\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]=\left[\begin{array}{lll}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]-\left[\begin{array}{lll}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right] }$

$=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$

Ans: Simplifying $\cos \theta\left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]+\sin \theta\left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right]$

$ \begin{aligned} &\Rightarrow\left[\begin{array}{cc} \cos ^{2} \theta & \cos \theta \sin \theta \\ -\sin \theta \cos \theta & \cos ^{2} \theta \end{array}\right]+\left[\begin{array}{cc} \sin ^{2} \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin ^{2} \theta \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} \cos ^{2} \theta+\sin ^{2} \theta & \cos \theta \sin \theta-\sin \theta \cos \theta \\ -\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^{2} \theta+\sin ^{2} \theta \end{array}\right] \end{aligned} $

$ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \quad\left(\because \cos ^{2} \theta+\sin ^{2} \theta=1\right) $

Ans: Given:

$ \begin{aligned} &X+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] \ldots(1) \\ &X-Y=\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] \ldots(2) \end{aligned} $ Adding these two equations, we get $ 2 X=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]+\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] $ $ \begin{aligned} &2 X=\left[\begin{array}{cc} 10 & 0 \\ 2 & 8 \end{array}\right] \\ &X=\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right] \end{aligned} $ Putting $X$ in (1) $ \begin{aligned} &{\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right]+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] Y=\left[\begin{array}{ll} 7-5 & 0-0 \\ 2-1 & 5-4 \end{array}\right]} \\ &Y=\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right] \end{aligned} $

Ans: Given:

$\begin{aligned} &2 X+3 Y=\left[\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}\right] \ldots(1) \\ &3 X+2 Y=\left[\begin{array}{cc} 2 & -2 \\ -1 & 5 \end{array}\right] \ldots \end{aligned} $ Multiplying (1) with 2, we have $4 X+6 Y=\left[\begin{array}{ll}4 & 6 \\ 8 & 0\end{array}\right] \ldots(3)$ Multiplying (2) with 3, we have $9 X+6 Y=\left[\begin{array}{cc} 6 & -6 \\ -3 & 15 \end{array}\right] \ldots(4) $ From (3) and (4), we have $-5 X=\left[\begin{array}{cc}4-6 & 6-(-6) \\ 8-(-3) & 0-15\end{array}\right] \therefore X=\left[\begin{array}{cc}\frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{array}\right]$ Putting $\mathrm{X}$ in (1) $ 2\left[\begin{array}{cc} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] 3 \mathrm{Y}=\left[\begin{array}{cc} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{array}\right] $ $ \therefore \mathrm{Y}=\left[\begin{array}{cc} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{array}\right] $

Ans: Given:

$2 \mathrm{X}+\mathrm{Y}=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]$, where $\mathrm{Y}=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$

$2 \mathrm{X}+\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right] 2 \mathrm{X}=\left[\begin{array}{cc}1-3 & 0-2 \\ -3-1 & 2-4\end{array}\right]$

$\therefore \mathrm{X}=\left[\begin{array}{rr}-1 & -1 \\ -2 & -1\end{array}\right]$

Ans: Given: $2\left[\begin{array}{ll}1 & 3 \\ 0 & \mathrm{x}\end{array}\right]+\left[\begin{array}{ll}\mathrm{y} & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}2 & 6 \\ 0 & 2 \mathrm{x}\end{array}\right]+\left[\begin{array}{cc}\mathrm{y} & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}5 & 6 \\ 1 & 8\end{array}\right] \Rightarrow\left[\begin{array}{cc}2+\mathrm{y} & 6+0 \\ 0+1 & 2 \mathrm{x}+2\end{array}\right]=\left[\begin{array}{cc}5 & 6 \\ 1 & 8\end{array}\right]$

Comparing the corresponding elements of these two matrices, we have:

Ans: Given:

$ 2\left[\begin{array}{cc} x & z \\ y & t \end{array}\right]+3\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]=3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right]$ $ \begin{aligned} &{\left[\begin{array}{cc} 2 x & 2 z \\ 2 y & 2 t \end{array}\right]+\left[\begin{array}{cc} 3 & -3 \\ 0 & 6 \end{array}\right]=\left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right]} \\ &{\left[\begin{array}{cc} 2 x+3 & 2 z-3 \\ 2 y & 2 t+6 \end{array}\right]=\left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right]} \end{aligned} $

Comparing the corresponding elements of these two matrices, we get:

$ \begin{aligned} &2 x+3=9 \\ &\Rightarrow x=3 \end{aligned} $ $ \begin{aligned} &2 \mathrm{y}=12 \\ &\Rightarrow \mathrm{y}=6 \end{aligned} $ $2 z-3=15 $ $ \Rightarrow \mathrm{z}=9 $ $ 2 t+6=18 $ $ \Rightarrow \mathrm{t}=6 $ $ \therefore \mathrm{x}=3, \mathrm{y}=6, \mathrm{z}=9 \text { and } \mathrm{t}=6 \text {. } $