Maths Solve the Following Question.(4 Marks)

$\begin{aligned} & \mathrm{B}+\mathrm{C}-\mathrm{A} \\ & =\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & -6 & 0 \\ 0 & 0 & -3\end{array}\right]+\left[\begin{array}{ccc}-3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right]-\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -5\end{array}\right] \\ & =\left[\begin{array}{ccc}4-3-2 & 0 & 0 \\ 0 & -6+4+3 & 0 \\ 0 & 0 & -3+1+5\end{array}\right]\end{aligned}$

$=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{array}\right]$

$=\operatorname{diag}[-1,1,3]$

2. $\begin{aligned} & 2 \mathrm{~A}+\mathrm{B}-5 \mathrm{C} \\ & =2\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -5\end{array}\right]+\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & -6 & 0 \\ 0 & 0 & -3\end{array}\right]-5\left[\begin{array}{ccc}-3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right]\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & -6 & 0 \\ 0 & 0 & -10\end{array}\right]+\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & -6 & 0 \\ 0 & 0 & -3\end{array}\right]-\left[\begin{array}{ccc}-15 & 0 & 0 \\ 0 & 20 & 0 \\ 0 & 0 & 5\end{array}\right] \\ & =\left[\begin{array}{ccc}4+4+15 & 0 & 0 \\ 0 & -6-6-20 & 0 \\ 0 & 0 & -10-3-5\end{array}\right]\end{aligned}$

$=\left[\begin{array}{ccc}23 & 0 & 0 \\ 0 & -32 & 0 \\ 0 & 0 & -18\end{array}\right]$

$=\operatorname{diag}[23,-32,-18]$

x – y + 2z = 7 3x + 4y – 5z = 5 2x – y + 3z = 12

$D=\left|\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right|$

= 1(12 – 5) – (-1)(9 + 10) + 2(-3 – 8) = 1(7) + 1(19) + 2(-11) = 7 + 19 – 22 = 4 ≠ 0

$D x=\left|\begin{array}{ccc}7 & -1 & 2 \\ 5 & 4 & -5 \\ 12 & -1 & 3\end{array}\right|$

= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48) = 7(7) + 1(75) + 2(-53) = 49 + 75 – 106 = 18

$D y=\left|\begin{array}{ccc}1 & 7 & 2 \\ 3 & 5 & -5 \\ 2 & 12 & 3\end{array}\right|$

= 1(15 + 60) – 7(9 + 10) + 2(36 – 10) = 1(75) – 7(19) + 2(26) = 75 – 133 + 52 = -6

$D z=\left|\begin{array}{ccc}1 & -1 & 7 \\ 3 & 4 & 5 \\ 2 & -1 & 12\end{array}\right|$

= 1(48 + 5) – (-1)(36 – 10) + 7(-3 – 8) = 1(53) + 1(26) + 7(-11) = 53 + 26 – 77 = 2 By Cramer’s Rule,

$x=\frac{D_x}{D}=\frac{18}{4}=\frac{9}{2}, y=\frac{D_y}{D}=\frac{-6}{4}=\frac{-3}{2}$,

$\mathrm{z}=\frac{\mathrm{D}_2}{\mathrm{D}}=\frac{2}{4}=\frac{1}{2}$

$\therefore x=\frac{9}{2}, y=\frac{-3}{2}$ and $z=\frac{1}{2}$ are the solutions of the given equations.

2x + 3y + 3z = 5 x – 2y + z = -4 3x – y – 2z = 3

$\mathrm{D}=\left|\begin{array}{ccc}2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2\end{array}\right|$

= 2(4 + 1) – 3(-2 – 3) + 3(-1 + 6) = 2(5) – 3(-5) + 3(5) = 10 + 15 + 15 = 40 ≠ 0

$D x=\left|\begin{array}{ccc}5 & 3 & 3 \\ -4 & -2 & 1 \\ 3 & -1 & -2\end{array}\right|$

= 5(4 + 1) – 3(8 – 3) + 3(4 + 6) = 5(5) – 3(5) + 3(10) = 25 – 15 + 30 = 40

$D y=\left|\begin{array}{ccc}2 & 5 & 3 \\ 1 & -4 & 1 \\ 3 & 3 & -2\end{array}\right|$

= 2(8 – 3) – 5(-2 – 3) + 3(3 + 12) = 2(5) – 5(-5) + 3(15) = 10 + 25 + 45 = 80

$D z=\left|\begin{array}{ccc}2 & 3 & 5 \\ 1 & -2 & -4 \\ 3 & -1 & 3\end{array}\right|$

= 2(-6 – 4) – 3(3 + 12) + 5(-1 + 6) = 2(-10) – 3(15) + 5(5) = -20 -45 + 25 = -40 By Cramer’s Rule,

$x=\frac{\mathrm{D}_y}{\mathrm{D}}=\frac{40}{40}=1, y=\frac{\mathrm{D}_y}{\mathrm{D}}=\frac{80}{40}=2$,

$z=\frac{D_x}{D}=\frac{-40}{40}=-1$

∴ x = 1, y = 2 and z = -1 are the solutions of the given equations.

∴ The given equations become

$p+q=\frac{3}{2}$

$\begin{aligned} & \text { i.e., } 2 p+2 q=3 \\ & \text { i.e., } 2 p+2 q+0=3\end{aligned}$

$q+r=\frac{5}{6}$

$\begin{aligned} & \text { i.e., } 6 q+6 r=5 \\ & \text { i.e., } 0 p+6 q+6 r=5\end{aligned}$

$r+p=\frac{4}{3}$

i.e., 3r + 3p = 4, i.e., 3p + 0q + 3r = 4

$D=\left|\begin{array}{lll}2 & 2 & 0 \\ 0 & 6 & 6 \\ 3 & 0 & 3\end{array}\right|$

= 2(18 – 0) -2(0 – 18) + 0 = 2(18) – 2(-18) = 36 + 36 = 72 ≠ 0

$D p=\left|\begin{array}{lll}3 & 2 & 0 \\ 5 & 6 & 6 \\ 4 & 0 & 3\end{array}\right|$

= 3(18 – 0) – 2(15 – 24) + 0 = 3(18) – 2(-9) = 54 + 18 = 72

$\mathrm{Dq}=\left|\begin{array}{lll}2 & 3 & 0 \\ 0 & 5 & 6 \\ 3 & 4 & 3\end{array}\right|$

= 2(15 – 24) – 3(0 – 18) + 0 = 2(-9) – 3(-18) = -18 + 54 = 36

$\mathrm{Dr}=\left|\begin{array}{lll}2 & 2 & 3 \\ 0 & 6 & 5 \\ 3 & 0 & 4\end{array}\right|$

= 2(24 – 0) – 2(0 – 15) + 3(0 – 18) = 2(24) – 2(-15) + 3(-18) = 48 + 30 – 54 = 24

By Cramer’s Rule,

$\mathrm{p}=\frac{\mathrm{D}_{\mathrm{p}}}{\mathrm{D}}=\frac{72}{72}=1, \mathrm{q}=\frac{\mathrm{D}_{\mathrm{q}}}{\mathrm{D}}=\frac{36}{72}=\frac{1}{2}$,

$\mathrm{r}=\frac{\mathrm{D}_{\mathrm{r}}}{\mathrm{D}}=\frac{24}{72}=\frac{1}{3}$

$\frac{1}{x}=1, \frac{1}{v}=\frac{1}{2}, \frac{1}{z}=\frac{1}{3}$

∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.

2x – y + z = 1 x + 2y + 3z = 8 3x + y – 4z = 1

$D=\left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & -4\end{array}\right|$

= 2(-8 – 3) – (-1)(-4 – 9) + 1(1 – 6) = 2(-11) + 1(-13) + 1(-5) = -22 – 13 – 5 = -40 ≠ 0

Dx $=\left|\begin{array}{ccc}1 & -1 & 1 \\ 8 & 2 & 3 \\ 1 & 1 & -4\end{array}\right|$

= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2) = 1(-11) + 1(-35) + 1(6) = -11 – 35 + 6 = -40

$D y=\left|\begin{array}{ccc}2 & 1 & 1 \\ 1 & 8 & 3 \\ 3 & 1 & -4\end{array}\right|$

= 2(-32 – 3) -1(-4 – 9) + 1(1 – 24) = 2(-35) – 1(-13) + 1(-23) = -70 + 13 – 23 = -80

$\mathrm{D} z=\left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 2 & 8 \\ 3 & 1 & 1\end{array}\right|$

= 2(2 – 8) – (-1)(1 – 24) + 1(1 – 6) = 2(-6) + 1(-23) + 1(-5) = -12 – 23 – 5 = -40

By Cramer's Rule,

$x=\frac{\mathrm{D}_x}{\mathrm{D}}=\frac{-40}{-40}=1, y=\frac{\mathrm{D}_z}{\mathrm{D}}=\frac{-80}{-40}=2$,

$z=\frac{D_z}{D}=\frac{-40}{-40}=$

∴ x = 1, y = 2 and z = 1 are the solutions of the given equations.

$\mathrm{M}_{11}=\left|\begin{array}{cc}0 & -2 \\ 0 & 3\end{array}\right|=0-0=0$

$\begin{aligned} \therefore & \mathrm{C}_{11}=(-1)^{1+1} \mathrm{M}_{11}=1(0)=0 \\ & \mathrm{M}_{12}=\left|\begin{array}{cc}3 & -2 \\ 1 & 3\end{array}\right|=9+2=11\end{aligned}$

$\begin{aligned} \therefore \quad & \mathrm{C}_{12}=(-1)^{1+2} \mathrm{M}_{12}=(-1)(11)=-11 \\ & \mathrm{M}_{13}=\left|\begin{array}{ll}3 & 0 \\ 1 & 0\end{array}\right|=0-0=0\end{aligned}$

$\begin{aligned} \therefore \quad & \mathrm{C}_{13}=(-1)^{1+3} \mathrm{M}_{13}=1(0)=0 \\ & \mathrm{M}_{21}=\left|\begin{array}{cc}-1 & 2 \\ 0 & 3\end{array}\right|=-3-0=-3\end{aligned}$

$\begin{aligned} \therefore & \mathrm{C}_{21}=(-1)^{2+1} \mathrm{M}_{21}=(-1)(-3)=3 \\ & \mathrm{M}_{22}=\left|\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right|=3-2=1\end{aligned}$

$\begin{aligned} \therefore \quad & \mathrm{C}_{22}=(-1)^{2+2} \mathrm{M}_{22}=1(1)=1 \\ & \mathrm{M}_{23}=\left|\begin{array}{cc}1 & -1 \\ 1 & 0\end{array}\right|=0+1=1\end{aligned}$

$\begin{aligned} \therefore \quad & C_{23}=(-1)^{2+3} M_{23}=(-1)(1)=-1 \\ & M_{31}=\left|\begin{array}{cc}-1 & 2 \\ 0 & -2\end{array}\right|=2-0=2\end{aligned}$

$\begin{aligned} \therefore \quad C_{31} & =(-1)^{3+1} M_{31}=1(2)=2 \\ M_{32} & =\left|\begin{array}{cc}1 & 2 \\ 3 & -2\end{array}\right|=-2-6=-8\end{aligned}$

$\begin{aligned} \therefore \quad & C_{32}=(-1)^{3+2} M_{32}=(-1)(-8)=8 \\ & M_{33}=\left|\begin{array}{cc}1 & -1 \\ 3 & 0\end{array}\right|=0+3=3\end{aligned}$

$\therefore \quad C_{43}=(-1)^{3+3} M_{33}=1(3)=3$

$\mathrm{M}_{11}=\left|\begin{array}{cc}1 & 3 \\ -4 & 2\end{array}\right|=2+12=14$

$\begin{aligned} \therefore \quad & C_{11}=(-1)^{1+1} M_{11}=1(14)=14 \\ & M_{12}=\left|\begin{array}{cc}-2 & 3 \\ 0 & 2\end{array}\right|=-4-0=-4\end{aligned}$

$\begin{array}{ll}\therefore \quad & C_{12}=(-1)^{1+2} M_{12}=(-1)(-4)=4 \\ & \mathrm{M}_{13}=\left|\begin{array}{cc}-2 & 1 \\ 0 & -4\end{array}\right|=8-0=8\end{array}$

$\begin{aligned} \therefore \quad C_{13} & =(-1)^{1+3} M_{13}=1(8)=8 \\ & M_{21}=\left|\begin{array}{cc}0 & 4 \\ -4 & 2\end{array}\right|=0+16=16\end{aligned}$

$\begin{aligned} \therefore \quad & C_{21}=(-1)^{2+1} M_{21}=(-1)(16)=-16 \\ & M_{22}=\left|\begin{array}{cc}-1 & 4 \\ 0 & 2\end{array}\right|=-2-0=-2\end{aligned}$

$\begin{aligned} \therefore \quad & \mathrm{C}_{22}=(-1)^{2+2} \mathrm{M}_{22}=1(-2)=-2 \\ & \mathrm{M}_{23}=\left|\begin{array}{cc}-1 & 0 \\ 0 & -4\end{array}\right|=4-0=4\end{aligned}$

$\begin{aligned} \therefore \quad C_{23} & =(-1)^{2+3} M_{23}=(-1)(4)=-4 \\ M_{31} & =\left|\begin{array}{ll}0 & 4 \\ 1 & 3\end{array}\right|=0-4=-4\end{aligned}$

$\begin{array}{ll}\therefore \quad & C_{31}=(-1)^{3+1} M_{31}=1(-4)=-4 \\ & M_{32}=\left|\begin{array}{cc}-1 & 4 \\ -2 & 3\end{array}\right|=-3+8=5\end{array}$

$\begin{aligned} \therefore \quad & C_{32}=(-1)^{3+2} M_{32}=(-1)(5)= \\ & M_{33}=\left|\begin{array}{cc}-1 & 0 \\ -2 & \end{array}\right|=-1+0=-1\end{aligned}$

$\therefore \quad C_{33}=(-1)^{3+3} \mathrm{M}_{33}=1(-1)=-1$

$\begin{aligned} & \mathrm{M}_{23}=\left|\begin{array}{rr}1 & 2 \\ 5 & -1\end{array}\right|=-1-10=-11 \\ & \mathrm{C}_{23}=(-1)^{2+3} \mathrm{M}_{23}=(-1)^{2+3} \cdot(-11)=11 \\ & \mathrm{M}_{31}=\left|\begin{array}{rr}2 & -3 \\ 0 & 4\end{array}\right|=8-0=8 \\ & \mathrm{C}_{31}=(-1)^{3+1} \mathrm{M}_{31}=(-1)^{3+1} \cdot 8=8 \\ & \mathrm{M}_{32}=\left|\begin{array}{lr}1 & -3 \\ -2 & 4\end{array}\right|=4-6=-2 \\ & \mathrm{C}_{32}=(-1)^{3+2} \mathrm{M}_{32}=(-1)^{3+2} \cdot(-2)=2 \\ & \mathrm{M}_{33}=\left|\begin{array}{lr}1 & 2 \\ -2 & 0\end{array}\right|=0+4=4 \\ & \mathrm{C}_{33}=(-1)^{3+3} \mathrm{M}_{33}=(-1)^{3+3} \cdot 4=4\end{aligned}$

$\therefore \quad A^T=\left[\begin{array}{ccc}5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3\end{array}\right]$

$\begin{aligned} \therefore \quad \mathrm{A}+\mathrm{A}^{\mathrm{T}} & =\left[\begin{array}{ccc}5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3\end{array}\right]+\left[\begin{array}{ccc}5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3\end{array}\right] \\ & =\left[\begin{array}{ccc}5+5 & 2+3 & -4+4 \\ 3+2 & -7-7 & 2-5 \\ 4-4 & -5+2 & -3-3\end{array}\right]\end{aligned}$

$\therefore \quad A+A^T=\left[\begin{array}{ccc}10 & 5 & 0 \\ 5 & -14 & -3 \\ 0 & -3 & -6\end{array}\right]$

$\therefore \quad\left(A+A^T\right)^T=\left[\begin{array}{ccc}10 & 5 & 0 \\ 5 & -14 & -3 \\ 0 & -3 & -6\end{array}\right]$

$\therefore\left(A+A^{\top}\right)^{\top}=A+A^{\top}$, i.e., $A+A^{\top}=\left(A+A^{\top}\right)^{\top}$

$\therefore \mathrm{A}+\mathrm{A}^{\top}$ is a symmetric matrix.

$A-A^T=\left[\begin{array}{ccc}5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3\end{array}\right]-\left[\begin{array}{ccc}5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{ccc}5-5 & 2-3 & -4-4 \\ 3-2 & -7+7 & 2+5 \\ 4+4 & -5-2 & -3+3\end{array}\right] \\ & =\left[\begin{array}{ccc}0 & -1 & -8 \\ 1 & 0 & 7 \\ 8 & -7 & 0\end{array}\right]\end{aligned}$

$\therefore \quad\left(A-A^T\right)^T=\left[\begin{array}{ccc}0 & 1 & 8 \\ -1 & 0 & -7 \\ -8 & 7 & 0\end{array}\right]=-\left[\begin{array}{ccc}0 & -1 & -8 \\ 1 & 0 & 7 \\ 8 & -7 & 0\end{array}\right]$

$ \therefore\left(A-A^{\top}\right)^{\top}=-\left(A-A^{\top}\right) $ i.e. $A-A^{\top}=-\left(A-A^{\top}\right)^{\top}$

$\therefore \mathrm{A}-\mathrm{A}^{\top}$ is skew symmetric matrix.

$\therefore \quad A^T=\left[\begin{array}{ccc}1 & 3 & -2 \\ 2 & 2 & -3 \\ 4 & 1 & 2\end{array}\right]$

$\therefore \quad \mathrm{A}+\mathbf{A}^{\mathrm{T}}=\left[\begin{array}{ccc}1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2\end{array}\right]+\left[\begin{array}{ccc}1 & 3 & -2 \\ 2 & 2 & -3 \\ 4 & -1 & 2\end{array}\right]$

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

$\therefore \quad \mathrm{A}+\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 5 & 2 \\ 5 & 4 & -2 \\ 2 & -2 & 4\end{array}\right]$

$\therefore \quad\left(\mathrm{A}+\mathrm{A}^{\mathrm{T}}\right)^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 5 & 2 \\ 5 & 4 & -2 \\ 2 & -2 & 4\end{array}\right]$

$\therefore\left(A+A^{\top}\right)^{\top}=A+A^{\top}$, i.e., $A+A^{\top}=\left(A+A^{\top}\right)^{\top}$

∴ A + AT is a symmetric matrix.

$A-A^T=\left[\begin{array}{ccc}1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2\end{array}\right]-\left[\begin{array}{ccc}1 & 3 & -2 \\ 2 & 2 & -3 \\ 4 & 1 & 2\end{array}\right]$

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

$\therefore \quad A-A^T=\left[\begin{array}{ccc}0 & -1 & 6 \\ 1 & 0 & 4 \\ -6 & -4 & 0\end{array}\right]$

$\therefore \quad\left(A-A^T\right)^T=\left[\begin{array}{ccc}0 & 1 & -6 \\ -1 & 0 & -4 \\ 6 & 4 & 0\end{array}\right]=-\left[\begin{array}{ccc}0 & -1 & 6 \\ 1 & 0 & 4 \\ -6 & -4 & 0\end{array}\right]$

$\begin{aligned} & \therefore\left(A-A^{\top}\right)^{\top}=-\left(A-A^{\top}\right) \\ & \text { i.e., } A-A^{\top}=-\left(A-A^{\top}\right)^{\top}\end{aligned}$

$\therefore A-A^{\top}$ is skew symmetric matrix.

$\therefore \quad A^T=\left[\begin{array}{cc}-1 & -3 \\ 2 & 2 \\ 1 & -3\end{array}\right]$ and $B^T=\left[\begin{array}{ccc}2 & -3 & -1 \\ 1 & 2 & 3\end{array}\right]$

$\therefore \quad A+B^T=\left[\begin{array}{ccc}-1 & 2 & 1 \\ -3 & 2 & -3\end{array}\right]+\left[\begin{array}{ccc}2 & -3 & -1 \\ 1 & 2 & 3\end{array}\right]$

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

$\therefore \quad\left(A+B^T\right)^T=\left[\begin{array}{cc}1 & -2 \\ -1 & 4 \\ 0 & 0\end{array}\right]$

$\ldots$..(i)

Now, $A^{\mathrm{T}}+\mathrm{B}=\left[\begin{array}{cc}-1 & -3 \\ 2 & 2 \\ 1 & -3\end{array}\right]+\left[\begin{array}{cc}2 & 1 \\ -3 & 2 \\ -1 & 3\end{array}\right]$

$=\left[\begin{array}{cc}-1+2 & -3+1 \\ 2-3 & 2+2 \\ 1-1 & -3+3\end{array}\right]$

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

...(ii)

$\therefore \quad A+2 B+3 C=\left[\begin{array}{ccc}5 & 8 & 2 \\ 6 & 8 & -2\end{array}\right]$

$\therefore \quad[\mathrm{A}+2 \mathrm{~B}+3 \mathrm{C}]^{\mathrm{T}}=\left[\begin{array}{cc}3 & 0 \\ 8 & 8 \\ 2 & -2\end{array}\right]$

$\ldots(\mathrm{i})$

Now, $A^T=\left[\begin{array}{ll}1 & 3 \\ 0 & 1 \\ 1 & 2\end{array}\right], B^T=\left[\begin{array}{cc}2 & 3 \\ 1 & 5 \\ -4 & -2\end{array}\right]$

and $C^T=\left[\begin{array}{cc}0 & -1 \\ 2 & -1 \\ 3 & 0\end{array}\right]$

$\therefore \quad A^T+2 B^T+3 C^T$

$\begin{aligned} & =\left[\begin{array}{ll}1 & 3 \\ 0 & 1 \\ 1 & 2\end{array}\right]+2\left[\begin{array}{cc}2 & 3 \\ 1 & 5 \\ -4 & -2\end{array}\right]+3\left[\begin{array}{cc}0 & -1 \\ 2 & -1 \\ 3 & 0\end{array}\right] \\ & =\left[\begin{array}{ll}1 & 3 \\ 0 & 1 \\ 1 & 2\end{array}\right]+\left[\begin{array}{cc}4 & 6 \\ 2 & 10 \\ -8 & -4\end{array}\right]+\left[\begin{array}{cc}0 & -3 \\ 6 & -3 \\ 9 & 0\end{array}\right] \\ & =\left[\begin{array}{ll}1+4+0 & 3+6-3 \\ 0+2+6 & 1+10-3 \\ 1-8+9 & 2-4+0\end{array}\right]\end{aligned}$

$\therefore A^{\top}+2 B^{\top}+3 C^{\top}=\left[\begin{array}{cc}5 & 6 \\ 8 & 8 \\ 2 & -2\end{array}\right]$

From (i) and (ii), we get

$(A+2 B+3 C)^{\top}=A^{\top}+2 B^{\top}+3 C^{\top}$

$\begin{aligned} A-I & =\left[\begin{array}{ccc}1 & 2 & 0 \\ 5 & 4 & 2 \\ 0 & 7 & -3\end{array}\right]-\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \\ & =\left[\begin{array}{ccc}1-1 & 2-0 & 0-0 \\ 5-0 & 4-1 & 2-0 \\ 0-0 & 7-0 & -3-1\end{array}\right]=\left[\begin{array}{ccc}0 & 2 & 0 \\ 5 & 3 & 2 \\ 0 & 7 & -4\end{array}\right]\end{aligned}$

$\begin{aligned} \therefore \quad & (A+I)(A-I) \\ \quad & {\left[\begin{array}{ccc}2 & 2 & 0 \\ 5 & 5 & 2 \\ 0 & 7 & -2\end{array}\right]\left[\begin{array}{lll}0 & 2 & 0 \\ 5 & 3 & 2 \\ 0 & 7 & -4\end{array}\right] }\end{aligned}$

$=\left[\begin{array}{ccc}0+10+0 & 4+6+0 & 0+4-0 \\ 0+25+0 & 10+15+14 & 0+10-8 \\ 0+35+0 & 0+21-14 & 0+14+8\end{array}\right]$

$=\left[\begin{array}{ccc}10 & 10 & 4 \\ 25 & 39 & 2 \\ 35 & 7 & 22\end{array}\right]$

[Note: Answer given in the textbook is $\left[\begin{array}{ccc}9 & 6 & 4 \\ 15 & 32 & -2 \\ 35 & -7 & 29\end{array}\right]$

However, as per our calculation it is $\left[\begin{array}{ccc}10 & 10 & 4 \\ 25 & 39 & 2 \\ 35 & 7 & 22\end{array}\right]$.]

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

$\therefore \quad A(B C)=\left[\begin{array}{lll}1 & 0 & 1 \\ 2 & 3 & 0 \\ 0 & 4 & 5\end{array}\right]\left[\begin{array}{ccc}2 & 4 & 2 \\ -1 & -2 & -1 \\ 6 & 0 & -6\end{array}\right]$

$=\left[\begin{array}{ccc}2-0+6 & 4-0+0 & 2-0-6 \\ 4-3+0 & 8-6+0 & 4-3-0 \\ 0-4+30 & 0-8+0 & 0-4-30\end{array}\right]$

$\therefore \quad A(B C)=\left[\begin{array}{ccc}8 & 4 & -4 \\ 1 & 2 & 1 \\ 26 & -8 & -34\end{array}\right]$

$\ldots$.(i)

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

$=\left[\begin{array}{ll}2+0+0 & -2+0+3 \\ 4-3+0 & -4+3+0 \\ 0-4+0 & 0+4+15\end{array}\right]=\left[\begin{array}{cc}2 & 1 \\ 1 & -1 \\ -4 & 19\end{array}\right]$

$\therefore \quad(A B) C=\left[\begin{array}{cc}2 & 1 \\ 1 & -1 \\ -4 & 19\end{array}\right]\left[\begin{array}{ccc}3 & 2 & -1 \\ 2 & 0 & -2\end{array}\right]$

$=\left[\begin{array}{ccc}6+2 & 4+0 & -2-2 \\ 3-2 & 2-0 & -1+2 \\ -12+38 & -8+0 & 4-38\end{array}\right]$

$\therefore \quad(A B) C=\left[\begin{array}{ccc}8 & 4 & -4 \\ 1 & 2 & 1 \\ 26 & -8 & -34\end{array}\right]$

...(ii)

$\therefore \quad A(B C)=\left[\begin{array}{ccc}8 & 4 & -4 \\ 1 & 2 & 1 \\ 26 & -8 & -34\end{array}\right]$

$\ldots$..(i)

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

$=\left[\begin{array}{ll}2+0+0 & -2+0+3 \\ 4-3+0 & -4+3+0 \\ 0-4+0 & 0+4+15\end{array}\right]=\left[\begin{array}{cc}2 & 1 \\ 1 & -1 \\ -4 & 19\end{array}\right]$

$\therefore \quad(A B) C=\left[\begin{array}{cc}2 & 1 \\ 1 & -1 \\ -4 & 19\end{array}\right]\left[\begin{array}{ccc}3 & 2 & -1 \\ 2 & 0 & -2\end{array}\right]$

$=\left[\begin{array}{ccc}6+2 & 4+0 & -2-2 \\ 3-2 & 2-0 & -1+2 \\ -12+38 & -8+0 & 4-38\end{array}\right]$

$\therefore \quad(\mathrm{AB}) \mathrm{C}=\left[\begin{array}{ccc}8 & 4 & -4 \\ 1 & 2 & 1 \\ 26 & -8 & -34\end{array}\right]$

From (i) and (ii), we get A(BC) = (AB)C.

$2 A-B=\left[\begin{array}{ccc}6 & -6 & 0 \\ -4 & 2 & 1\end{array}\right]$

…………… (i)

and $A-2 B=\left[\begin{array}{ccc}3 & 2 & 8 \\ -2 & 1 & -7\end{array}\right] \ldots \ldots \ldots \ldots \ldots$..(ii)

By (i) – (ii) x 2, we get

$\begin{aligned} 3 B & =\left[\begin{array}{ccc}6 & -6 & 0 \\ -4 & 2 & 1\end{array}\right]-2\left[\begin{array}{ccc}3 & 2 & 8 \\ -2 & 1 & -7\end{array}\right] \\ & =\left[\begin{array}{ccc}6 & -6 & 0 \\ -4 & 2 & 1\end{array}\right]-\left[\begin{array}{ccc}6 & 4 & 16 \\ -4 & 2 & -14\end{array}\right] \\ & =\left[\begin{array}{lll}6-6 & -6-4 & 0-16 \\ -4+4 & 2-2 & 1+14\end{array}\right]\end{aligned}$

$\begin{array}{ll}\therefore & 3 B=\left[\begin{array}{ccc}0 & -10 & -16 \\ 0 & 0 & 15\end{array}\right] \\ \therefore & B=\frac{1}{3}\left[\begin{array}{ccc}0 & -10 & -16 \\ 0 & 0 & 15\end{array}\right] \\ \therefore & B=\left[\begin{array}{ccc}0 & \frac{-10}{3} & \frac{-16}{3} \\ 0 & 0 & 5\end{array}\right] \mathrm{M}\end{array}$

By (i) $\times 2-$ (ii), we get

$\begin{aligned} 3 A & =2\left[\begin{array}{ccc}6 & -6 & 0 \\ -4 & 2 & 1\end{array}\right]-\left[\begin{array}{ccc}3 & 2 & 8 \\ -2 & 1 & -7\end{array}\right] \\ & =\left[\begin{array}{ccc}12 & -12 & 0 \\ -8 & 4 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & 2 & 8 \\ -2 & 1 & -7\end{array}\right] \\ & =\left[\begin{array}{ccc}12-3 & -12-2 & 0-8 \\ -8+2 & 4-1 & 2+7\end{array}\right]\end{aligned}$

$\begin{array}{ll}\therefore & 3 A=\left[\begin{array}{ccc}9 & -14 & -8 \\ -6 & 3 & 9\end{array}\right] \\ \therefore & A=\frac{1}{3}\left[\begin{array}{ccc}9 & -14 & -8 \\ -6 & 3 & 9\end{array}\right] \\ \therefore & A=\left[\begin{array}{ccc}3 & \frac{-14}{3} & \frac{-8}{3} \\ -2 & 1 & 3\end{array}\right]\end{array}$

$=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$

(i)

and $X-3 Y=\left[\begin{array}{ll}0 & -1 \\ 0 & -1\end{array}\right] \ldots \ldots \ldots \ldots \ldots$ (ii)

By (i) x 3 – (ii) we get

$8 X=3\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -1 \\ 0 & -1\end{array}\right]$

$=\left[\begin{array}{cc}3 & -3 \\ -3 & 3\end{array}\right]-\left[\begin{array}{cc}0 & -1 \\ 0 & -1\end{array}\right]$

$=\left[\begin{array}{cc}3-0 & -3+1 \\ -3-0 & 3+1\end{array}\right]$

$\therefore \quad 8 X=\left[\begin{array}{cc}3 & -2 \\ -3 & 4\end{array}\right]$

$\therefore \quad X=\frac{1}{8}\left[\begin{array}{cc}3 & -2 \\ -3 & 4\end{array}\right]$

$\therefore \quad X=\left[\begin{array}{cc}\frac{3}{8} & \frac{-2}{8} \\ \frac{-3}{8} & \frac{4}{8}\end{array}\right]=\left[\begin{array}{cc}\frac{3}{8} & \frac{-1}{4} \\ \frac{-3}{8} & \frac{1}{2}\end{array}\right]$

By (i) - (ii) $\times 3$, we get

$8 Y=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]-3\left[\begin{array}{cc}0 & -1 \\ 0 & -1\end{array}\right]$

$=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -3 \\ 0 & -3\end{array}\right]$

$=\left[\begin{array}{cc}1-0 & -1+3 \\ -1-0 & 1+3\end{array}\right]$

$\therefore \quad 8 Y=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right]$

$\therefore \quad Y=\frac{1}{8}\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right]=\left[\begin{array}{cc}\frac{1}{8} & \frac{2}{8} \\ \frac{-1}{8} & \frac{4}{8}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{8} & \frac{1}{4} \\ \frac{-1}{8} & \frac{1}{2}\end{array}\right]$

A(□ ABDC) = A(ΔABD) + A(ΔADC)

Area of triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$A(\Delta \mathrm{ABD})=\frac{1}{2}\left|\begin{array}{ccc}-3 & 1 & 1 \\ -2 & -2 & 1 \\ 3 & -1 & 1\end{array}\right|$

$\begin{aligned} & =\frac{1}{2}[-3(-2+1)-1(-2-3)+1(2+6) \\ & =\frac{1}{2}[-3(-1)-1(-5)+1(8)] \\ & =\frac{1}{2}(3+5+8) \\ & =\frac{1}{2}(16)\end{aligned}$

∴ A(ΔABD) = 8 sq. units

$\mathrm{A}(\Delta \mathrm{ADC})=\frac{1}{2}\left|\begin{array}{ccc}-3 & 1 & 1 \\ 3 & -1 & 1 \\ 1 & 4 & 1\end{array}\right|$

$\begin{aligned} & =\frac{1}{2}[-3(-1-4)-1(3-1)+1(12+1)] \\ & =\frac{1}{2}[-3(-5)-1(2)+1(13)] \\ & =\frac{1}{2}[15-2+13] \\ & =\frac{1}{2}(26)\end{aligned}$

∴ A(ΔADC) = 13 sq. units ∴ A(□ ABDC) = A(ΔABD) + A(ΔADC) = 8 + 13 = 21 sq. units