Solve the following system of equations by matrix method:

Solution:

(i)

Here,

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

$|A|=\left|\begin{array}{ccc}1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7\end{array}\right|$

$=1(-21+1)-1(-14-3)-1(-2-9)$

$=-20+17+11$

$=8$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right] .$ Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}3 & 1 \\ -1 & -7\end{array}\right|=-20, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}2 & 1 \\ 3 & -7\end{array}\right|=17, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}2 & 3 \\ 3 & -1\end{array}\right|=-11$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}1 & -1 \\ -1 & -7\end{array}\right|=8, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}1 & -1 \\ 3 & -7\end{array}\right|=-4, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc}1 & 1 \\ 3 & -1\end{array}\right|=4$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}1 & -1 \\ 3 & 1\end{array}\right|=4, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}1 & -1 \\ 2 & 1\end{array}\right|=-3, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right|=1$

$\operatorname{adj} A=\left[\begin{array}{ccc}-20 & 17 & -11 \\ 8 & -4 & 4 \\ 4 & -3 & 1\end{array}\right]^{T}$

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{8}\left[\begin{array}{ccc}-20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{8}\left[\begin{array}{ccc}-20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1\end{array}\right]\left[\begin{array}{c}3 \\ 10 \\ 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{8}\left[\begin{array}{c}-60+80+4 \\ 51-40-3 \\ -33+40+1\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{8}\left[\begin{array}{c}24 \\ 8 \\ 8\end{array}\right]$

$\Rightarrow x=\frac{24}{8}, y=\frac{8}{8}$ and $z=\frac{8}{8}$

$\therefore x=3, y=1$ and $z=1$

(ii)

Here,

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

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

$=1(3-1)-1(-6-2)+1(2+2)$

$=2+8+4$

$=14$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}-1 & 1 \\ 1 & -3\end{array}\right|=2, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}2 & 1 \\ 2 & -3\end{array}\right|=8, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}2 & -1 \\ 2 & 1\end{array}\right|=4$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}1 & 1 \\ 1 & -3\end{array}\right|=4, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}1 & 1 \\ 2 & -3\end{array}\right|=-5, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right|=1$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right|=2, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right|=1, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}1 & 1 \\ 2 & -1\end{array}\right|=-3$

$\operatorname{adj} A=\left[\begin{array}{ccc}2 & 8 & 4 \\ 4 & -5 & 1 \\ 2 & 1 & -3\end{array}\right]^{T}$

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

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{14}\left[\begin{array}{ccc}2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{14}\left[\begin{array}{ccc}2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3\end{array}\right]\left[\begin{array}{c}3 \\ -1 \\ -9\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{14}\left[\begin{array}{c}6-4-18 \\ 24+5-9 \\ 12-1+27\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{14}\left[\begin{array}{c}-16 \\ 20 \\ 38\end{array}\right]$

$\Rightarrow x=\frac{-16}{14}, y=\frac{20}{14}$ and $z=\frac{38}{14}$

$\therefore x=\frac{-8}{7}, y=\frac{10}{7}$ and $z=\frac{19}{7}$

(iii)

Here,

$A=\left[\begin{array}{ccc}6 & -12 & 25 \\ 4 & 15 & -20 \\ 2 & 18 & 15\end{array}\right]$

$|A|=\left|\begin{array}{ccc}6 & -12 & 25 \\ 4 & 15 & -20 \\ 2 & 18 & 15\end{array}\right|$

$=6(225+360)+12(60+40)+25(72-30)$

$=3510+1200+1050$

$=5760$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}15 & -20 \\ 18 & 15\end{array}\right|=585, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}4 & -20 \\ 2 & 15\end{array}\right|=-100, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{ll}4 & 15 \\ 2 & 18\end{array}\right|=42$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}-12 & 25 \\ 18 & 15\end{array}\right|=630, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}6 & 25 \\ 2 & 15\end{array}\right|=40, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc|c}6 & -12 \\ 2 & 18\end{array}\right|=-132$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-12 & 25 \\ 15 & -20\end{array}\right|=-135, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}6 & 25 \\ 4 & -20\end{array}\right|=220, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}6 & -12 \\ 4 & 15\end{array}\right|=138$

$\operatorname{adj} A=\left[\begin{array}{ccc}585 & -100 & 42 \\ 630 & 40 & -132 \\ -135 & 220 & 138\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{5760}\left[\begin{array}{ccc}585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{5760}\left[\begin{array}{ccc}585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138\end{array}\right]\left[\begin{array}{c}4 \\ 3 \\ 10\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{5760}\left[\begin{array}{c}2340+1890-1350 \\ -400+120+2200 \\ 168-396+1380\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{5760}\left[\begin{array}{l}2880 \\ 1920 \\ 1152\end{array}\right]$

$\Rightarrow x=\frac{2880}{5760}, y=\frac{1920}{5760}$ and $z=\frac{1152}{5760}$

$\therefore x=\frac{1}{2}, y=\frac{1}{3}$ and $z=\frac{1}{5}$

(iv)

Here,

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

$|A|=\left|\begin{array}{ccc}3 & 4 & 7 \\ 2 & -1 & 3 \\ 2 & 1 & -3\end{array}\right|$

$=3(3-3)-4(-6-6)+7(2+2)$

$=0+48+28$

$=76$

Let $\mathrm{C}_{i j}$ be the cofactors of elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}-1 & 3 \\ 1 & -3\end{array}\right|=0, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}2 & 3 \\ 2 & -3\end{array}\right|=12, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}2 & -1 \\ 2 & 1\end{array}\right|=4$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}4 & 7 \\ 1 & -3\end{array}\right|=19, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}3 & 7 \\ 2 & -3\end{array}\right|=-23, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}3 & 4 \\ 2 & 1\end{array}\right|=5$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}4 & 7 \\ -1 & 3\end{array}\right|=19, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}3 & 7 \\ 2 & 3\end{array}\right|=5, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}3 & 4 \\ 2 & -1\end{array}\right|=-11$

$\operatorname{adj} A=\left[\begin{array}{ccc}0 & 12 & 4 \\ 19 & -23 & 5 \\ 19 & 5 & -11\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}0 & 19 & 19 \\ 12 & -23 & 5 \\ 4 & 5 & -11\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{76}\left[\begin{array}{ccc}0 & 19 & 19 \\ 12 & -23 & 5 \\ 4 & 5 & -11\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{76}\left[\begin{array}{ccc}0 & 19 & 19 \\ 12 & -23 & 5 \\ 4 & 5 & -11\end{array}\right]\left[\begin{array}{c}14 \\ 4 \\ 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{76}\left[\begin{array}{c}0+76+0 \\ 168-92+0 \\ 56+20+0\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{76}\left[\begin{array}{l}76 \\ 76 \\ 76\end{array}\right]$

$\Rightarrow x=\frac{76}{76}, y=\frac{76}{76}$ and $z=\frac{76}{76}$

$\therefore x=1, y=1$ and $z=1$

(v)

Let $\frac{1}{x}$ be $a, \frac{1}{y}$ be $b$ and $\frac{1}{z}$ be $c$.

Here,

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

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

$=2(2+1)+3(2-3)+3(-1-3)$

$=6-3-12$

$=-9$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}1 & 1 \\ -1 & 2\end{array}\right|=3, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll}1 & 1 \\ 3 & 2\end{array}\right|=1, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}1 & 1 \\ 3 & -1\end{array}\right|=-4$

$C_{21}=(-1)^{2+1}\left|\begin{array}{ll}-3 & 3 \\ -1 & 2\end{array}\right|=3, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}2 & 3 \\ 3 & 2\end{array}\right|=-5, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}2 & -3 \\ 3 & -1\end{array}\right|=-7$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-3 & 3 \\ 1 & 1\end{array}\right|=-6, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}2 & 3 \\ 1 & 1\end{array}\right|=1, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}2 & -3 \\ 1 & 1\end{array}\right|=5$

$\operatorname{adj} A=\left[\begin{array}{ccc}3 & 1 & -4 \\ 3 & -5 & -7 \\ -6 & 1 & 5\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{-9}\left[\begin{array}{ccc}3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}a \\ b \\ c\end{array}\right]=\frac{1}{-9}\left[\begin{array}{ccc}3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5\end{array}\right]\left[\begin{array}{l}10 \\ 10 \\ 13\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}a \\ b \\ c\end{array}\right]=\frac{1}{-9}\left[\begin{array}{c}30+30-78 \\ 10-50+13 \\ -40-70+65\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}a \\ b \\ c\end{array}\right]=\frac{1}{-9}\left[\begin{array}{l}-18 \\ -27 \\ -45\end{array}\right]$

$\Rightarrow x=\frac{1}{a}=\frac{-9}{-18}, y=\frac{1}{b}=\frac{-9}{-27}$ and $z=\frac{1}{c}=\frac{-9}{-45}$

$\therefore x=\frac{1}{a}=\frac{1}{2}, y=\frac{1}{b}=\frac{1}{3}$ and $z=\frac{1}{c}=\frac{1}{5}$

(vi)

Here,

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

$|A|=\left|\begin{array}{lll}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{array}\right|=5(4-6)-3(8-3)+1(4-1)$

$=-10-15+3$

$=-22$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right|=-2, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right|=-5, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right|=3$

$C_{21}=(-1)^{2+1}\left|\begin{array}{ll}3 & 1 \\ 2 & 4\end{array}\right|=-10, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}5 & 1 \\ 1 & 4\end{array}\right|=19, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right|=-7$

$C_{31}=(-1)^{3+1}\left|\begin{array}{ll}3 & 1 \\ 1 & 3\end{array}\right|=8, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}5 & 1 \\ 2 & 3\end{array}\right|=-13, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{ll}5 & 3 \\ 2 & 1\end{array}\right|=-1$

$\operatorname{adj} A=\left[\begin{array}{ccc}-2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}-2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{-22}\left[\begin{array}{ccc}-2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-22}\left[\begin{array}{ccc}-2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1\end{array}\right]\left[\begin{array}{l}16 \\ 19 \\ 25\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-22}\left[\begin{array}{l}-32-190+200 \\ -80+361-325 \\ 48-133-25\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-22}\left[\begin{array}{c}-22 \\ -44 \\ -110\end{array}\right]$

$\Rightarrow x=\frac{-22}{-22}, y=\frac{-44}{-22}$ and $z=\frac{-110}{-22}$

$\therefore x=1, y=2$ and $z=5$

(vii)

Here,

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

$|A|=\left|\begin{array}{ccc}3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6\end{array}\right|$

$=3(12-6)-4(0+3)+2(0-2)$

$=18-12-4$

$=2$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}2 & -3 \\ -2 & 6\end{array}\right|=6, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}0 & -3 \\ 1 & 6\end{array}\right|=-3, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}0 & 2 \\ 1 & -2\end{array}\right|=-2$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}4 & 2 \\ -2 & 6\end{array}\right|=-28, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}3 & 2 \\ 1 & 6\end{array}\right|=16, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc}3 & 4 \\ 1 & -2\end{array}\right|=10$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}4 & 2 \\ 2 & -3\end{array}\right|=-16, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}3 & 2 \\ 0 & -3\end{array}\right|=9, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{ll}3 & 4 \\ 0 & 2\end{array}\right|=6$

$\operatorname{adj} A=\left[\begin{array}{ccc}6 & -3 & -2 \\ -28 & 16 & 10 \\ -16 & 9 & 6\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{2}\left[\begin{array}{ccc}6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{2}\left[\begin{array}{ccc}6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6\end{array}\right]\left[\begin{array}{c}8 \\ 3 \\ -2\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}48-84+32 \\ -24+48-18 \\ -16+30-12\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}-4 \\ 6 \\ 2\end{array}\right]$

$\Rightarrow x=\frac{-4}{2}, y=\frac{6}{2}$ and $z=\frac{2}{2}$

$\therefore x=-2, y=3$ and $z=1$

$($ viiii $)$

Here,

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

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

$=2(-6+1)-1(-2+3)+1(1-9)$

$=-10-1-8$

$=-19$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right] .$ Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}3 & -1 \\ 1 & -2\end{array}\right|=-5, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll}1 & -1 \\ 3 & -2\end{array}\right|=-1, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right|=-8$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}1 & 1 \\ 1 & -2\end{array}\right|=3, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}2 & -1 \\ 3 & -2\end{array}\right|=-7, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}2 & 1 \\ 3 & 1\end{array}\right|=1$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}1 & 1 \\ 3 & -1\end{array}\right|=-4, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}2 & 1 \\ 1 & -1\end{array}\right|=3, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right|=5$

$\operatorname{adj} A=\left[\begin{array}{ccc}-5 & -1 & -8 \\ 3 & -7 & 1 \\ -4 & 3 & 5\end{array}\right]^{T}$

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

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{-19}\left[\begin{array}{ccc}-5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-19}\left[\begin{array}{ccc}-5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5\end{array}\right]\left[\begin{array}{l}2 \\ 5 \\ 6\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-19}\left[\begin{array}{l}-10+15-24 \\ -2-35+18 \\ -16+5+30\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-19}\left[\begin{array}{c}-19 \\ 19 \\ 19\end{array}\right]$

$\Rightarrow x=\frac{-19}{-19}, y=\frac{19}{-19}$ and $z=\frac{19}{-19}$

$\therefore x=1, y=3$ and $z=-1$

$(\mathrm{ix})$

Here,

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

$|A|=\left|\begin{array}{ccc}2 & 6 & 0 \\ 3 & 0 & -1 \\ 2 & -1 & 1\end{array}\right|$

$=2(0-1)-6(3+2)+0(-3+0)$

$=-2-30$

$=-32$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}0 & -1 \\ -1 & 1\end{array}\right|=-1, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}3 & -1 \\ 2 & 1\end{array}\right|=-5, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}3 & 0 \\ 2 & -1\end{array}\right|=-3$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}6 & 0 \\ -1 & 1\end{array}\right|=-6, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}2 & 0 \\ 2 & 1\end{array}\right|=2, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc}2 & 6 \\ 2 & -1\end{array}\right|=14$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}6 & 0 \\ 0 & -1\end{array}\right|=-6, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}2 & 0 \\ 3 & -1\end{array}\right|=2, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}2 & 6 \\ 3 & 0\end{array}\right|=-18$

$\operatorname{adj} A=\left[\begin{array}{ccc}-1 & -5 & -3 \\ -6 & 2 & 14 \\ -6 & 2 & -18\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}-1 & -6 & -6 \\ -5 & 2 & 2 \\ -3 & 14 & -18\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{-32}\left[\begin{array}{ccc}-1 & -6 & -6 \\ -5 & 2 & 2 \\ -3 & 14 & -18\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-32}\left[\begin{array}{ccc}-1 & -6 & -6 \\ -5 & 2 & 2 \\ -3 & 14 & -18\end{array}\right]\left[\begin{array}{c}2 \\ -8 \\ -3\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-32}\left[\begin{array}{l}-2+48+18 \\ -10-16-6 \\ -6-112+54\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-32}\left[\begin{array}{c}64 \\ -32 \\ -64\end{array}\right]$

$\Rightarrow x=\frac{64}{-32}, y=\frac{-32}{-32}$ and $z=\frac{-64}{-32}$

$\therefore x=-2, y=1$ and $z=2$

$(x)$

Here,

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

$|A|=\left|\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & -1\end{array}\right|$

$=1(1-0)+1(-2-0)+1(4-0)$

$=1-2+4$

$=3$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}-1 & 0 \\ 2 & -1\end{array}\right|=1, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}2 & 0 \\ 0 & -1\end{array}\right|=2, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}\right|=4$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}-1 & 1 \\ 2 & -1\end{array}\right|=1, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}1 & 1 \\ 0 & -1\end{array}\right|=-1, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right|=-2$

$C_{31}=(-1)^{3+1}\left|\begin{array}{ll}-1 & 1 \\ -1 & 0\end{array}\right|=1, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}1 & 1 \\ 2 & 0\end{array}\right|=2, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right|=1$

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

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

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

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

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 2 \\ 4 & -2 & 1\end{array}\right]\left[\begin{array}{l}2 \\ 0 \\ 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{3}\left[\begin{array}{l}2+1 \\ 4+2 \\ 8+1\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{1}\left[\begin{array}{l}3 \\ 6 \\ 9\end{array}\right]$

$\Rightarrow x=\frac{3}{3}, y=\frac{6}{3}$ and $z=\frac{9}{3}$

$\therefore x=1, y=2$ and $z=3$

$(\mathrm{xi})$

Here,

$A=\left[\begin{array}{lll}8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1\end{array}\right]$

$|A|=\left|\begin{array}{lll}8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1\end{array}\right|$

$=8(1-2)-4(2-1)+3(4-1)$

$=-8-4+9$

$=-3$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right] .$ Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right|=-1, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll}2 & 1 \\ 1 & 1\end{array}\right|=-1, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right|=3$

$C_{21}=(-1)^{2+1}\left|\begin{array}{ll}4 & 3 \\ 2 & 1\end{array}\right|=2, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}8 & 3 \\ 1 & 1\end{array}\right|=5, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}8 & 4 \\ 1 & 2\end{array}\right|=-12$

$C_{31}=(-1)^{3+1}\left|\begin{array}{ll}4 & 3 \\ 1 & 1\end{array}\right|=1, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}8 & 3 \\ 2 & 1\end{array}\right|=-2, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{ll}8 & 4 \\ 2 & 1\end{array}\right|=0$

$\operatorname{adj} A=\left[\begin{array}{ccc}-1 & -1 & 3 \\ 2 & 5 & -12 \\ 1 & -2 & 0\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}-1 & 2 & 1 \\ -1 & 5 & -2 \\ 3 & -12 & 0\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{-3}\left[\begin{array}{ccc}-1 & 2 & 1 \\ -1 & 5 & -2 \\ 3 & -12 & 0\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-3}\left[\begin{array}{ccc}-1 & 2 & 1 \\ -1 & 5 & -2 \\ 3 & -12 & 0\end{array}\right]\left[\begin{array}{c}18 \\ 5 \\ 5\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-3}\left[\begin{array}{c}-18+10+5 \\ -18+25-10 \\ 54-60\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-3}\left[\begin{array}{l}-3 \\ -3 \\ -6\end{array}\right]$

$\Rightarrow x=\frac{-3}{-3}, y=\frac{-3}{-3}$ and $z=\frac{-6}{-3}$

$\therefore x=1, y=1$ and $z=2$

$($ xii $)$

Here,

$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{array}\right]$

$|A|=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{array}\right|$

$=1(0-2)-1(1-6)+1(1-0)$

$=-2+5+1$

$=4$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}0 & 2 \\ 1 & 1\end{array}\right|=-2, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right|=5, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right|=1$

$C_{21}=(-1)^{2+1}\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|=0, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}1 & 1 \\ 3 & 1\end{array}\right|=-2, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}1 & 1 \\ 3 & 1\end{array}\right|=2$

$C_{31}=(-1)^{3+1}\left|\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right|=2, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right|=-1, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right|=-1$

$\operatorname{adj} A=\left[\begin{array}{ccc}-2 & 5 & 1 \\ 0 & -2 & 2 \\ 2 & -1 & -1\end{array}\right]^{T}$

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

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

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

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{ccc}-2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1\end{array}\right]\left[\begin{array}{c}6 \\ 7 \\ 12\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}-12+0+24 \\ 30-14-12 \\ 6-14-12\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}12 \\ 4 \\ -20\end{array}\right]$

$\Rightarrow x=\frac{12}{4}, y=\frac{4}{4}$ and $z=\frac{-20}{4}$

$\therefore x=3, y=1$ and $z=-5$

(xiii)

Let $\frac{1}{x}$ be $a, \frac{1}{y}$ be $b$ and $\frac{1}{z}$ be $c .$

Here,

$A=\left[\begin{array}{ccc}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right]$

$|A|=\left|\begin{array}{ccc}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right|$

$=2(120-45)-3(-80-30)+10(36+36)$

$=150+330+720$

$=1200$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}-6 & 5 \\ 9 & -20\end{array}\right|=75, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}4 & 5 \\ 6 & -20\end{array}\right|=110, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}4 & -6 \\ 6 & 9\end{array}\right|=72$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}3 & 10 \\ 9 & -20\end{array}\right|=150, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}2 & 10 \\ 6 & -20\end{array}\right|=-100, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}2 & 3 \\ 6 & 9\end{array}\right|=0$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}3 & 10 \\ -6 & 5\end{array}\right|=75, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}2 & 10 \\ 4 & 5\end{array}\right|=30, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}2 & 3 \\ 4 & -6\end{array}\right|=-24$

$\operatorname{adj} A=\left[\begin{array}{ccc}75 & 110 & 72 \\ 150 & -100 & 0 \\ 75 & 30 & -24\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\left[\begin{array}{ccc}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{1200}\left[\begin{array}{ccc}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}a \\ b \\ c\end{array}\right]=\frac{1}{1200}\left[\begin{array}{ccc}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]\left[\begin{array}{l}4 \\ 1 \\ 2\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}a \\ b \\ c\end{array}\right]=\frac{1}{1200}\left[\begin{array}{c}300+150+150 \\ 440-100+60 \\ 288-48\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}a \\ b \\ c\end{array}\right]=\frac{1}{1200}\left[\begin{array}{l}600 \\ 400 \\ 240\end{array}\right]$

$\Rightarrow x=\frac{1}{a}=\frac{1200}{600}, y=\frac{1}{b}=\frac{1200}{400}$ and $z=\frac{1}{c}=\frac{1200}{240}$

$\therefore x=2, y=3$ and $z=5$

$($ xiv $)$ Here,

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

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

$=1(12-5)+1(9+10)+2(-3-8)$

$=7+19-22$

$=4$

Let $C_{i j}$ be the cofactors of elements $a_{i j}$ in $A=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}4 & -5 \\ -1 & 3\end{array}\right|=7, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}3 & -5 \\ 2 & 3\end{array}\right|=-19, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}3 & 4 \\ 2 & -1\end{array}\right|=-11$

$C_{21}=(-1)^{2+1}\left|\begin{array}{ll}-1 & 2 \\ -1 & 3\end{array}\right|=1, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|=-1, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right|=-1$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-1 & 2 \\ 4 & -5\end{array}\right|=-3, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}1 & 2 \\ 3 & -5\end{array}\right|=11, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}1 & -1 \\ 3 & 4\end{array}\right|=7$

$\operatorname{adj} A=\left[\begin{array}{ccc}7 & -19 & -11 \\ 1 & -1 & -1 \\ -3 & 11 & 7\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{4}\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}49-5-36 \\ -133+5+132 \\ -77+5+84\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}8 \\ 4 \\ 12\end{array}\right]$

$\Rightarrow x=\frac{8}{4}, y=\frac{4}{4}$ and $z=\frac{12}{4}$

$\therefore x=2, y=1$ and $z=3 .$