Question

Solve the following system of equations by matrix method:
(i) x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1

(ii) x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9

(iii) 6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10

(iv) 3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0

(v)
$$\begin{gathered} \frac{2}{x}-\frac{3}{y}+\frac{3}{z}=10 \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=10 \\ \frac{3}{x}-\frac{1}{y}+\frac{2}{z}=13 \end{gathered}$$

(vi) 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25

(vii) 3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2

(viii) 2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6

(ix) 2x + 6y = 2
3x − z = −8
2x − y + z = −3

(x) x − y + z = 2
2x − y = 0
2y − z = 1

(xi) 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5

(xii) x + y + z = 6
x + 2z = 7
3x + y + z = 12

(xiii) $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4,\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1,\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2;x,y,z\ne 0$

(xiv) x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12

Answer 1

(i)
$$\begin{gathered} \text{Here,} \\ A=\left[\begin{array}{ccc}1& 1& -1\\ 2& 3& 1\\ 3& -1& -7\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}1& 1& -1\\ 2& 3& 1\\ 3& -1& -7\end{array}\right| \\ =1\left(-21+1\right)-1\left(-14-3\right)-1(-2-9) \\ =-20+17+11 \\ =8 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}3& 1\\ -1& -7\end{array}\right|=-20,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 1\\ 3& -7\end{array}\right|=17,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& 3\\ 3& -1\end{array}\right|=-11 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}1& -1\\ -1& -7\end{array}\right|=8,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& -1\\ 3& -7\end{array}\right|=-4,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 1\\ 3& -1\end{array}\right|=4 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}1& -1\\ 3& 1\end{array}\right|=4,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& -1\\ 2& 1\end{array}\right|=-3,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 1\\ 2& 3\end{array}\right|=1 \\ \mathrm{adj}A={\left[\begin{array}{ccc}-20& 17& -11\\ 8& -4& 4\\ 4& -3& 1\end{array}\right]}^T \\ =\left[\begin{array}{ccc}-20& 8& 4\\ 17& -4& -3\\ -11& 4& 1\end{array}\right] \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}adjA \\ =\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}{c}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}{c}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}{c}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}\mathrm{and}z=\frac{8}{8} \\ \therefore x=3,y=1\mathrm{and}z=1 \end{gathered}$$

$$\begin{gathered} \text{(ii)} \\ \text{Here,} \\ A=\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& 1\\ 2& 1& -3\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}1& 1& 1\\ 2& -1& 1\\ 2& 1& -3\end{array}\right| \\ =1\left(3-1\right)-1\left(-6-2\right)+1(2+2) \\ =2+8+4 \\ =14 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}-1& 1\\ 1& -3\end{array}\right|=2,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 1\\ 2& -3\end{array}\right|=8,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& -1\\ 2& 1\end{array}\right|=4 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|=4,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 1\\ 2& -3\end{array}\right|=-5,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 1\\ 2& 1\end{array}\right|=1 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}1& 1\\ -1& 1\end{array}\right|=2,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 1\\ 2& 1\end{array}\right|=1,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 1\\ 2& -1\end{array}\right|=-3 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}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}{c}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}\mathrm{and}z=\frac{38}{14} \\ \therefore x=\frac{-8}{7},y=\frac{10}{7}\mathrm{and}z=\frac{19}{7} \end{gathered}$$

$$\begin{gathered} \text{(iii)} \\ \text{Here,} \\ A=\left[\begin{array}{ccc}6& -12& 25\\ 4& 15& -20\\ 2& 18& 15\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}6& -12& 25\\ 4& 15& -20\\ 2& 18& 15\end{array}\right| \\ =6\left(225+360\right)+12\left(60+40\right)+25(72-30) \\ =3510+1200+1050 \\ =5760 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}15& -20\\ 18& 15\end{array}\right|=585,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}4& -20\\ 2& 15\end{array}\right|=-100,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}4& 15\\ 2& 18\end{array}\right|=42 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-12& 25\\ 18& 15\end{array}\right|=630,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}6& 25\\ 2& 15\end{array}\right|=40,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}6& -12\\ 2& 18\end{array}\right|=-132 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-12& 25\\ 15& -20\end{array}\right|=-135,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}6& 25\\ 4& -20\end{array}\right|=220,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}6& -12\\ 4& 15\end{array}\right|=138 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}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}{c}x\\ y\\ z\end{array}\right]=\frac{1}{5760}\left[\begin{array}{c}2880\\ 1920\\ 1152\end{array}\right] \\ \Rightarrow x=\frac{2880}{5760},y=\frac{1920}{5760}\mathrm{and}z=\frac{1152}{5760} \\ \therefore x=\frac{1}{2},y=\frac{1}{3}\mathrm{and}z=\frac{1}{5} \end{gathered}$$

$$\begin{gathered} \text{(iv)} \\ \text{Here,} \\ A=\left[\begin{array}{ccc}3& 4& 7\\ 2& -1& 3\\ 2& 1& -3\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}3& 4& 7\\ 2& -1& 3\\ 2& 1& -3\end{array}\right| \\ =3\left(3-3\right)-4\left(-6-6\right)+7(2+2) \\ =0+48+28 \\ =76 \\ {\text{Let C}}_{ij}{\text{be the cofactors of elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}-1& 3\\ 1& -3\end{array}\right|=0,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 3\\ 2& -3\end{array}\right|=12,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& -1\\ 2& 1\end{array}\right|=4 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}4& 7\\ 1& -3\end{array}\right|=19,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}3& 7\\ 2& -3\end{array}\right|=-23,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}3& 4\\ 2& 1\end{array}\right|=5 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}4& 7\\ -1& 3\end{array}\right|=19,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}3& 7\\ 2& 3\end{array}\right|=5,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}3& 4\\ 2& -1\end{array}\right|=-11 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}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}{c}x\\ y\\ z\end{array}\right]=\frac{1}{76}\left[\begin{array}{c}76\\ 76\\ 76\end{array}\right] \\ \Rightarrow x=\frac{76}{76},y=\frac{76}{76}\mathrm{and}z=\frac{76}{76} \\ \therefore x=1,y=1\mathrm{and}z=1 \end{gathered}$$


(v)
$$\begin{gathered} \text{Let}\frac{1}{x}\text{be}\mathit{\text{a}}\text{,}\frac{1}{y}\text{be}\mathit{\text{b}}\mathrm{and}\frac{1}{z}\text{be}\mathit{\text{c.}} \\ \text{Here,} \\ A=\left[\begin{array}{ccc}2& -3& 3\\ 1& 1& 1\\ 3& -1& 2\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}2& -3& 3\\ 1& 1& 1\\ 3& -1& 2\end{array}\right| \\ =2\left(2+1\right)+3\left(2-3\right)+3(-1-3) \\ =6-3-12 \\ =-9 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}1& 1\\ -1& 2\end{array}\right|=3,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& 1\\ 3& 2\end{array}\right|=1,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}1& 1\\ 3& -1\end{array}\right|=-4 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-3& 3\\ -1& 2\end{array}\right|=3,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}2& 3\\ 3& 2\end{array}\right|=-5,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}2& -3\\ 3& -1\end{array}\right|=-7 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-3& 3\\ 1& 1\end{array}\right|=-6,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}2& 3\\ 1& 1\end{array}\right|=1,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}2& -3\\ 1& 1\end{array}\right|=5 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}10\\ 10\\ 13\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}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}{c}a\\ b\\ c\end{array}\right]=\frac{1}{-9}\left[\begin{array}{c}-18\\ -27\\ -45\end{array}\right] \\ \Rightarrow x=\frac{1}{a}=\frac{-9}{-18},y=\frac{1}{b}=\frac{-9}{-27}\mathrm{and}z=\frac{1}{c}=\frac{-9}{-45} \\ \therefore x=\frac{1}{a}=\frac{1}{2},y=\frac{1}{b}=\frac{1}{3}\mathrm{and}z=\frac{1}{c}=\frac{1}{5} \end{gathered}$$

(vi)

$$\begin{gathered} \text{Here,} \\ A=\left[\begin{array}{ccc}5& 3& 1\\ 2& 1& 3\\ 1& 2& 4\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}5& 3& 1\\ 2& 1& 3\\ 1& 2& 4\end{array}\right|=5\left(4-6\right)-3\left(8-3\right)+1(4-1) \\ =-10-15+3 \\ =-22 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}1& 3\\ 2& 4\end{array}\right|=-2,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 3\\ 1& 4\end{array}\right|=-5,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& 1\\ 1& 2\end{array}\right|=3 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}3& 1\\ 2& 4\end{array}\right|=-10,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}5& 1\\ 1& 4\end{array}\right|=19,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}5& 3\\ 1& 2\end{array}\right|=-7 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}3& 1\\ 1& 3\end{array}\right|=8,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}5& 1\\ 2& 3\end{array}\right|=-13,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}5& 3\\ 2& 1\end{array}\right|=-1 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}16\\ 19\\ 25\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-22}\left[\begin{array}{c}-32-190+200\\ -80+361-325\\ 48-133-25\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}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}\mathrm{and}z=\frac{-110}{-22} \\ \therefore x=1,y=2\mathrm{and}z=5 \end{gathered}$$

(vii)
$$\begin{gathered} \text{Here,} \\ A=\left[\begin{array}{ccc}3& 4& 2\\ 0& 2& -3\\ 1& -2& 6\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}3& 4& 2\\ 0& 2& -3\\ 1& -2& 6\end{array}\right| \\ =3\left(12-6\right)-4\left(0+3\right)+2(0-2) \\ =18-12-4 \\ =2 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}2& -3\\ -2& 6\end{array}\right|=6,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}0& -3\\ 1& 6\end{array}\right|=-3,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}0& 2\\ 1& -2\end{array}\right|=-2 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}4& 2\\ -2& 6\end{array}\right|=-28,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}3& 2\\ 1& 6\end{array}\right|=16,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}3& 4\\ 1& -2\end{array}\right|=10 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}4& 2\\ 2& -3\end{array}\right|=-16,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}3& 2\\ 0& -3\end{array}\right|=9,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}3& 4\\ 0& 2\end{array}\right|=6 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}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}{c}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}\mathrm{and}z=\frac{2}{2} \\ \therefore x=-2,y=3\mathrm{and}z=1 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{viiii}\right) \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}2& 1& 1\\ 1& 3& -1\\ 3& 1& -2\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}2& 1& 1\\ 1& 3& -1\\ 3& 1& -2\end{array}\right| \\ =2\left(-6+1\right)-1\left(-2+3\right)+1(1-9) \\ =-10-1-8 \\ =-19 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}3& -1\\ 1& -2\end{array}\right|=-5,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& -1\\ 3& -2\end{array}\right|=-1,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}1& 3\\ 3& 1\end{array}\right|=-8 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}1& 1\\ 1& -2\end{array}\right|=3,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}2& -1\\ 3& -2\end{array}\right|=-7,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}2& 1\\ 3& 1\end{array}\right|=1 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}1& 1\\ 3& -1\end{array}\right|=-4,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}2& 1\\ 1& -1\end{array}\right|=3,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}2& 1\\ 1& 3\end{array}\right|=5 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}2\\ 5\\ 6\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-19}\left[\begin{array}{c}-10+15-24\\ -2-35+18\\ -16+5+30\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}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}\mathrm{and}z=\frac{19}{-19} \\ \therefore x=1,y=3\mathrm{and}z=-1 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ix}\right) \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}2& 6& 0\\ 3& 0& -1\\ 2& -1& 1\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}2& 6& 0\\ 3& 0& -1\\ 2& -1& 1\end{array}\right| \\ =2\left(0-1\right)-6\left(3+2\right)+0(-3+0) \\ =-2-30 \\ =-32 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}0& -1\\ -1& 1\end{array}\right|=-1,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}3& -1\\ 2& 1\end{array}\right|=-5,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}3& 0\\ 2& -1\end{array}\right|=-3 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}6& 0\\ -1& 1\end{array}\right|=-6,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}2& 0\\ 2& 1\end{array}\right|=2,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}2& 6\\ 2& -1\end{array}\right|=14 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}6& 0\\ 0& -1\end{array}\right|=-6,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}2& 0\\ 3& -1\end{array}\right|=2,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}2& 6\\ 3& 0\end{array}\right|=-18 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-32}\left[\begin{array}{c}-2+48+18\\ -10-16-6\\ -6-112+54\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}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}\mathrm{and}z=\frac{-64}{-32} \\ \therefore x=-2,y=1\mathrm{and}z=2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{x}\right) \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 0& 2& -1\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 0& 2& -1\end{array}\right| \\ =1\left(1-0\right)+1\left(-2-0\right)+1(4-0) \\ =1-2+4 \\ =3 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}-1& 0\\ 2& -1\end{array}\right|=1,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 0\\ 0& -1\end{array}\right|=2,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& -1\\ 0& 2\end{array}\right|=4 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-1& 1\\ 2& -1\end{array}\right|=1,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 1\\ 0& -1\end{array}\right|=-1,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& -1\\ 0& 2\end{array}\right|=-2 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-1& 1\\ -1& 0\end{array}\right|=1,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 1\\ 2& 0\end{array}\right|=2,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& -1\\ 2& -1\end{array}\right|=1 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}2\\ 0\\ 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}2+1\\ 4+2\\ 8+1\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{1}\left[\begin{array}{c}3\\ 6\\ 9\end{array}\right] \\ \Rightarrow x=\frac{3}{3},y=\frac{6}{3}\mathrm{and}z=\frac{9}{3} \\ \therefore x=1,y=2\mathrm{and}z=3 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xi}\right) \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}8& 4& 3\\ 2& 1& 1\\ 1& 2& 1\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}8& 4& 3\\ 2& 1& 1\\ 1& 2& 1\end{array}\right| \\ =8\left(1-2\right)-4\left(2-1\right)+3(4-1) \\ =-8-4+9 \\ =-3 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}1& 1\\ 2& 1\end{array}\right|=-1,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|=-1,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& 1\\ 1& 2\end{array}\right|=3 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}4& 3\\ 2& 1\end{array}\right|=2,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}8& 3\\ 1& 1\end{array}\right|=5,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}8& 4\\ 1& 2\end{array}\right|=-12 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}4& 3\\ 1& 1\end{array}\right|=1,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}8& 3\\ 2& 1\end{array}\right|=-2,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}8& 4\\ 2& 1\end{array}\right|=0 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}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}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-3}\left[\begin{array}{c}-3\\ -3\\ -6\end{array}\right] \\ \Rightarrow x=\frac{-3}{-3},y=\frac{-3}{-3}\mathrm{and}z=\frac{-6}{-3} \\ \therefore x=1,y=1\mathrm{and}z=2 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xii}\right) \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 2\\ 3& 1& 1\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& 0& 2\\ 3& 1& 1\end{array}\right| \\ =1\left(0-2\right)-1\left(1-6\right)+1(1-0) \\ =-2+5+1 \\ =4 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}0& 2\\ 1& 1\end{array}\right|=-2,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& 2\\ 3& 1\end{array}\right|=5,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|=1 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}1& 1\\ 1& 1\end{array}\right|=0,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 1\\ 3& 1\end{array}\right|=-2,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 1\\ 3& 1\end{array}\right|=2 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}1& 1\\ 0& 2\end{array}\right|=2,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|=-1,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 1\\ 1& 0\end{array}\right|=-1 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}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}{c}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}\mathrm{and}z=\frac{-20}{4} \\ \therefore x=3,y=1\mathrm{and}z=-5 \end{gathered}$$


$$\begin{gathered} \text{(xiii)} \\ \text{Let}\frac{1}{x}\text{be}\mathit{\text{a,}}\frac{1}{y}\text{be}\mathit{\text{b}}\mathrm{and}\frac{1}{z}\mathrm{be}\mathit{\text{c.}} \\ \text{Here,} \\ A=\left[\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right| \\ =2\left(120-45\right)-3\left(-80-30\right)+10(36+36) \\ =150+330+720 \\ =1200 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}-6& 5\\ 9& -20\end{array}\right|=75,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}4& 5\\ 6& -20\end{array}\right|=110,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}4& -6\\ 6& 9\end{array}\right|=72 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}3& 10\\ 9& -20\end{array}\right|=150,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}2& 10\\ 6& -20\end{array}\right|=-100,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}2& 3\\ 6& 9\end{array}\right|=0 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}3& 10\\ -6& 5\end{array}\right|=75,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}2& 10\\ 4& 5\end{array}\right|=30,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}2& 3\\ 4& -6\end{array}\right|=-24 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}4\\ 1\\ 2\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}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}{c}a\\ b\\ c\end{array}\right]=\frac{1}{1200}\left[\begin{array}{c}600\\ 400\\ 240\end{array}\right] \\ \Rightarrow x=\frac{1}{a}=\frac{1200}{600},y=\frac{1}{b}=\frac{1200}{400}\mathrm{and}z=\frac{1}{c}=\frac{1200}{240} \\ \therefore x=2,y=3\mathrm{and}z=5 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xiv}\right)\mathrm{Here}, \\ A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right| \\ =1\left(12-5\right)+1\left(9+10\right)+2(-3-8) \\ =7+19-22 \\ =4 \\ \mathrm{Let}{C}_{ij}\mathrm{be}\mathrm{the}\mathrm{cofactors}\mathrm{of}\mathrm{elements}{a}_{ij}\mathrm{in}A=\left[a_{ij}\right].\mathrm{Then}, \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}4& -5\\ -1& 3\end{array}\right|=7,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}3& -5\\ 2& 3\end{array}\right|=-19,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}3& 4\\ 2& -1\end{array}\right|=-11 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-1& 2\\ -1& 3\end{array}\right|=1,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 2\\ 2& 3\end{array}\right|=-1,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& -1\\ 2& -1\end{array}\right|=-1 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-1& 2\\ 4& -5\end{array}\right|=-3,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 2\\ 3& -5\end{array}\right|=11,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& -1\\ 3& 4\end{array}\right|=7 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}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}{c}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}{c}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}\mathrm{and}z=\frac{12}{4} \\ \therefore x=2,y=1\mathrm{and}z=3. \end{gathered}$$