Question

(i) If $A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right]$, find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7

(ii) $A=\left[\begin{array}{ccc}3& -4& 2\\ 2& 3& 5\\ 1& 0& 1\end{array}\right]$, find A⁻¹ and hence solve the following system of equations:
3x − 4y + 2z = −1, 2x + 3y + 5z = 7, x + z = 2

(iii) $A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right]$, find AB. Hence, solve the system of equations:
x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

(iv) If $A=\left[\begin{array}{ccc}1& 2& 0\\ -2& -1& -2\\ 0& -1& 1\end{array}\right]$, find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x − y − z = 8, −2y + z = 7

(v) Given $A=\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right],B=\left[\begin{array}{ccc}1& -1& 0\\ 2& 3& 4\\ 0& 1& 2\end{array}\right]$, find BA and use this to solve the system of equations
y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17

(vi) If $\mathrm{A}=\left(\begin{array}{rrr}2& 3& 1\\ 1& 2& 2\\ –3& 1& -1\end{array}\right)$, find A^–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8.
(vii) Use product $\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]$ to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.

Answer 1

​$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Here}, \\ A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right] \\ \left|A\right|\text{=1}\left(1+6\right)+2\left(2-0\right)+0\left(-4-0\right) \\ =7+4+0 \\ =11 \\ {\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& 1\end{array}\right|=7,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 3\\ 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}-2& 0\\ -2& 1\end{array}\right|=2,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|=1,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& -2\\ 0& -2\end{array}\right|=2 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-2& 0\\ 1& 3\end{array}\right|=-6,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 0\\ 2& 3\end{array}\right|=-3,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& -2\\ 2& 1\end{array}\right|=5 \\ \therefore \mathrm{adj}A={\left[\begin{array}{ccc}7& -2& -4\\ 2& 1& 2\\ -6& -3& 5\end{array}\right]}^T \\ =\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right] \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{11}\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right] \\ \mathrm{or},AX=B \\ \mathrm{where},A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right] \\ \mathrm{Now}, \\ \therefore X=A^{-1}B \\ \Rightarrow X=\frac{1}{11}\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right]\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right] \\ \Rightarrow X=\frac{1}{11}\left[\begin{array}{c}70+16-42\\ -20+8-21\\ -40+16+35\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}44\\ -33\\ 11\end{array}\right] \\ \therefore x=4,y=-3\mathrm{and}z=1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\text{Here,} \\ A=\left[\begin{array}{ccc}3& -4& 2\\ 2& 3& 5\\ 1& 0& 1\end{array}\right] \\ \left|A\right|\text{=3}\left(3-0\right)+4\left(2-5\right)+2\left(0-3\right) \\ =9-12-6 \\ =-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}3& 5\\ 0& 1\end{array}\right|=3,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 5\\ 1& 1\end{array}\right|=3,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& 3\\ 1& 0\end{array}\right|=-3 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-4& 2\\ 0& 1\end{array}\right|=4,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}3& 2\\ 1& 1\end{array}\right|=1,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}3& -4\\ 1& 0\end{array}\right|=-4 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-4& 2\\ 3& 5\end{array}\right|=-26,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}3& 2\\ 2& 5\end{array}\right|=-11,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}3& -4\\ 2& 3\end{array}\right|=17 \\ \mathrm{adj}A={\left[\begin{array}{ccc}3& 3& -3\\ 4& 1& -4\\ -26& -11& 17\end{array}\right]}^{\mathrm{T}} \\ =\left[\begin{array}{ccc}3& 4& -26\\ 3& 1& -11\\ -3& -4& 17\end{array}\right] \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{-9}\left[\begin{array}{ccc}3& 4& -26\\ 3& 1& -11\\ -3& -4& 17\end{array}\right] \\ AX\mathit{=}B \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}3& -4& 2\\ 2& 3& 5\\ 1& 0& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}-1\\ 7\\ 2\end{array}\right] \\ X=A^{-1}B \\ \Rightarrow X=\frac{1}{-9}\left[\begin{array}{ccc}3& 4& -26\\ 3& 1& -11\\ -3& -4& 17\end{array}\right]\left[\begin{array}{c}-1\\ 7\\ 2\end{array}\right] \\ \Rightarrow X=\frac{1}{-9}\left[\begin{array}{c}-3+28-52\\ -3+7-22\\ 3-28+34\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-9}\left[\begin{array}{c}-27\\ -18\\ 9\end{array}\right] \\ \therefore x=3,y=2\mathrm{and}z=-1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\text{Here,} \\ A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right] \\ AB=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right]\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}7+4+0& 2-2+0& -6+6+0\\ 14-2-12& 4+1+6& -12-3+15\\ 0+4-4& 0-2+2& 0+6+5\end{array}\right] \\ =\left[\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& 11\end{array}\right] \\ \Rightarrow AB=11\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ AB=11I_3 \\ \Rightarrow \frac{1}{11}AB=I_3 \\ \Rightarrow \left(\frac{1}{11}B\right)A=I_3 \\ \Rightarrow A^{-1}=\frac{1}{11}B \\ \Rightarrow A^{-1}=\frac{1}{11}\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right] \\ X=A^{-1}B \\ X=\frac{1}{11}\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right]\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}70+16-42\\ -20+9-21\\ -40+16+35\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}44\\ -33\\ 11\end{array}\right] \\ \therefore x=4,y=-3\mathrm{and}z=1 \end{gathered}$$

​$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{Here}, \\ A=\left[\begin{array}{ccc}1& 2& 0\\ -2& -1& -2\\ 0& -1& 1\end{array}\right] \\ \left|A\right|=1\left(-1-2\right)+2\left(2\right) \\ =-3+4 \\ =1 \\ \mathrm{Let}{C}_{ij}\mathrm{be}\mathrm{the}\mathrm{cofactors}\mathrm{of}\mathrm{the}\mathrm{elements}{a}_{ij}\mathrm{in}A=\left[a_{ij}\right].\mathrm{Then}, \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}-1& -2\\ -1& 1\end{array}\right|=-3,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}-2& -2\\ 0& 1\end{array}\right|=2,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}-2& -1\\ 0& -1\end{array}\right|=2 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}2& 0\\ -1& 1\end{array}\right|=-2,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|=1,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 2\\ 0& -1\end{array}\right|=1 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}2& 0\\ -1& -2\end{array}\right|=-4,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 0\\ -2& -2\end{array}\right|=2,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 2\\ -2& -1\end{array}\right|=3 \\ \therefore \mathrm{adj}A={\left[\begin{array}{ccc}-3& 2& 2\\ -2& 1& 1\\ -4& 2& 3\end{array}\right]}^T \\ =\left[\begin{array}{ccc}-3& -2& -4\\ 2& 1& 2\\ 2& 1& 3\end{array}\right] \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{1}\left[\begin{array}{ccc}-3& -2& -4\\ 2& 1& 2\\ 2& 1& 3\end{array}\right] \\ =\left[\begin{array}{ccc}-3& -2& -4\\ 2& 1& 2\\ 2& 1& 3\end{array}\right] \\ \mathrm{We}\mathrm{know}\mathrm{that},{\left(A^T\right)}^{-1}={\left(A^{-1}\right)}^T. \\ \mathrm{Here},C=A^T \\ \mathrm{i}.\mathrm{e}.,C=\left[\begin{array}{ccc}1& -2& 0\\ 2& -1& -1\\ 0& -2& 1\end{array}\right] \\ \therefore C^{-1}=\left[\begin{array}{ccc}-3& 2& 2\\ -2& 1& 1\\ -4& 2& 3\end{array}\right] \\ \mathrm{or},CX=B \\ \mathrm{where},C=\left[\begin{array}{ccc}1& -2& 0\\ 2& -1& -1\\ 0& -2& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right] \\ \mathrm{Now}, \\ \therefore X=C^{-1}B \\ \Rightarrow X=\left[\begin{array}{ccc}-3& 2& 2\\ -2& 1& 1\\ -4& 2& 3\end{array}\right]\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right] \\ \Rightarrow X=\left[\begin{array}{c}-30+16+14\\ -20+8+7\\ -40+16+21\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ -5\\ -3\end{array}\right] \\ \therefore x=0,y=-5\mathrm{and}z=-3. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{Here}, \\ A=\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}1& -1& 0\\ 2& 3& 4\\ 0& 1& 2\end{array}\right] \\ BA=\left[\begin{array}{ccc}1& -1& 0\\ 2& 3& 4\\ 0& 1& 2\end{array}\right]\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{ccc}2+4+0& 2-2+0& -4+4+0\\ 4-12+8& 4+6-4& -8-12+20\\ 0-4+4& 0+2-2& 0-4+10\end{array}\right] \\ =\left[\begin{array}{ccc}6& 0& 0\\ 0& 6& 0\\ 0& 0& 6\end{array}\right] \\ \Rightarrow BA=6\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ \Rightarrow BA=6I_3 \\ \Rightarrow B\left(\frac{1}{6}A\right)=I_3 \\ \Rightarrow B^{-1}=\frac{1}{6}A \\ \Rightarrow B^{-1}=\frac{1}{6}\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right] \\ \mathrm{Now},BX=C \\ \mathrm{where},B=\left[\begin{array}{ccc}1& -1& 0\\ 2& 3& 4\\ 0& 1& 2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}C=\left[\begin{array}{c}3\\ 17\\ 7\end{array}\right] \\ \therefore X=B^{-1}C \\ \Rightarrow X=\frac{1}{6}\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right]\left[\begin{array}{c}3\\ 17\\ 7\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}6+34-28\\ -12+34-28\\ 6-17+35\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}12\\ -6\\ 24\end{array}\right] \\ \therefore x=2,y=-1\mathrm{and}z=4. \end{gathered}$$

(vi) We have, $A=\left[\begin{array}{rrr}2& 3& 1\\ 1& 2& 2\\ –3& 1& -1\end{array}\right]$.
$$\begin{gathered} \therefore \left|A\right|=\left|\begin{array}{ccc}2& 3& 1\\ 1& 2& 2\\ -3& 1& -1\end{array}\right| \\ =2\left(-2-2\right)-3\left(-1+6\right)+1\left(1+6\right) \\ =-8-15+7 \\ =-16\ne 0 \end{gathered}$$
So, A is invertible.
Let C_ij be the co-factors of the elements a_ij in A[a_ij]. Then,
$$\begin{gathered} C_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}2& 2\\ 1& -1\end{array}\right|=-2-2=-4 \\ {C}_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& 2\\ -3& -1\end{array}\right|=-1\left(-1+6\right)=-5 \\ {C}_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}1& 2\\ -3& 1\end{array}\right|=1+6=7 \end{gathered}$$
$$\begin{gathered} C_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}3& 1\\ 1& -1\end{array}\right|=3+1=4 \\ {C}_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}2& 1\\ -3& -1\end{array}\right|=-2+3=1 \\ {C}_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}2& 3\\ -3& 1\end{array}\right|=-1\left(2+9\right)=-11 \end{gathered}$$
$$\begin{gathered} C_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}3& 1\\ 2& 2\end{array}\right|=6-2=4 \\ {C}_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}2& 1\\ 1& 2\end{array}\right|=-1\left(4-1\right)=-3 \\ {C}_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}2& 3\\ 1& 2\end{array}\right|=4-3=1 \end{gathered}$$
$$\begin{gathered} \therefore \mathrm{Adj}A={\left[\begin{array}{ccc}-4& -5& 7\\ 4& 1& -11\\ 4& -3& 1\end{array}\right]}^T=\left[\begin{array}{ccc}-4& 4& 4\\ -5& 1& -3\\ 7& -11& 1\end{array}\right] \\ \Rightarrow A^{-1}=\frac{\mathrm{Adj}A}{\left|A\right|}=\frac{1}{-16}\left[\begin{array}{ccc}-4& 4& 4\\ -5& 1& -3\\ 7& -11& 1\end{array}\right] \end{gathered}$$
Now, the given system of equations is expressible as
$\left[\begin{array}{ccc}2& 1& -3\\ 3& 2& 1\\ 1& 2& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}13\\ 4\\ 8\end{array}\right]$
Or A^T X = B, where $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}13\\ 4\\ 8\end{array}\right]$
Now, $\left|A^T\right|=\left|A\right|=-16\ne 0$
So, the given system of equations is consistent with a unique solution given by
$X={\left(A^T\right)}^{-1}B={\left(A^{-1}\right)}^{T}B$
$$\begin{gathered} \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=-\frac{1}{16}{\left[\begin{array}{ccc}-4& 4& 4\\ -5& 1& -3\\ 7& -11& 1\end{array}\right]}^T\left[\begin{array}{c}13\\ 4\\ 8\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{ccc}-4& -5& 7\\ 4& 1& -11\\ 4& -3& 1\end{array}\right]\left[\begin{array}{c}13\\ 4\\ 8\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{c}-52-20+56\\ 52+4-88\\ 52-12+8\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{c}-16\\ -32\\ 48\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right] \end{gathered}$$
Hence, x = 1, y = 2 and z = −3 is the required solution.

(vii) Suppose, A = $\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]$ $B=\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]$

$$\begin{gathered} A\times B=\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right] \\ =\left[\begin{array}{ccc}-2-9+12& 0-2+2& 1+3-4\\ 0+18-18& 0+4-3& 0-6+6\\ -6-18+24& 0-4+4& 3+6-8\end{array}\right] \\ =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

Since, A × B = I,
$\therefore$ B = A⁻¹ .....(1)

Now, the given system of equations is

x + 3z = 9
−x + 2y − 2z = 4
2x − 3y + 4z = −3

This can also be represented as,

$\left[\begin{array}{ccc}1& 0& 3\\ -1& 2& -2\\ 2& -3& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right]$
Here, we can observe that $\left[\begin{array}{ccc}1& 0& 3\\ -1& 2& -2\\ 2& -3& 4\end{array}\right]=A^T$
So, $A^T\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right]$

Multiply the above expression by ${\left(A^T\right)}^{-1}$.

$$\begin{gathered} \left[\begin{array}{c}x\\ y\\ z\end{array}\right]={\left(A^T\right)}^{-1}\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]={\left(A^{-1}\right)}^T\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=B^T\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right]\left[\mathrm{Using}\left(1\right)\right] \end{gathered}$$

$$\begin{gathered} \left[\begin{array}{c}x\\ y\\ z\end{array}\right]={\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]}^T\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right] \\ =\left[\begin{array}{ccc}-2& 9& 6\\ 0& 2& 1\\ 1& -3& -2\end{array}\right]\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right] \\ =\left[\begin{array}{c}-18+36-18\\ 0+8-3\\ 9-12+6\end{array}\right] \\ =\left[\begin{array}{c}0\\ 5\\ 3\end{array}\right] \end{gathered}$$

Hence, x = 0, y = 5 and z = 3.