Given the equations

x + y + z = 6

y + 3z = 11

x – 2y + z = 0

using matrix method.

            $\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 3\\ 1& -2& 1\end{array}\right]\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\left[\begin{array}{l}6\\ 11\\ 0\end{array}\right]$

let                 A                    X           B

Ax = B

$\mathrm{X}=\frac{\mathrm{B}}{\mathrm{A}},$   X = A^–1B      ....(1)

we know that

${\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{adj}\left(\mathrm{A}\right)$

then cofactor of matrix A

${\mathrm{C}}_{11}=\left[\begin{array}{cc}1& 3\\ -2& 1\end{array}\right]=7$

${\mathrm{C}}_{12}=\left[\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]=-3$

${\mathrm{C}}_{13}=\left[\begin{array}{cc}0& 1\\ 1& -2\end{array}\right]=-1$

${\mathrm{C}}_{21}=\left[\begin{array}{cc}1& 1\\ -2& 1\end{array}\right]=3$

${\mathrm{C}}_{22}=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]=0$

${\mathrm{C}}_{23}=\left[\begin{array}{cc}1& 1\\ 1& -2\end{array}\right]=-3$

${\mathrm{C}}_{31}=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\end{array}\right]=2$

${\mathrm{C}}_{32}=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\end{array}\right]=3$

${\mathrm{C}}_{33}=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]=1$

sign change

C₁₁ = 7        C₁₂ = +3       C₁₃ = – 1

C₂₁ = –3      C₂₂ = 0         C₂₃ = +3

C₃₁ = 2        C₃₂ = –3       C₃₃ = 1

then

$\mathrm{C}=\left[\begin{array}{l}{\mathrm{c}}_{11}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{c}}_{12}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{c}}_{13}\\ {\mathrm{c}}_{21}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{c}}_{22}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{c}}_{23}\\ {\mathrm{c}}_{31}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{c}}_{32}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{c}}_{33}\end{array}\right]=\left[\begin{array}{ccc}7& 3& -1\\ -3& 0& 3\\ 2& -3& 1\end{array}\right]$

adj(A) = C^T   (transpose of c)

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

|A| a₁₁c₁₁ + a₁₂c₁₂ + a₁₃c₁₃

= 1 × 7 + 1 × 3 + 1 × (–1)

= 7 + 3 – 1 = 9

${\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{ady}\left(\mathrm{A}\right)$

${\mathrm{A}}^{-1}=\frac{1}{9}\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 1\end{array}\right]$

by equation (1) and we get

x = A^–1B

$\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{9}\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 2\end{array}\right]\left[\begin{array}{c}6\\ 11\\ 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{9}\left[\begin{array}{c}42-33+0\\ 18+0+0\\ -6+33+0\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{9}\left[\begin{array}{c}9\\ 18\\ 27\end{array}\right]$

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

then x = 1, y = 2, z = 3   (by matrix method)

Hence, this is the answer.