Question

Solve by matrix inversion method the following system of equations:
​​x + y – z + 3 = 0, 2x + 3y + z = 2, 8y + 3z = 1
[Answer : x = 1, y = -1, z = 3]

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}, \\ \mathrm{x}+\mathrm{y}-\mathrm{z}=-3 \\ 2\mathrm{x}+3\mathrm{y}+\mathrm{z}=2 \\ 0\mathrm{x}+8\mathrm{y}+3\mathrm{z}=1 \\ \mathrm{Now},\mathrm{the}\mathrm{above}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{matrix}\mathrm{form}\mathrm{as} \\ \mathrm{AX}=\mathrm{B} \\ \mathrm{where}\mathrm{A}=\left[\begin{array}{ccc}1& 1& -1\\ 2& 3& 1\\ 0& 8& 3\end{array}\right];\mathrm{X}=\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right];\mathrm{B}=\left[\begin{array}{c}-3\\ 2\\ 1\end{array}\right] \\ \mathrm{Now},\left|\mathrm{A}\right|=\left|\begin{array}{ccc}1& 1& -1\\ 2& 3& 1\\ 0& 8& 3\end{array}\right|=1\left(9-8\right)-1\left(6-0\right)-1\left(16-0\right)=1-6-16=-21 \\ \mathrm{Since},\left|\mathrm{A}\right|\ne 0,\mathrm{so}{\mathrm{A}}^{-1}\mathrm{exists}. \\ \mathrm{Let}{\mathrm{A}}_{\mathrm{ij}}\mathrm{be}\mathrm{the}\mathrm{cofactors}\mathrm{of}{\mathrm{a}}_{\mathrm{ij}}\mathrm{in}\left|\mathrm{A}\right|.\mathrm{Then} \\ {\mathrm{A}}_{11}={\left(-1\right)}^2\left(9-8\right)=1 \\ {\mathrm{A}}_{12}={\left(-1\right)}^3\left(6-0\right)=-6 \\ {\mathrm{A}}_{13}={\left(-1\right)}^4\left(16-0\right)=16 \\ {\mathrm{A}}_{21}={\left(-1\right)}^3\left(3+8\right)=-11 \\ {\mathrm{A}}_{22}={\left(-1\right)}^4\left(3-0\right)=3 \\ {\mathrm{A}}_{23}={\left(-1\right)}^5\left(8-0\right)=-8 \\ {\mathrm{A}}_{31}={\left(-1\right)}^4\left(1+3\right)=4 \\ {\mathrm{A}}_{32}={\left(-1\right)}^5\left(1+2\right)=-3 \\ {\mathrm{A}}_{33}={\left(-1\right)}^6\left(3-2\right)=1 \\ \mathrm{Now},\mathrm{adj}\left(\mathrm{A}\right)=\left[\begin{array}{ccc}1& -11& 4\\ -6& 3& -3\\ 16& -8& 1\end{array}\right] \\ \mathrm{Now},{\mathrm{A}}^{-1}=\frac{\mathrm{adj}\left(\mathrm{A}\right)}{\left|\mathrm{A}\right|}=-\frac{1}{21}\left[\begin{array}{ccc}1& -11& 4\\ -6& 3& -3\\ 16& -8& 1\end{array}\right] \\ \mathrm{Now},\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B} \\ \Rightarrow \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=-\frac{1}{21}\left[\begin{array}{ccc}1& -11& 4\\ -6& 3& -3\\ 16& -8& 1\end{array}\right]\left[\begin{array}{c}-3\\ 2\\ 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=-\frac{1}{21}\left[\begin{array}{c}-3-22+4\\ 18+6-3\\ -48-16+1\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=-\frac{1}{21}\left[\begin{array}{c}-21\\ 21\\ -63\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 3\end{array}\right] \\ \Rightarrow \mathrm{x}=1;\mathrm{y}=-1;\mathrm{z}=3 \end{gathered}$$