Question

If $A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$, find A⁻¹ and hence solve the system of linear equations
2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3

Answer 1

$$\begin{gathered} \text{Here,} \\ A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right| \\ =2\left(-4+4\right)+3\left(-6+4\right)+5(3-2) \\ =0-6+5 \\ =-1 \\ {\text{Let C}}_{ij}{\text{be the co factors 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& -4\\ 1& -2\end{array}\right|=0,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}3& -4\\ 1& -2\end{array}\right|=2,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}3& 2\\ 1& 1\end{array}\right|=1 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-3& 5\\ 1& -2\end{array}\right|=-1,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}2& 5\\ 1& -2\end{array}\right|=-9,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}2& -3\\ 1& 1\end{array}\right|=-5 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-3& 5\\ 2& -4\end{array}\right|=2,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}2& 5\\ 3& -4\end{array}\right|=23,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}2& -3\\ 3& 2\end{array}\right|=13 \\ \mathrm{adj}A={\left[\begin{array}{ccc}0& 2& 1\\ -1& -9& -5\\ 2& 23& 13\end{array}\right]}^{\mathrm{T}} \\ =\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right] \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}adjA \\ =\frac{1}{-1}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right] \\ \text{The given system of equations can be written in matrix form as follows:} \\ \left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right] \\ X=A^{-1}B \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-1}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-1}\left[\begin{array}{c}0+5-6\\ 22+45-69\\ 11+25-39\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-1}\left[\begin{array}{c}-1\\ -2\\ -3\end{array}\right] \\ \Rightarrow x=\frac{-1}{-1},y=\frac{-2}{-1}\mathrm{and}z=\frac{-3}{-1} \\ \therefore x=1,y=2\mathrm{and}z=3 \end{gathered}$$