Question

Find A⁻¹, if $A=\left[\begin{array}{ccc}1& 2& 5\\ 1& -1& -1\\ 2& 3& -1\end{array}\right]$. Hence solve the following system of linear equations:
x + 2y + 5z = 10, x − y − z = −2, 2x + 3y − z = −11

Answer 1

$$\begin{gathered} \text{Here,} \\ A=\left[\begin{array}{ccc}1& 2& 5\\ 1& -1& -1\\ 2& 3& -1\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}1& 2& 5\\ 1& -1& -1\\ 2& 3& -1\end{array}\right| \\ =1\left(1+3\right)-2\left(-1+2\right)+5(3+2) \\ =4-2+25 \\ =27 \\ {\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}-1& -1\\ 3& -1\end{array}\right|=4,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& -1\\ 2& -1\end{array}\right|=-1,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}1& -1\\ -2& 3\end{array}\right|=5 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}2& 5\\ 3& -1\end{array}\right|=17,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 5\\ 2& -1\end{array}\right|=-11,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 2\\ 2& 3\end{array}\right|=1 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}2& 5\\ -1& -1\end{array}\right|=3,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 5\\ 1& -1\end{array}\right|=6,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 2\\ 1& -1\end{array}\right|=-3 \\ \mathrm{adj}A={\left[\begin{array}{ccc}4& -1& 5\\ 17& -11& 1\\ 3& 6& -3\end{array}\right]}^{\mathrm{T}} \\ =\left[\begin{array}{ccc}4& 17& 3\\ -1& -11& 6\\ 5& 1& -3\end{array}\right] \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{27}\left[\begin{array}{ccc}4& 17& 3\\ -1& -11& 6\\ 5& 1& -3\end{array}\right] \\ \text{The given system of equations can be written in matrix form as follows:} \\ \left[\begin{array}{ccc}1& 2& 5\\ 1& -1& -1\\ 2& 3& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}10\\ -2\\ -11\end{array}\right] \\ X=A^{-1}B \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{27}\left[\begin{array}{ccc}4& 17& 3\\ -1& -11& 6\\ 5& 1& -3\end{array}\right]\left[\begin{array}{c}10\\ -2\\ -11\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{27}\left[\begin{array}{c}40-34-33\\ -10+22-66\\ 50-2+33\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{27}\left[\begin{array}{c}-27\\ -54\\ 81\end{array}\right] \\ \therefore x=\frac{-27}{27},y=\frac{-54}{27}\mathrm{and}z=\frac{81}{27} \\ \Rightarrow x=-1,y=-2\mathrm{and}z=3 \end{gathered}$$