Question

If \[A = \begin{bmatrix}2 & 3 & 1 \\ 1 & 2 & 2 \\ 3 & 1 & - 1\end{bmatrix}\] , find A^–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8.

Answer 1

We have,
\[A = \begin{bmatrix}2 & 3 & 1 \\ 1 & 2 & 2 \\ -3 & 1 & - 1\end{bmatrix}\]
\[\therefore \left| A \right| = \begin{vmatrix}2 & 3 & 1 \\ 1 & 2 & 2 \\ - 3 & 1 & - 1\end{vmatrix}\]
\[ = 2\left( - 2 - 2 \right) - 3\left( - 1 + 6 \right) + 1\left( 1 + 6 \right)\]
\[ = - 8 - 15 + 7\]
\[ = - 16 \neq 0\]
So, A is invertible.
Let C_ij be the co-factors of the elements a_ij in A[a_ij]. Then,
\[C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 2 \\ 1 & - 1\end{vmatrix} = - 2 - 2 = - 4\]
\[ C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 2 \\ - 3 & - 1\end{vmatrix} = - 1\left( - 1 + 6 \right) = - 5\]
\[ C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 2 \\ - 3 & 1\end{vmatrix} = 1 + 6 = 7\]
\[C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 1 \\ 1 & - 1\end{vmatrix} = 3 + 1 = 4\]
\[ C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 1 \\ - 3 & - 1\end{vmatrix} = - 2 + 3 = 1\]
\[ C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 3 \\ - 3 & 1\end{vmatrix} = - 1\left( 2 + 9 \right) = - 11\]
\[C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 1 \\ 2 & 2\end{vmatrix} = 6 - 2 = 4\]
\[ C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix} = - 1\left( 4 - 1 \right) = - 3\]
\[ C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 3 \\ 1 & 2\end{vmatrix} = 4 - 3 = 1\]
\[\therefore Adj A = \begin{bmatrix}- 4 & - 5 & 7 \\ 4 & 1 & - 11 \\ 4 & - 3 & 1\end{bmatrix}^T = \begin{bmatrix}- 4 & 4 & 4 \\ - 5 & 1 & - 3 \\ 7 & - 11 & 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{Adj A}{\left| A \right|} = \frac{1}{- 16}\begin{bmatrix}- 4 & 4 & 4 \\ - 5 & 1 & - 3 \\ 7 & - 11 & 1\end{bmatrix}\]
Now, the given system of equations is expressible as
Or A^T X = B, where 

\[X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}, B = \begin{bmatrix}13 \\ 4 \\ 8\end{bmatrix}\]
Now,

\[\left| A^T \right| = \left| A \right| = - 16 \neq 0\]
So, the given system of equations is consistent with a unique solution given by

\[X = \left( A^T \right)^{- 1} B = \left( A^{- 1} \right)^T B\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{16} \begin{bmatrix}- 4 & 4 & 4 \\ - 5 & 1 & - 3 \\ 7 & - 11 & 1\end{bmatrix}^T \begin{bmatrix}13 \\ 4 \\ 8\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{16}\begin{bmatrix}- 4 & - 5 & 7 \\ 4 & 1 & - 11 \\ 4 & - 3 & 1\end{bmatrix}\begin{bmatrix}13 \\ 4 \\ 8\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{16}\begin{bmatrix}- 52 - 20 + 56 \\ 52 + 4 - 88 \\ 52 - 12 + 8\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{16}\begin{bmatrix}- 16 \\ - 32 \\ 48\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 2 \\ - 3\end{bmatrix}\]
Hence, x = 1, y = 2 and z = −3 is the required solution.