Question

Find \( A^{-1} \), if \( A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right] \) then hence solve the system of linear equations \[ x+y+2 z=0, x+2 y-z=9, x-3 y+3 z=-14 \]

Answer 1

\(A = \begin{bmatrix}1 &1&1\\1&2&-3\\2&-1&3\end{bmatrix}\)

∵ c_ij = (-1)^ij is cofactor of element a_ij of matrix A where m_ij is minor of element a_ij of matrix A which equals to the determinant of matrix obtained by hiding i^th row and j^th column of matrix A

∴ Cofactor matrix

\(C = \begin{bmatrix} C_{11}& C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{33} \end{bmatrix}= \begin{bmatrix}3&-9&-5\\-4&1&3\\-5&4&1 \end{bmatrix}\)

\(Adj(A) = C^T= \begin{bmatrix}3&-4&-5\\-9&1&4\\-5&3&1\end{bmatrix}\)

\(|A| = 1(6 -3) - 1(3 + 6) + 1( -1 -4)\)

= 3 - 9 - 5 

= -11

\(\therefore A^{-1} = \frac{AdjA}{|A|} = \frac{-1}{11} \begin{bmatrix}3&-4&-5\\ -9&1&4\\-5&3&1\end {bmatrix}\)

Matrix form of given system of equations is A^Tx = B

where

\(A^T = \begin{bmatrix}1&1&2\\1&2&-1\\1&-3&3\end{bmatrix}\)

\(x= \begin{bmatrix}x\\y\\z\end{bmatrix}\)

\(B = \begin {bmatrix}0\\9\\-14 \end{bmatrix}\)

∴ \(x = (A^T)^{-1}B\)

\(= (A^{-1})^TB\)

\(= \frac{-1}{11} \begin{bmatrix}3 &-9&-5\\-4&1&3\\-5&4&1 \end{bmatrix} \begin{bmatrix}0\\9\\-14\end{bmatrix} \)

\(= \frac{-1}{11} \begin{bmatrix}-11\\-33\\22\end{bmatrix}\)

\(\therefore x = \begin{bmatrix}1\\3\\-2\end{bmatrix}\)

Hence, 

\(x= \begin{bmatrix}x\\y\\z\end{bmatrix}= \begin{bmatrix}1\\3\\-2\end{bmatrix}\)

∴ x = 1, y = 3 & z = -2 is solution of given system.