Question

If $A=\left[ \begin{matrix} 5 & 8 & 1 \\\ 0 & 2 & 1 \\\ 4 & 3 & -1 \\\ \end{matrix} \right],\ B=\left[ \begin{matrix} 2 \\\ -1 \\\ 3 \\\ \end{matrix} \right]$ and AX = B, then find X.

Answer 1

Hint: Use simple matrix multiplication to understand the relation given to us in the question. Then, use the formula for calculating the inverse of a given matrix, to calculate the matrix X. This is because, if we multiply ${{A}^{-1}}$ on both sides, then the LHS becomes only X, and further calculation will get us the answer.

Complete step-by-step answer:

Given to us is the equation AX = B.
Let’s multiply both sides by ${{A}^{-1}}$.
$\begin{aligned} & \Rightarrow {{A}^{-1}}AX={{A}^{-1}}B \\\ & \Rightarrow IX={{A}^{-1}}B\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{Since }A.{{A}^{-1}}={{A}^{-1}}A=I \right) \\\ & \Rightarrow X={{A}^{-1}}B \\\ \end{aligned}$
We can find the matrix ${{A}^{-1}}$ by using the formula ${{A}^{-1}}=\dfrac{adjA}{\left| A \right|}$.
To find the adjoint of a matrix, first find the cofactor matrix of the given matrix. Then find the transpose of the cofactor matrix.
Cofactor of 5 $\text{ = }{{\text{A}}_{11}}=\left| \begin{matrix} 2 & 1 \\\ 3 & -1 \\\ \end{matrix} \right|=-5$
Cofactor of 8 $\text{= }{{\text{A}}_{12}}=-\left| \begin{matrix} 0 & 1 \\\ 4 & -1 \\\ \end{matrix} \right|=4$
Cofactor of 1 $\text{ = }{{\text{A}}_{13}}=\left| \begin{matrix} 0 & 2 \\\ 4 & 3 \\\ \end{matrix} \right|=-8$
Cofactor of 0 $\text{= }{{\text{A}}_{21}}=-\left| \begin{matrix} 8 & 1 \\\ 3 & -1 \\\ \end{matrix} \right|=11$
Cofactor of 2 $\text{= }{{\text{A}}_{22}}=\left| \begin{matrix} 5 & 1 \\\ 4 & -1 \\\ \end{matrix} \right|=-9$
Cofactor of 1 $\text{= }{{\text{A}}_{23}}=-\left| \begin{matrix} 5 & 8 \\\ 4 & 3 \\\ \end{matrix} \right|=17$
Cofactor of 4 $\text{= }{{\text{A}}_{31}}=\left| \begin{matrix} 8 & 1 \\\ 2 & 1 \\\ \end{matrix} \right|=6$
Cofactor of 3 $\text{= }{{\text{A}}_{32}}=-\left| \begin{matrix} 5 & 1 \\\ 0 & 1 \\\ \end{matrix} \right|=-5$
Cofactor of -1 $\text{= }{{\text{A}}_{33}}=\left| \begin{matrix} 5 & 8 \\\ 0 & 2 \\\ \end{matrix} \right|=10$
So, the cofactor matrix of A is $\left[ \begin{matrix} -5 & 4 & -8 \\\ 11 & -9 & 17 \\\ 6 & -5 & 10 \\\ \end{matrix} \right]$ .
Now, the adjoint of matrix A is the transpose of the cofactor matrix of A, which is given as $\left[ \begin{matrix} -5 & 11 & 6 \\\ 4 & -9 & -5 \\\ -8 & 17 & 10 \\\ \end{matrix} \right]$ .
Now, the determinant of matrix A is given as $\left| A \right|=\left| \begin{matrix} 5 & 8 & 1 \\\ 0 & 2 & 1 \\\ 4 & 3 & -1 \\\ \end{matrix} \right|$ .
$=5\left( -5 \right)-8\left( -4 \right)+1\left( -8 \right)$
$=-25-8+32=-1$
Now, we know the inverse of a matrix given by dividing the adjoint of the matrix by its determinant. So, the inverse of matrix A is given as ${{A}^{-1}}=\dfrac{\left( adjA \right)}{\left| A \right|}$ .
$=\dfrac{\left[ \begin{matrix} -5 & 11 & 6 \\\ 4 & -9 & -5 \\\ -8 & 17 & 10 \\\ \end{matrix} \right]}{-1}$
$=\left[ \begin{matrix} 5 & -11 & -6 \\\ -4 & 9 & 5 \\\ 8 & -17 & -10 \\\ \end{matrix} \right]$
Now, we have $X={{A}^{-1}}B$ .
$\Rightarrow X=\left[ \begin{matrix} 5 & -11 & -6 \\\ -4 & 9 & 5 \\\ 8 & -17 & -10 \\\ \end{matrix} \right]\left[ \begin{matrix} 2 \\\ -1 \\\ 3 \\\ \end{matrix} \right]$
$\Rightarrow X=\left[ \begin{matrix} \left( 5\times 2 \right)+\left( -11\times \left( -1 \right) \right)+\left( -6\times 3 \right) \\\ \left( -4\times 2 \right)+\left( 9\times \left( -1 \right) \right)+\left( 5\times 3 \right) \\\ \left( 8\times 2 \right)+\left( -17\times \left( -1 \right) \right)+\left( -10\times 3 \right) \\\ \end{matrix} \right]$
$\Rightarrow X=\left[ \begin{matrix} 10+11-18 \\\ -8-9+15 \\\ 16+17-30 \\\ \end{matrix} \right]$
$\Rightarrow X=\left[ \begin{matrix} 3 \\\ -2 \\\ 3 \\\ \end{matrix} \right]$
Therefore, $X=\left[ \begin{matrix} 3 \\\ -2 \\\ 3 \\\ \end{matrix} \right]$.

Note: You can also find the inverse of the matrix by carrying out row or column transformations.
Be careful about the fact when (i + j) is odd in ${{A}_{ij}}$ the cofactor is the negative of the minor value of ${{A}_{ij}}$.