Question

If $AX=B$, where $A=\left[ \begin{matrix} 1 & 2 & 3 \\\ -1 & 1 & 2 \\\ 1 & 2 & 4 \\\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]$, what is X?

Answer 1

In this problem, we have to find the value of X, with the given matrix A and B. We know that $AX=B$, where we can write it as, $X={{A}^{-1}}B$, we also know that ${{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adjA \right)$, here we have to find the determinant and the adjoint of A, and finally multiply it with B to get the value of X.

Complete step by step solution:
We know that the given matrices are,

1 & 2 & 3 \\\ -1 & 1 & 2 \\\ 1 & 2 & 4 \\\ \end{matrix} \right]$$, $$B=\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]$$. We also given that $$AX=B$$, we can now write it as, $$X={{A}^{-1}}B$$……. (1) Where $${{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adjA \right),\left| A \right|\ne 0$$…… (2) We can now find the inverse of A by finding the determinant and the adjoint of A. We can now find the determinant, we get $$\Rightarrow \left| A \right|=1\left( 4-4 \right)-2\left( -4-2 \right)+3\left( -2-1 \right)=3\ne 0$$ The determinant of A is 3. We can now find the adjoint by finding the cofactor. We can now find the cofactor of 1 in A, we get $$\Rightarrow {{A}_{11}}=\left| \begin{matrix} 1 & 2 \\\ 2 & 4 \\\ \end{matrix} \right|=4-4=0$$ Similarly, we can find the remaining cofactors, we get $$\begin{aligned} & \Rightarrow {{A}_{12}}=\left| \begin{matrix} -1 & 2 \\\ 1 & 4 \\\ \end{matrix} \right|=-4-2=-6 \\\ & \Rightarrow {{A}_{13}}=\left| \begin{matrix} -1 & 1 \\\ 1 & 2 \\\ \end{matrix} \right|=-2-1=-3 \\\ \end{aligned}$$ We can now find the cofactors for the second row, we get $$\begin{aligned} & \Rightarrow {{A}_{21}}=\left| \begin{matrix} 2 & 3 \\\ 2 & 4 \\\ \end{matrix} \right|=8-6=2 \\\ & \Rightarrow {{A}_{22}}=\left| \begin{matrix} 1 & 3 \\\ 1 & 4 \\\ \end{matrix} \right|=4-3=1 \\\ & \Rightarrow {{A}_{23}}=\left| \begin{matrix} 1 & 2 \\\ 1 & 2 \\\ \end{matrix} \right|=2-2=0 \\\ \end{aligned}$$ We can now find the cofactors for the third row, we get $$\begin{aligned} & \Rightarrow {{A}_{31}}=\left| \begin{matrix} 2 & 3 \\\ 1 & 2 \\\ \end{matrix} \right|=4-3=1 \\\ & \Rightarrow {{A}_{32}}=\left| \begin{matrix} 1 & 3 \\\ -1 & 2 \\\ \end{matrix} \right|=2+3=5 \\\ & \Rightarrow {{A}_{33}}=\left| \begin{matrix} 1 & 2 \\\ -1 & 1 \\\ \end{matrix} \right|=1+2=3 \\\ \end{aligned}$$ We can now arrange the cofactors, we get $$Cofactor=\left[ \begin{matrix} 0 & 6 & -3 \\\ -2 & 1 & 0 \\\ 1 & -5 & 3 \\\ \end{matrix} \right]$$ We can now find the transpose of the cofactor which is the adjoint, we get $$Adj=\left[ \begin{matrix} 0 & -2 & 1 \\\ 6 & 1 & -5 \\\ -3 & 0 & 3 \\\ \end{matrix} \right]$$ We can now write the inverse of A, we get $$\Rightarrow {{A}^{-1}}=\dfrac{1}{3}\times \left[ \begin{matrix} 0 & -2 & 1 \\\ 6 & 1 & -5 \\\ -3 & 0 & 3 \\\ \end{matrix} \right]$$ We can now substitute the above step in (1), we get $$\Rightarrow X=\dfrac{1}{3}\times \left[ \begin{matrix} 0 & -2 & 1 \\\ 6 & 1 & -5 \\\ -3 & 0 & 3 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]$$ We can now simplify the above step by matrix multiplication, we get $$\Rightarrow X=\dfrac{1}{3}\left[ \begin{matrix} 0-4+3 \\\ 6+2-15 \\\ -3+0+9 \\\ \end{matrix} \right]=\dfrac{1}{3}\left[ \begin{matrix} -1 \\\ -7 \\\ 6 \\\ \end{matrix} \right]$$ We can now simplify the above step, we get $$\Rightarrow X=\dfrac{1}{3}\left[ \begin{matrix} -1 \\\ -7 \\\ 6 \\\ \end{matrix} \right]=\left[ \begin{matrix} -\dfrac{1}{3} \\\ -\dfrac{7}{3} \\\ 2 \\\ \end{matrix} \right]$$ Therefore, the answer is $$X=\left[ \begin{matrix} -\dfrac{1}{3} \\\ -\dfrac{7}{3} \\\ 2 \\\ \end{matrix} \right]$$. **Note:** Students make mistakes while finding the cofactor value in the symbol part. We should remember that the inverse of the matrix can be found by the formula $${{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adjA \right),\left| A \right|\ne 0$$, where, we should remember that the adjoint is the transpose of the cofactor.