Question

Let we are given the matrices as $X=\left[ \begin{matrix} {{x}_{1}} \\\ {{x}_{2}} \\\ {{x}_{3}} \\\ \end{matrix} \right]$, $A=\left[ \begin{matrix} 1 & -1 & 2 \\\ 2 & 0 & 1 \\\ 3 & 2 & 1 \\\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 3 \\\ 1 \\\ 4 \\\ \end{matrix} \right]$. If $AX=B$ then $X$ is equal to $$$$
A. \left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]$$$$$ B. \left[ \begin{matrix}
-1 \\
-2 \\
-3 \\
\end{matrix} \right] C. $\left[ \begin{matrix} -1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]
D. $\left[ \begin{matrix}
0 \\
2 \\
1 \\
\end{matrix} \right]$$$$$

Answer 1

We multiply ${{A}^{-1}}$ at both sides of the given equation on the left and get $X={{A}^{-1}}B$. We find the inverse matrix ${{A}^{-1}}=\dfrac{\text{adj}\text{.}A}{\det \left( A \right)}$ where $\text{adj}\text{.}A$ is the adjoint matrix and $\det \left( A \right)$ is the determinant value of $A$. We find the adjoint by taking the transpose of the cofactor matrix of $A$ which means $text{adj}.A={{\left( \text{cof}.A \right)}^{T}}$ , the determinant value and then ${{A}^{-1}}$. We multiply ${{A}^{-1}}B$ to get $X$.

Complete step-by-step solution:
We are given in the question a column vector $X$ of order $3\times 1$ with three unknowns${{x}_{1}},{{x}_{2}},{{x}_{3}}$, a square matrix $A$ of order $3\times 3$ with integers as entries and a column vector $B$of order $3\times 1$ with integer as entries. We have

{{x}_{1}} \\\ {{x}_{2}} \\\ {{x}_{3}} \\\ \end{matrix} \right],A=\left[ \begin{matrix} 1 & -1 & 2 \\\ 2 & 0 & 1 \\\ 3 & 2 & 1 \\\ \end{matrix} \right],B=\left[ \begin{matrix} 3 \\\ 1 \\\ 4 \\\ \end{matrix} \right]$$ We are also given an equation of matrices $$AX=B$$ Let us multiply the inverse of the matrix $A$ that is ${{A}^{-1}}$ on both sides of the equation from the left. We have, $$\Rightarrow {{A}^{-1}}\left( AX \right)={{A}^{-1}}B$$ We know that matrix multiplication is associative. So we have $$\begin{aligned} & \Rightarrow \left( {{A}^{-1}}A \right)X={{A}^{-1}}B \\\ & \Rightarrow IX={{A}^{-1}}B \\\ & \Rightarrow X={{A}^{-1}}B......\left( 1 \right) \\\ \end{aligned}$$ Here “I” is the identity matrix of order $3\times 3$. So we have to find ${{A}^{-1}}$. We know that $${{A}^{-1}}=\dfrac{1}{\det \left( A \right)}\text{adj}A$$ Here $\text{adj}A$ is the adjoint matrix of A and $\det \left( A \right)$ is the determinant value of $A$, Let us find the determinant value first by expanding by first row. We have, $$\det \left( A \right)=\left| \begin{matrix} 1 & -1 & 2 \\\ 2 & 0 & 1 \\\ 3 & 2 & 1 \\\ \end{matrix} \right|=1\left( 0-2 \right)-\left( -1 \right)\left( 2-3 \right)+2\left( 4-0 \right)=5$$ We know that the adjoint matrix of $A$ is the transpose of cofactor matrix of which means $$\text{adj}A={{\left( \text{cof}A \right)}^{T}}$$ We know that the cofactor matrix is made up of entries as the cofactors of each element from the original matrix. The cofactor of the entry ${{a}_{ij}}$ (where $i$ is row position and $j$ is the column position) is equal to the determinant value of the square matrix (order 1 less than original matrix) t formed by rows and columns excluding the ${{i}^{\text{th}}}$row and ${{j}^{\text{th}}}$column and then multiplied by ${{\left( -1 \right)}^{i+j}}$. The co-factors of the entries in first row are $$\begin{aligned} & C\left( 1 \right)={{\left( -1 \right)}^{1+1}}\left| \begin{matrix} 0 & 1 \\\ 2 & 1 \\\ \end{matrix} \right|=1\left( 0-2 \right)=-2 \\\ & C\left( -1 \right)={{\left( -1 \right)}^{1+2}}\left| \begin{matrix} 2 & 1 \\\ 3 & 1 \\\ \end{matrix} \right|=-1\left( 2-3 \right)=1 \\\ & C\left( 2 \right)={{\left( -1 \right)}^{1+3}}\left| \begin{matrix} 2 & 0 \\\ 3 & 2 \\\ \end{matrix} \right|=1\left( 4-0 \right)=4 \\\ \end{aligned}$$ The cofactors of the elements in the second row are $$\begin{aligned} & C\left( 2 \right)={{\left( -1 \right)}^{2+1}}\left| \begin{matrix} -1 & 2 \\\ 2 & 1 \\\ \end{matrix} \right|=-1\left( -1-4 \right)=5 \\\ & C\left( 0 \right)={{\left( -1 \right)}^{2+2}}\left| \begin{matrix} 1 & 2 \\\ 3 & 1 \\\ \end{matrix} \right|=1\left( 1-6 \right)=-5 \\\ & C\left( 1 \right)={{\left( -1 \right)}^{2+3}}\left| \begin{matrix} 1 & -1 \\\ 3 & 2 \\\ \end{matrix} \right|=-1\left( 2-\left( -3 \right) \right)=-5 \\\ \end{aligned}$$ The cofactors of the elements in the third row are $$\begin{aligned} & C\left( 3 \right)={{\left( -1 \right)}^{3+1}}\left| \begin{matrix} -1 & 2 \\\ 0 & 1 \\\ \end{matrix} \right|=1\left( -1-0 \right)=-1 \\\ & C\left( 2 \right)={{\left( -1 \right)}^{3+2}}\left| \begin{matrix} 1 & 2 \\\ 2 & 1 \\\ \end{matrix} \right|=-1\left( 1-4 \right)=3 \\\ & C\left( 1 \right)={{\left( -1 \right)}^{3+3}}\left| \begin{matrix} 1 & -1 \\\ 2 & 0 \\\ \end{matrix} \right|=1\left( 0-\left( -2 \right) \right)=2 \\\ \end{aligned}$$ So the co-factor matrix $A$ is, $$\text{cof}\text{.}A=\left[ \begin{matrix} C\left( 1 \right) & C\left( -1 \right) & C\left( 2 \right) \\\ C\left( 2 \right) & C\left( 0 \right) & C\left( 1 \right) \\\ C\left( 3 \right) & C\left( 2 \right) & C\left( 1 \right) \\\ \end{matrix} \right]=\left[ \begin{matrix} -2 & 1 & 4 \\\ 5 & -5 & 5 \\\ -1 & 3 & 2 \\\ \end{matrix} \right]$$ Let us take the transpose of the cofactor matrix of $A$ by writing rows as column to get adjoint matrix of $A$ We have $$\text{adj}\text{.}A={{\left( \text{cof}\text{.}A \right)}^{T}}={{\left[ \begin{matrix} -2 & 1 & 4 \\\ 5 & -5 & -5 \\\ -1 & 3 & 2 \\\ \end{matrix} \right]}^{T}}=\left[ \begin{matrix} -2 & 5 & -1 \\\ 1 & -5 & 3 \\\ 4 & -5 & 2 \\\ \end{matrix} \right]$$ So the inverse of matrix of $A$ is, $${{A}^{-1}}=\dfrac{adj.A}{\det \left( A \right)}=\dfrac{\left[ \begin{matrix} -2 & 5 & -1 \\\ 1 & -5 & 3 \\\ 4 & -5 & 2 \\\ \end{matrix} \right]}{5}=\left[ \begin{matrix} \dfrac{-2}{5} & 1 & \dfrac{-1}{5} \\\ \dfrac{1}{5} & -1 & \dfrac{3}{5} \\\ \dfrac{4}{5} & -1 & \dfrac{2}{5} \\\ \end{matrix} \right]$$ We put ${{A}^{-1}}$ and $B$ in equation (1) and find the unknown vector $X$ as, $$X={{A}^{-1}}B=\left[ \begin{matrix} \dfrac{-2}{5} & 1 & \dfrac{-1}{5} \\\ \dfrac{1}{5} & -1 & \dfrac{3}{5} \\\ \dfrac{4}{5} & -1 & \dfrac{2}{5} \\\ \end{matrix} \right]\times \left[ \begin{matrix} 3 \\\ 1 \\\ 4 \\\ \end{matrix} \right]=\left[ \begin{matrix} -1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]$$ So the correct choice is C.$$$$ **Note:** We note that we could multiply ${{A}^{-1}}$ from the left only not from the right as multiplication in matrix is not commutative but ${{A}^{-1}}A=A{{A}^{-1}}=I$. We can alternatively find the inverse by using Gauss-Jordan elimination of the augmented matrix$\left( A, B \right)$. The matrix equation $AX=B$ is a system of equations which here in this problem has a unique solution. . We note that we cannot find an inverse if when$\det \left( A \right)=0$.