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Question: If \(AX=B\), where \(A=\left[ \begin{matrix} 3 & 1 \\\ -1 & 2 \\\ \end{matrix} \right]\)...

If AX=BAX=B, where A=[31 −12 ]A=\left[ \begin{matrix} 3 & 1 \\\ -1 & 2 \\\ \end{matrix} \right], B=[73 06 ]B=\left[ \begin{matrix} 7 & 3 \\\ 0 & 6 \\\ \end{matrix} \right]. Then X=X=?
A. [10 23 ]\left[ \begin{matrix} 1 & 0 \\\ 2 & 3 \\\ \end{matrix} \right]
B. [03 12 ]\left[ \begin{matrix} 0 & 3 \\\ 1 & 2 \\\ \end{matrix} \right]
C. [32 01 ]\left[ \begin{matrix} 3 & 2 \\\ 0 & 1 \\\ \end{matrix} \right]
D. [20 13 ]\left[ \begin{matrix} 2 & 0 \\\ 1 & 3 \\\ \end{matrix} \right]

Explanation

Solution

We first form the multiplication form of matrices to find the coefficient matrix AX=BAX=B. Then we use the inverse matrix to find the variable matrix with the form of X=A−1BX={{A}^{-1}}B. We find the inverse after finding the matrix being singular or not.

Complete step-by-step solution:
We multiply the equation AX=BAX=B with A−1{{A}^{-1}} to get
A−1.AX=A−1.B ⇒IX=X=A−1B \begin{aligned} & {{A}^{-1}}.AX={{A}^{-1}}.B \\\ & \Rightarrow IX=X={{A}^{-1}}B \\\ \end{aligned}
The variables matrix will be in the form of X=A−1BX={{A}^{-1}}B. We have to multiply the inverse matrix of A with the solution matrix.
The inverse of any matrix A=[aij]A=\left[ {{a}_{ij}} \right] will be A−1=adj(A)∣A∣{{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}. Here aij{{a}_{ij}} denotes the element of ith{{i}^{th}} row and jth{{j}^{th}} column.
∣A∣\left| A \right| is defined as the determinant value of the matrix A.
The adj(A)adj\left( A \right) is defined by the adj(A)=[Aij]T=[Aji]adj\left( A \right)={{\left[ {{A}_{ij}} \right]}^{T}}=\left[ {{A}_{ji}} \right]. Here, Aij{{A}_{ij}} denotes the cofactor of the element aij{{a}_{ij}}. The term ‘T’ denotes the transpose of the matrix.
We find the cofactors of the matrix A and get [Aij]=[21 −13 ]\left[ {{A}_{ij}} \right]=\left[ \begin{matrix} 2 & 1 \\\ -1 & 3 \\\ \end{matrix} \right] which gives
adj(A)=[Aij]T=[Aij]=[2−1 13 ]adj\left( A \right)={{\left[ {{A}_{ij}} \right]}^{T}}=\left[ {{A}_{ij}} \right]=\left[ \begin{matrix} 2 & -1 \\\ 1 & 3 \\\ \end{matrix} \right].
Now we find the determinant which is ∣A∣=3×2−1×(−1)=6+1=7\left| A \right|=3\times 2-1\times \left( -1 \right)=6+1=7.
As the matrix A is a non-singular matrix the inverse of the matrix exists.
We have A−1=adj(A)∣A∣=17[2−1 13 ]{{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}=\dfrac{1}{7}\left[ \begin{matrix} 2 & -1 \\\ 1 & 3 \\\ \end{matrix} \right].
We get X=A−1BX={{A}^{-1}}B.
Therefore, X=A−1B=17[2−1 13 ][73 06 ]=17[140 721 ]=[20 13 ]X={{A}^{-1}}B=\dfrac{1}{7}\left[ \begin{matrix} 2 & -1 \\\ 1 & 3 \\\ \end{matrix} \right]\left[ \begin{matrix} 7 & 3 \\\ 0 & 6 \\\ \end{matrix} \right]=\dfrac{1}{7}\left[ \begin{matrix} 14 & 0 \\\ 7 & 21 \\\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 0 \\\ 1 & 3 \\\ \end{matrix} \right].
The correct option is D.

Note: If the determinant value of the matrix of which we are finding the inverse is 0 then the matrix is non-invertible. Those types of matrices are called singular matrices.
Also, we need to remember that the multiplications X=A−1BX={{A}^{-1}}B and X=BA−1X=B{{A}^{-1}} are different.