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Mathematics Question on Matrices

The matrix 'X' in the equation AX=BAX = B, such that A=[13 01]A = \begin{bmatrix}1&3\\\ 0&1\end{bmatrix} and B=[1−1 01] B = \begin{bmatrix}1&-1\\\ 0&1\end{bmatrix} is given by

A

[10 −31]\begin{bmatrix}1&0\\\ -3&1\end{bmatrix}

B

[1−4 01]\begin{bmatrix}1& -4 \\\ 0 & 1\end{bmatrix}

C

[1−3 01]\begin{bmatrix}1& -3 \\\ 0 &1\end{bmatrix}

D

[0−1 −31]\begin{bmatrix} 0 & -1 \\\ -3&1\end{bmatrix}

Answer

[1−4 01]\begin{bmatrix}1& -4 \\\ 0 & 1\end{bmatrix}

Explanation

Solution

Given A=[13 01]⇒detA=1≠0.A = \begin{bmatrix}1&3\\\ 0&1\end{bmatrix} \Rightarrow det A = 1 \ne 0. So, A−1A^{-1} exists. Hence AX=BAX = B ⇒X=A−1B\Rightarrow X = A^{-1} B ⇒X=[1−3 01][1−1 01]=[1−4 01]\Rightarrow X = \begin{bmatrix}1&-3\\\ 0&1\end{bmatrix}\begin{bmatrix}1&-1\\\ 0&1\end{bmatrix} = \begin{bmatrix}1&-4\\\ 0&1\end{bmatrix}