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Mathematics Question on Determinants

If Matrix A=[12 43]A = \begin{bmatrix}1&2\\\ 4&3\end{bmatrix} such that Ax=IAx = I, then X=X = _______

A

15[13 2−1]\frac{1}{5}\begin{bmatrix}1&3\\\ 2&-1\end{bmatrix}

B

15[42 4−1]\frac{1}{5}\begin{bmatrix}4&2\\\ 4&-1\end{bmatrix}

C

15[−32 4−1]\frac{1}{5}\begin{bmatrix}-3&2\\\ 4&-1\end{bmatrix}

D

15[−12 −14]\frac{1}{5}\begin{bmatrix}-1&2\\\ -1&4\end{bmatrix}

Answer

15[−32 4−1]\frac{1}{5}\begin{bmatrix}-3&2\\\ 4&-1\end{bmatrix}

Explanation

Solution

Given , A=[12 43]A=\begin{bmatrix}1&2\\\ 4&3\end{bmatrix}
and AX=IAX=I
⇒X=A−1I\Rightarrow X=A^{-1}I
⇒X=A−1\Rightarrow X=A^{-1}
Now, A−1=1∣A∣[3−2 −41]A^{-1}=\frac{1}{\left|A\right|}\begin{bmatrix}3&-2\\\ -4&1\end{bmatrix}
=13−8[3−2 −41]=\frac{1}{3-8}\begin{bmatrix}3&-2\\\ -4&1\end{bmatrix}
=1−5[3−2 −41]=\frac{1}{-5}\begin{bmatrix}3&-2\\\ -4&1\end{bmatrix}
=15[−32 4−1]=\frac{1}{5}\begin{bmatrix}-3&2\\\ 4&-1\end{bmatrix}