Solveeit Logo
Login

Question

Mathematics Question on Matrices

If A=[123 −112 124]A = \begin{bmatrix}1&2&3\\\ -1&1&2\\\ 1&2&4\end{bmatrix} then (A2−5A)A−1=\left(A^{2} - 5A\right)A^{-1} =

A

[423 −142 121]\begin{bmatrix}4&2&3\\\ -1&4&2\\\ 1&2&1\end{bmatrix}

B

[−423 −1−42 12−1]\begin{bmatrix} - 4&2&3\\\ -1& -4&2\\\ 1&2& -1\end{bmatrix}

C

[−4−11 2−42 32−1]\begin{bmatrix}-4&-1&1\\\ 2&-4&2\\\ 3&2&-1\end{bmatrix}

D

[−1−21 4−2−3 14−2]\begin{bmatrix}-1&-2&1\\\ 4&-2&-3\\\ 1&4&-2\end{bmatrix}

Answer

[−423 −1−42 12−1]\begin{bmatrix} - 4&2&3\\\ -1& -4&2\\\ 1&2& -1\end{bmatrix}

Explanation

Solution

we have
A=[123 −112 124]A = \begin{bmatrix}1&2&3\\\ -1&1&2\\\ 1&2&4\end{bmatrix}
Now,(A2−5A)A−1 \left(A^{2} -5A\right)A^{-1}
=A2⋅A−1−5A⋅A−1=A−5I= A^{2} \cdot A^{-1} - 5A \cdot A^{-1} = A - 5I
=[123 −112 124]−[500 050 005]= \begin{bmatrix}1&2&3\\\ -1&1&2\\\ 1&2&4\end{bmatrix}-\begin{bmatrix}5&0&0\\\ 0&5&0\\\ 0&0&5\end{bmatrix}
=[−423 −1−42 12−1]= \begin{bmatrix}-4&2&3\\\ -1&-4&2\\\ 1&2&-1\end{bmatrix}