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Mathematics Question on Invertible Matrices

The inverse of the matrix [250 011 −103]\begin{bmatrix}2&5&0\\\ 0&1&1\\\ -1&0&3\end{bmatrix} is

A

[3−155 −16−2 1−52]\begin{bmatrix}3&-15&5\\\ -1&6&-2\\\ 1&-5&2\end{bmatrix}

B

[3−11 −156−5 5−22]\begin{bmatrix}3&-1&1\\\ -15&6&-5\\\ 5&-2&2\end{bmatrix}

C

[3−155 −16−2 1−5−2]\begin{bmatrix}3&-15&5\\\ -1&6&-2\\\ 1&-5&-2\end{bmatrix}

D

[3−55 −1−6−2 1−52]\begin{bmatrix}3&-5&5\\\ -1&-6&-2\\\ 1&-5&2\end{bmatrix}

Answer

[3−155 −16−2 1−52]\begin{bmatrix}3&-15&5\\\ -1&6&-2\\\ 1&-5&2\end{bmatrix}

Explanation

Solution

LetA=[250 011 −103]Let A=\left[\begin{matrix}2&5&0\\\ 0&1&1\\\ -1&0&3\end{matrix}\right]
∣A∣=2(3−0)−5(0+1)\left|A\right|=2\left(3-0\right)-5\left(0+1\right)
=6−5=1= 6 - 5 = 1
∴ ∣A∣=1\therefore\, \left|A\right|=1
adj A=[3−11 −156−5 5−22]T=[3−155 −16−2 1−52]adj \, A=\left[\begin{matrix}3&-1&1\\\ -15&6&-5\\\ 5&-2&2\end{matrix}\right]^{T}=\left[\begin{matrix}3&-15&5\\\ -1&6&-2\\\ 1&-5&2\end{matrix}\right]
A−1=adj A∣A∣=[3−155 −16−2 1−52]A^{-1}=\frac{adj\, A}{\left|A\right|}=\left[\begin{matrix}3&-15&5\\\ -1&6&-2\\\ 1&-5&2\end{matrix}\right]