Question

The inverse of the matrix $\begin{bmatrix}2&5&0\\ 0&1&1\\ -1&0&3\end{bmatrix} $ is

• A. $\begin{bmatrix}3&-15&5\\ -1&6&-2\\ 1&-5&2\end{bmatrix}$
• B. $\begin{bmatrix}3&-1&1\\ -15&6&-5\\ 5&-2&2\end{bmatrix}$
• C. $\begin{bmatrix}3&-15&5\\ -1&6&-2\\ 1&-5&-2\end{bmatrix}$
• D. $\begin{bmatrix}3&-5&5\\ -1&-6&-2\\ 1&-5&2\end{bmatrix}$

Answer 1

Answer: (C)

$Let A=\left[\begin{matrix}2&5&0\\ 0&1&1\\ -1&0&3\end{matrix}\right]$
$\left|A\right|=2\left(3-0\right)-5\left(0+1\right)$
$ = 6 - 5 = 1$
$\therefore \, \left|A\right|=1$
$adj \, A=\left[\begin{matrix}3&-1&1\\ -15&6&-5\\ 5&-2&2\end{matrix}\right]^{T}=\left[\begin{matrix}3&-15&5\\ -1&6&-2\\ 1&-5&2\end{matrix}\right]$
$A^{-1}=\frac{adj\, A}{\left|A\right|}=\left[\begin{matrix}3&-15&5\\ -1&6&-2\\ 1&-5&2\end{matrix}\right]$