Question

select the correct answer.
consider matrix v.
$v = \

$$\begin{bmatrix}5 & 2 \\\\ 8 & 3\\end{bmatrix}$$

$
which equation shows $v^{-1}$?
\\(\boldsymbol{\circ}\\) a. $v^{-1} = -1 \

$$\begin{bmatrix}-5 & 8 \\\\ 2 & -3\\end{bmatrix}$$

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\\(\boldsymbol{\circ}\\) b. $v^{-1} = -1 \

$$\begin{bmatrix}3 & -2 \\\\ -8 & 5\\end{bmatrix}$$

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\\(\boldsymbol{\circ}\\) c. $v^{-1} = -1 \

$$\begin{bmatrix}5 & -8 \\\\ -2 & 3\\end{bmatrix}$$

$
\\(\boldsymbol{\circ}\\) d. $v^{-1} = -1 \

$$\begin{bmatrix}-3 & 2 \\\\ 8 & -5\\end{bmatrix}$$

$

Explanation:

Step1: Find determinant of V

For $V=

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

$, $\det(V)=ad-bc$
$\det(V)=(5)(3)-(2)(8)=15-16=-1$

Step2: Find adjugate of V

Swap diagonal, off-diagonal sign flip:

$$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\begin{bmatrix}3&-2\\-8&5\end{bmatrix}$$

Step3: Compute inverse matrix

$V^{-1}=\frac{1}{\det(V)}\times\text{adjugate}(V)$
$V^{-1}=\frac{1}{-1}

$$\begin{bmatrix}3&-2\\-8&5\end{bmatrix}$$

=-1

$$\begin{bmatrix}3&-2\\-8&5\end{bmatrix}$$

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Answer:

B. $V^{-1} = -1

$$\begin{bmatrix} 3 & -2 \\ -8 & 5 \end{bmatrix}$$

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