Question

I need this practice problem answered I will provide the answer options in another pic

Answer 1

The inverse of a matrix can be calculated as:

`\begin{gathered} \text{When} \\ A=\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & {} \\ {} & {} & \end{bmatrix} \\ \text{Then A\textasciicircum-1 is:} \\ A^(-1)=(1)/(ad-bc)\begin{bmatrix}{d} & -{b} & {} \\ {-c} & {a} & {} \\ {} & {} & \end{bmatrix} \end{gathered}`

Then, let's start by calculating the inverse of the given matrix:

`\begin{gathered} \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^(-1)=(1)/(4\cdot3-1\cdot(-2))\begin{bmatrix}{3} & -{1} & {} \\ {-(-2)} & {4} & {} \\ {} & {} & \end{bmatrix} \\ \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^(-1)=(1)/(14)\begin{bmatrix}{3} & -{1} & {} \\ {2} & {4} & {} \\ {} & {} & \end{bmatrix} \end{gathered}`

The problem says he multiplies the left side of the coefficient matrix by the inverse matrix, thus:

*These matrices will be the options to put on the first and second boxes.

Then: