Question

Which equation can be used to solve the matrix equation?

Answer 1

Answer:

The correct option is 4.

Step-by-step explanation:

The given matrix equation is

$\begin{bmatrix}\frac{1}{2} & \frac{-1}{4}\\ 2 & -\frac{3}{4}\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}2 & -4\\ 1 & 6\end{bmatrix}$

Let $A=\begin{bmatrix}\frac{1}{2} & \frac{-1}{4}\\ 2 & -\frac{3}{4}\end{bmatrix}$, $X=\begin{bmatrix}x\\ y\end{bmatrix}$ and $B=\begin{bmatrix}2 & -4\\ 1 & 6\end{bmatrix}$

The given equation can be written as

$AX=B$

If $AX=B$, then $X=A^{-1}B$

It means we have to find the matrix A⁻¹ .

$A=\begin{bmatrix}\frac{1}{2} & \frac{-1}{4}\\ 2 & -\frac{3}{4}\end{bmatrix}$

$|A|=\frac{1}{2}\times \frac{-3}{4}-\frac{-1}{4}\times 2=\frac{-3}[8}+\frac{1}{2}=\frac{1}{8}$

If $P=\begin{bmatrix}a & b\\ c & d\end{bmatrix}$, then $P^{-1}=\frac{1}{|P|}\begin{bmatrix}d &-b\\ -c & a\end{bmatrix}$

$A^{-1}=\frac{1}{\frac{1}{8}}\begin{bmatrix}-\frac{3}{4}& \frac{1}{4}\\ -2 &\frac{1}{2}\end{bmatrix}$

$A^{-1}=8\begin{bmatrix}-\frac{3}{4}& \frac{1}{4}\\ -2 &\frac{1}{2}\end{bmatrix}$

$A^{-1}=\begin{bmatrix}-6& 2\\ -16 &4\end{bmatrix}$

$X=A^{-1}B$

$\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-6& 2\\ -16 &4\end{bmatrix}\begin{bmatrix}2 & -4\\ 1 & 6\end{bmatrix}$

Therefore the correct option is 4.

Answer 2

Answer:

A matrix equation is an equation in which a variable stands for a matrix . You can solve the simpler matrix equations using matrix addition and scalar multiplication .

Step-by-step explanation: