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:
Answer:
The correct option is 4.
Step-by-step explanation:
The given matrix equation is
[tex]\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}[/tex]
Let [tex]A=\begin{bmatrix}\frac{1}{2} & \frac{-1}{4}\\ 2 & -\frac{3}{4}\end{bmatrix}[/tex], [tex]X=\begin{bmatrix}x\\ y\end{bmatrix}[/tex] and [tex]B=\begin{bmatrix}2 & -4\\ 1 & 6\end{bmatrix}[/tex]
The given equation can be written as
[tex]AX=B[/tex]
If [tex]AX=B[/tex], then [tex]X=A^{-1}B[/tex]
It means we have to find the matrix A⁻¹ .
[tex]A=\begin{bmatrix}\frac{1}{2} & \frac{-1}{4}\\ 2 & -\frac{3}{4}\end{bmatrix}[/tex]
[tex]|A|=\frac{1}{2}\times \frac{-3}{4}-\frac{-1}{4}\times 2=\frac{-3}[8}+\frac{1}{2}=\frac{1}{8}[/tex]
If [tex]P=\begin{bmatrix}a & b\\ c & d\end{bmatrix}[/tex], then [tex]P^{-1}=\frac{1}{|P|}\begin{bmatrix}d &-b\\ -c & a\end{bmatrix}[/tex]
[tex]A^{-1}=\frac{1}{\frac{1}{8}}\begin{bmatrix}-\frac{3}{4}& \frac{1}{4}\\ -2 &\frac{1}{2}\end{bmatrix}[/tex]
[tex]A^{-1}=8\begin{bmatrix}-\frac{3}{4}& \frac{1}{4}\\ -2 &\frac{1}{2}\end{bmatrix}[/tex]
[tex]A^{-1}=\begin{bmatrix}-6& 2\\ -16 &4\end{bmatrix}[/tex]
[tex]X=A^{-1}B[/tex]
[tex]\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-6& 2\\ -16 &4\end{bmatrix}\begin{bmatrix}2 & -4\\ 1 & 6\end{bmatrix}[/tex]
Therefore the correct option is 4.
