Which equation can be used to solve the matrix equation

[tex]\[ \left[\begin{array}{rr}\frac{1}{2} & -\frac{1}{4} \\ 2 & -\frac{3}{4}\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{ll}2 & -4 \\ 1 & 6\end{array}\right] ? \][/tex]

A. [tex]\( \left[\begin{array}{rr}\frac{1}{2} & -\frac{1}{4} \\ 2 & -\frac{3}{4}\end{array}\right] \left[\begin{array}{l}x \\ y\end{array}\right] = \left[\begin{array}{l} 2 \\ 1 \end{array}\right] \)[/tex]

B. [tex]\( \left[\begin{array}{rr}\frac{1}{2} & -\frac{1}{4} \\ 2 & -\frac{3}{4}\end{array}\right] \left[\begin{array}{l}x \\ y\end{array}\right] = \left[\begin{array}{l} -4 \\ 6 \end{array}\right] \)[/tex]

C. [tex]\( \left[\begin{array}{rr}\frac{1}{2} & -\frac{1}{4} \\ 2 & -\frac{3}{4}\end{array}\right] \left[\begin{array}{l}x \\ y\end{array}\right] = \left[\begin{array}{l} 2 \\ -4 \end{array}\right] \)[/tex]

D. [tex]\( \left[\begin{array}{rr}\frac{1}{2} & -\frac{1}{4} \\ 2 & -\frac{3}{4}\end{array}\right] \left[\begin{array}{l}x \\ y\end{array}\right] = \left[\begin{array}{l} -4 \\ 1 \end{array}\right] \)[/tex]



Answer :

To solve the matrix equation

[tex]\[ \begin{pmatrix} \frac{1}{2} & -\frac{1}{4} \\ 2 & -\frac{3}{4} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 & -4 \\ 1 & 6 \end{pmatrix}, \][/tex]

we need to find the inverse of the matrix

[tex]\[ A = \begin{pmatrix} \frac{1}{2} & -\frac{1}{4} \\ 2 & -\frac{3}{4} \end{pmatrix}. \][/tex]

Let's denote the matrix on the right-hand side as [tex]\( B \)[/tex]:

[tex]\[ B = \begin{pmatrix} 2 & -4 \\ 1 & 6 \end{pmatrix}. \][/tex]

### Step 1: Calculate the inverse of matrix [tex]\(A\)[/tex]

Given [tex]\( A = \begin{pmatrix} \frac{1}{2} & -\frac{1}{4} \\ 2 & -\frac{3}{4} \end{pmatrix} \)[/tex],

the inverse of [tex]\( A \)[/tex] is

[tex]\[ A^{-1} = \begin{pmatrix} -6 & 2 \\ -16 & 4 \end{pmatrix}. \][/tex]

### Step 2: Find the matrix solution [tex]\(X\)[/tex]

To solve for [tex]\(X\)[/tex], we use the equation:

[tex]\[ A^{-1} A \begin{pmatrix} x \\ y \end{pmatrix} = A^{-1} B. \][/tex]

Since [tex]\( A^{-1} A = I \)[/tex] (where [tex]\( I \)[/tex] is the identity matrix), the left-hand side simplifies to

[tex]\[ I \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}. \][/tex]

Therefore,

[tex]\[ \begin{pmatrix} x \\ y \end{pmatrix} = A^{-1} B. \][/tex]

### Step 3: Calculate [tex]\( A^{-1} B \)[/tex]

Multiplying the inverse of [tex]\( A \)[/tex] by [tex]\( B \)[/tex]:

[tex]\[ A^{-1} B = \begin{pmatrix} -6 & 2 \\ -16 & 4 \end{pmatrix} \begin{pmatrix} 2 & -4 \\ 1 & 6 \end{pmatrix} = \begin{pmatrix} -10 & 36 \\ -28 & 88 \end{pmatrix}. \][/tex]

So the matrix solution [tex]\(X\)[/tex] is

[tex]\[ X = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -10 & 36 \\ -28 & 88 \end{pmatrix}. \][/tex]

Thus, we can summarize our solution to the matrix equation as

[tex]\[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -10 & 36 \\ -28 & 88 \end{pmatrix}. \][/tex]