Question 2 (Essay Worth 4 points)

The matrix equation represents a system of equations:

[tex]\[
\left[\begin{array}{cc}
2 & 9 \\
2 & 8
\end{array}\right]\left[\begin{array}{c}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
8 \\
8
\end{array}\right]
\][/tex]

Solve for [tex]\( y \)[/tex] using matrices. Show or explain all necessary steps.



Answer :

To solve for [tex]\( y \)[/tex] in the given system of equations using matrices, we need to follow these steps:

### Step 1: Write the system of equations in matrix form
Given the system of linear equations:

[tex]\[ \begin{cases} 2x + 9y = 8 \\ 2x + 8y = 8 \end{cases} \][/tex]

We can represent this system as a matrix equation of the form [tex]\( A \mathbf{x} = \mathbf{B} \)[/tex], where:

[tex]\[ A = \begin{pmatrix} 2 & 9 \\ 2 & 8 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 8 \\ 8 \end{pmatrix} \][/tex]

### Step 2: Invert the coefficient matrix [tex]\( A \)[/tex]
To solve for [tex]\( \mathbf{x} \)[/tex], we need to find the inverse of the matrix [tex]\( A \)[/tex], denoted [tex]\( A^{-1} \)[/tex]. The inverse of a matrix [tex]\( A \)[/tex] satisfies the equation [tex]\( A A^{-1} = I \)[/tex], where [tex]\( I \)[/tex] is the identity matrix.

The inverse of a matrix [tex]\( A \)[/tex], if it exists, can be found using several methods, such as Gaussian elimination, or using the formula for a 2x2 matrix:

[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \][/tex]

where:
- [tex]\(\text{det}(A)\)[/tex] is the determinant of [tex]\( A \)[/tex],
- [tex]\(\text{adj}(A)\)[/tex] is the adjugate of [tex]\( A \)[/tex].

First, calculate the determinant of [tex]\( A \)[/tex]:

[tex]\[ \text{det}(A) = (2)(8) - (2)(9) = 16 - 18 = -2 \][/tex]

Since the determinant is non-zero ([tex]\(\text{det}(A) = -2\)[/tex]), the matrix [tex]\( A \)[/tex] is invertible.

Next, find the adjugate of [tex]\( A \)[/tex]:

[tex]\[ \text{adj}(A) = \begin{pmatrix} 8 & -9 \\ -2 & 2 \end{pmatrix} \][/tex]

Now, calculate [tex]\( A^{-1} \)[/tex]:

[tex]\[ A^{-1} = \frac{1}{-2} \begin{pmatrix} 8 & -9 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} -4 & 4.5 \\ 1 & -1 \end{pmatrix} \][/tex]

### Step 3: Solve for [tex]\( \mathbf{x} \)[/tex]
We now have [tex]\( A^{-1} \)[/tex]. To find [tex]\( \mathbf{x} \)[/tex], we multiply both sides of the equation [tex]\( A \mathbf{x} = \mathbf{B} \)[/tex] by [tex]\( A^{-1} \)[/tex]:

[tex]\[ \mathbf{x} = A^{-1} \mathbf{B} \][/tex]

Substitute the values of [tex]\( A^{-1} \)[/tex] and [tex]\( \mathbf{B} \)[/tex]:

[tex]\[ \mathbf{x} = \begin{pmatrix} -4 & 4.5 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 8 \\ 8 \end{pmatrix} \][/tex]

Perform the matrix multiplication:

[tex]\[ \begin{pmatrix} -4 & 4.5 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 8 \\ 8 \end{pmatrix} = \begin{pmatrix} (-4 \cdot 8) + (4.5 \cdot 8) \\ (1 \cdot 8) + (-1 \cdot 8) \end{pmatrix} = \begin{pmatrix} -32 + 36 \\ 8 - 8 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} \][/tex]

Thus, we find:

[tex]\[ \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} \][/tex]

### Step 4: Extract the value of [tex]\( y \)[/tex]
From the solution vector [tex]\(\mathbf{x}\)[/tex], we see that [tex]\( x = 4 \)[/tex] and [tex]\( y = 0 \)[/tex].

Therefore, the value of [tex]\( y \)[/tex] is [tex]\( y = 0 \)[/tex].