Solution:
### Part (a)
To determine the value of [tex]\( y \)[/tex] when the given matrix [tex]\( A \)[/tex] is singular, we first need to find the determinant of the matrix [tex]\( A \)[/tex]. The matrix [tex]\( A \)[/tex] is given by:
[tex]\[
A = \begin{pmatrix}
3y - 1 & y + 1 \\
2 & 3
\end{pmatrix}
\][/tex]
For a matrix to be singular, its determinant must be zero. The determinant [tex]\(|A|\)[/tex] of a 2x2 matrix [tex]\(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\)[/tex] is calculated as:
[tex]\[
|A| = ad - bc
\][/tex]
Substituting the corresponding values from matrix [tex]\( A \)[/tex]:
[tex]\[
a = 3y - 1, \quad b = y + 1, \quad c = 2, \quad d = 3
\][/tex]
So, the determinant is:
[tex]\[
|A| = (3y - 1) \cdot 3 - (y + 1) \cdot 2
\][/tex]
Calculating this expression:
[tex]\[
|A| = 3(3y - 1) - 2(y + 1)
\][/tex]
[tex]\[
= 9y - 3 - 2y - 2
\][/tex]
[tex]\[
= 7y - 5
\][/tex]
For the matrix to be singular, the determinant must be 0:
[tex]\[
7y - 5 = 0
\][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[
7y = 5
\][/tex]
[tex]\[
y = \frac{5}{7}
\][/tex]
[tex]\[
y \approx 0.7142857142857143
\][/tex]
### Part (b)
To solve the given system of linear equations using the matrix method, the system of equations is:
[tex]\[
\begin{cases}
2x + y = 7 \\
3x + 5y = 20
\end{cases}
\][/tex]
First, we write the system in matrix form [tex]\( AX = B \)[/tex]:
[tex]\[
A = \begin{pmatrix}
2 & 1 \\
3 & 5
\end{pmatrix}, \quad
X = \begin{pmatrix}
x \\
y
\end{pmatrix}, \quad
B = \begin{pmatrix}
7 \\
20
\end{pmatrix}
\][/tex]
The solution to [tex]\( AX = B \)[/tex] is found by [tex]\( X = A^{-1}B \)[/tex], where [tex]\( A^{-1} \)[/tex] is the inverse of the matrix [tex]\( A \)[/tex].
Calculating the inverse of the matrix [tex]\( A \)[/tex]:
[tex]\[
A^{-1} = \frac{1}{|A|} \left( \begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix} \right)
\][/tex]
For the matrix [tex]\( A \)[/tex]:
[tex]\[
a = 2, \quad b = 1, \quad c = 3, \quad d = 5
\][/tex]
The determinant [tex]\(|A|\)[/tex] is:
[tex]\[
|A| = ad - bc = (2)(5) - (1)(3) = 10 - 3 = 7
\][/tex]
Thus, the inverse [tex]\( A^{-1} \)[/tex] is:
[tex]\[
A^{-1} = \frac{1}{7} \begin{pmatrix}
5 & -1 \\
-3 & 2
\end{pmatrix}
\][/tex]
So,
[tex]\[
X = A^{-1} B = \frac{1}{7} \begin{pmatrix}
5 & -1 \\
-3 & 2
\end{pmatrix} \begin{pmatrix}
7 \\
20
\end{pmatrix}
\][/tex]
Now, perform the matrix multiplication:
[tex]\[
X = \frac{1}{7} \begin{pmatrix}
(5 \cdot 7) + (-1 \cdot 20) \\
(-3 \cdot 7) + (2 \cdot 20)
\end{pmatrix}
\][/tex]
[tex]\[
= \frac{1}{7} \begin{pmatrix}
35 - 20 \\
-21 + 40
\end{pmatrix}
\][/tex]
[tex]\[
= \frac{1}{7} \begin{pmatrix}
15 \\
19
\end{pmatrix}
\][/tex]
[tex]\[
= \begin{pmatrix}
\frac{15}{7} \\
\frac{19}{7}
\end{pmatrix}
\][/tex]
[tex]\[
\approx \begin{pmatrix}
2.14285714 \\
2.71428571
\end{pmatrix}
\][/tex]
So, the solution [tex]\( (x, y) \)[/tex] is approximately:
[tex]\[
x \approx 2.14285714, \quad y \approx 2.71428571
\][/tex]
