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Sagot :
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]
### 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]
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