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Sagot :
To solve for [tex]\( y \)[/tex] using Cramer's Rule, follow these steps:
1. Write the system of equations:
[tex]\[ \begin{cases} 9x - 2y = 5 \\ -3x - 4y = -4 \end{cases} \][/tex]
2. Determine the coefficient matrix [tex]\( A \)[/tex] and the constants matrix:
[tex]\[ A = \begin{pmatrix} 9 & -2 \\ -3 & -4 \end{pmatrix} ,\quad B = \begin{pmatrix} 5 \\ -4 \end{pmatrix} \][/tex]
3. Calculate the determinant [tex]\( D \)[/tex] of matrix [tex]\( A \)[/tex]:
[tex]\[ D = \begin{vmatrix} 9 & -2 \\ -3 & -4 \end{vmatrix} = (9)(-4) - (-2)(-3) = -36 - 6 = -42 \][/tex]
4. Calculate the determinant [tex]\( D_y \)[/tex] for the [tex]\( y \)[/tex]-component by replacing the [tex]\( y \)[/tex]-column in [tex]\( A \)[/tex] with the constants matrix [tex]\( B \)[/tex]:
[tex]\[ D_y = \begin{vmatrix} 9 & 5 \\ -3 & -4 \end{vmatrix} = (9)(-4) - (5)(-3) = -36 + 15 = -21 \][/tex]
5. Calculate [tex]\( y \)[/tex] using Cramer's Rule:
[tex]\[ y = \frac{D_y}{D} = \frac{-21}{-42} = \frac{1}{2} \][/tex]
Therefore, the value of [tex]\( y \)[/tex] in the solution to the system of linear equations is [tex]\( \frac{1}{2} \)[/tex].
1. Write the system of equations:
[tex]\[ \begin{cases} 9x - 2y = 5 \\ -3x - 4y = -4 \end{cases} \][/tex]
2. Determine the coefficient matrix [tex]\( A \)[/tex] and the constants matrix:
[tex]\[ A = \begin{pmatrix} 9 & -2 \\ -3 & -4 \end{pmatrix} ,\quad B = \begin{pmatrix} 5 \\ -4 \end{pmatrix} \][/tex]
3. Calculate the determinant [tex]\( D \)[/tex] of matrix [tex]\( A \)[/tex]:
[tex]\[ D = \begin{vmatrix} 9 & -2 \\ -3 & -4 \end{vmatrix} = (9)(-4) - (-2)(-3) = -36 - 6 = -42 \][/tex]
4. Calculate the determinant [tex]\( D_y \)[/tex] for the [tex]\( y \)[/tex]-component by replacing the [tex]\( y \)[/tex]-column in [tex]\( A \)[/tex] with the constants matrix [tex]\( B \)[/tex]:
[tex]\[ D_y = \begin{vmatrix} 9 & 5 \\ -3 & -4 \end{vmatrix} = (9)(-4) - (5)(-3) = -36 + 15 = -21 \][/tex]
5. Calculate [tex]\( y \)[/tex] using Cramer's Rule:
[tex]\[ y = \frac{D_y}{D} = \frac{-21}{-42} = \frac{1}{2} \][/tex]
Therefore, the value of [tex]\( y \)[/tex] in the solution to the system of linear equations is [tex]\( \frac{1}{2} \)[/tex].
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