Solve the system by using the inverse of the coefficient matrix.
-$$\begin{array} { l }
4 x - 5 y = - 11 \
- 2 x + 3 y = 5
\end{array}$$
A) $\{ ( 4,1 ) \}$
B) $\{ ( - 1 , - 4 ) \}$
C) $\{ ( - 4 , - 1 ) \}$
D) $\{ ( 1,4 ) \}$

Question

Solve the system by using the inverse of the coefficient matrix.
-$$\begin{array} { l } 
4 x - 5 y = - 11 \
- 2 x + 3 y = 5
\end{array}$$
A) $\{ ( 4,1 ) \}$
B) $\{ ( - 1 , - 4 ) \}$
C) $\{ ( - 4 , - 1 ) \}$
D) $\{ ( 1,4 ) \}$

Quizplus · Accepted Answer

To solve the system using the inverse of the coefficient matrix, we first write the system in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the column matrix of variables, and $B$ is the column matrix of constants. For this system, we have:$$A = \begin{pmatrix} 4 & -5 \ -2 & 3 \end{pmatrix}, \quad X = \begin{pmatrix} x \ y \end{pmatrix}, \quad B = \begin{pmatrix} -11 \ 5 \end{pmatrix}$$To find $X$, we compute $X = A^{-1}B$. First, we find the inverse of $A$, $A^{-1}$:$$A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} 3 & 5 \ 2 & 4 \end{pmatrix}$$The determinant of $A$, $\text{det}(A)$, is $4(3) - (-5)(-2) = 12 - 10 = 2$. Thus,$$A^{-1} = \frac{1}{2} \begin{pmatrix} 3 & 5 \ 2 & 4 \end{pmatrix} = \begin{pmatrix} 1.5 & 2.5 \ 1 & 2 \end{pmatrix}$$Multiplying $A^{-1}$ by $B$, we get:$$X = \begin{pmatrix} 1.5 & 2.5 \ 1 & 2 \end{pmatrix} \begin{pmatrix} -11 \ 5 \end{pmatrix} = \begin{pmatrix} 1.5(-11) + 2.5(5) \ 1(-11) + 2(5) \end{pmatrix} = \begin{pmatrix} -16.5 + 12.5 \ -11 + 10 \end{pmatrix} = \begin{pmatrix} -4 \ -1 \end{pmatrix}$$Therefore, the solution is $x = -4$ and $y = -1$, which corresponds to choice C.

Solve the System by Using the Inverse of the Coefficient 4x−5y=−11−2x+3y=5\begin{array} { l } 4 x - 5 y = - 11 \\- 2 x + 3 y = 5\end{array}4x−5y=−11−2x+3y=5​

Solve the System by Using the Inverse of the Coefficient $\begin{array} { l } 4 x - 5 y = - 11 \\- 2 x + 3 y = 5\end{array}$