Solve using matrices.
-$$\begin{array} { l }
5 x + 6 y = 20 \
2 x + 8 y = 36
\end{array}$$
A) $( - 2,5 )$
B) $( 5,2 )$
C) $( - 2 , - 5 )$
D) $( 5 , - 2 )$

Question

Solve using matrices.
-$$\begin{array} { l } 
5 x + 6 y = 20 \
2 x + 8 y = 36
\end{array}$$
A) $( - 2,5 )$
B) $( 5,2 )$
C) $( - 2 , - 5 )$
D) $( 5 , - 2 )$

Quizplus · Accepted Answer

To solve the system of equations using matrices, we can use the matrix equation $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix. For the given system:$$A = \begin{pmatrix} 5 & 6 \ 2 & 8 \end{pmatrix}, X = \begin{pmatrix} x \ y \end{pmatrix}, B = \begin{pmatrix} 20 \ 36 \end{pmatrix}$$To find $X$, we calculate $A^{-1}B$:$$A^{-1} = \frac{1}{(5*8 - 6*2)} \begin{pmatrix} 8 & -6 \ -2 & 5 \end{pmatrix} = \frac{1}{40 - 12} \begin{pmatrix} 8 & -6 \ -2 & 5 \end{pmatrix} = \frac{1}{28} \begin{pmatrix} 8 & -6 \ -2 & 5 \end{pmatrix}$$$$X = A^{-1}B = \frac{1}{28} \begin{pmatrix} 8 & -6 \ -2 & 5 \end{pmatrix} \begin{pmatrix} 20 \ 36 \end{pmatrix} = \frac{1}{28} \begin{pmatrix} 160 - 216 \ -40 + 180 \end{pmatrix} = \frac{1}{28} \begin{pmatrix} -56 \ 140 \end{pmatrix} = \begin{pmatrix} -2 \ 5 \end{pmatrix}$$Therefore, the solution to the system of equations is $(-2, 5)$.

Solve Using Matrices 5x+6y=202x+8y=36\begin{array} { l } 5 x + 6 y = 20 \\2 x + 8 y = 36\end{array}5x+6y=202x+8y=36​

Solve Using Matrices $\begin{array} { l } 5 x + 6 y = 20 \\2 x + 8 y = 36\end{array}$