Solve the system by using the inverse of the coefficient matrix.
-$$\begin{array} { l }
2 x _ { 1 } - 6 x _ { 2 } = - 6 \
3 x _ { 1 } + 2 x _ { 2 } = 13
\end{array}$$
A) $( 2,3 )$
B) $( - 2 , - 3 )$
C) $( - 3 , - 2 )$
D) $( 3,2 )$

Question

Solve the system by using the inverse of the coefficient matrix.
-$$\begin{array} { l } 
2 x _ { 1 } - 6 x _ { 2 } = - 6 \
3 x _ { 1 } + 2 x _ { 2 } = 13
\end{array}$$
A) $( 2,3 )$
B) $( - 2 , - 3 )$
C) $( - 3 , - 2 )$
D) $( 3,2 )$

Quizplus · Accepted Answer

To solve the system using the inverse of the coefficient matrix, first, we write the system in matrix form $AX = B$, where $A$ is the coefficient matrix, $X$ is the column matrix of variables, and $B$ is the column matrix of constants. For the given system:$$A = \begin{pmatrix} 2 & -6 \ 3 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x_1 \ x_2 \end{pmatrix}, \quad B = \begin{pmatrix} -6 \ 13 \end{pmatrix}$$The solution is given by $X = A^{-1}B$. First, find the inverse of $A$, $A^{-1}$:$$A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} 2 & 6 \ -3 & 2 \end{pmatrix} = \frac{1}{(2*2 - (-6)*3)} \begin{pmatrix} 2 & 6 \ -3 & 2 \end{pmatrix} = \frac{1}{4 + 18} \begin{pmatrix} 2 & 6 \ -3 & 2 \end{pmatrix} = \frac{1}{22} \begin{pmatrix} 2 & 6 \ -3 & 2 \end{pmatrix}$$Then, multiply $A^{-1}$ by $B$:$$X = A^{-1}B = \frac{1}{22} \begin{pmatrix} 2 & 6 \ -3 & 2 \end{pmatrix} \begin{pmatrix} -6 \ 13 \end{pmatrix} = \frac{1}{22} \begin{pmatrix} (2*-6) + (6*13) \ (-3*-6) + (2*13) \end{pmatrix} = \frac{1}{22} \begin{pmatrix} -12 + 78 \ 18 + 26 \end{pmatrix} = \frac{1}{22} \begin{pmatrix} 66 \ 44 \end{pmatrix} = \begin{pmatrix} 3 \ 2 \end{pmatrix}$$Therefore, the solution is $X = (3, 2)$, which corresponds to choice D.

Solve the System by Using the Inverse of the Coefficient 2x1−6x2=−63x1+2x2=13\begin{array} { l } 2 x _ { 1 } - 6 x _ { 2 } = - 6 \\3 x _ { 1 } + 2 x _ { 2 } = 13\end{array}2x1​−6x2​=−63x1​+2x2​=13​

Solve the System by Using the Inverse of the Coefficient $\begin{array} { l } 2 x _ { 1 } - 6 x _ { 2 } = - 6 \\3 x _ { 1 } + 2 x _ { 2 } = 13\end{array}$