Use a matrix approach to solve the system. $$\left( \begin{array} { c c c c }
x - 4 y & = & 11 \
4 x + 3 y & = & - 32
\end{array} \right)$$
A) (-5, -2)
B) (-4, -4)
C) (-5, -4)
D) (-4, -2)
E) inconsistent

Question

Use a matrix approach to solve the system. $$\left( \begin{array} { c c c c } 
x - 4 y & = & 11 \
4 x + 3 y & = & - 32
\end{array} \right)$$
A) (-5, -2)
B) (-4, -4)
C) (-5, -4)
D) (-4, -2)
E) inconsistent

Quizplus · Accepted Answer

To solve the system using a matrix approach, we first write the system of equations 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:$$A = \left( \begin{array}{cc} 1 & -4 \ 4 & 3 \end{array} \right), X = \left( \begin{array}{c} x \ y \end{array} \right), B = \left( \begin{array}{c} 11 \ -32 \end{array} \right)$$To find X, we calculate $A^{-1}B$. The inverse of A, $A^{-1}$, is found using the formula for a 2x2 matrix:$$A^{-1} = \frac{1}{\text{det}(A)} \left( \begin{array}{cc} d & -b \ -c & a \end{array} \right)$$where $a, b, c, d$ are the elements of A, so in this case:$$A^{-1} = \frac{1}{(1)(3) - (-4)(4)} \left( \begin{array}{cc} 3 & 4 \ -4 & 1 \end{array} \right) = \frac{1}{19} \left( \begin{array}{cc} 3 & 4 \ -4 & 1 \end{array} \right)$$Multiplying $A^{-1}$ by B gives:$$X = A^{-1}B = \frac{1}{19} \left( \begin{array}{cc} 3 & 4 \ -4 & 1 \end{array} \right) \left( \begin{array}{c} 11 \ -32 \end{array} \right) = \left( \begin{array}{c} -5 \ -4 \end{array} \right)$$Therefore, the solution to the system is $x = -5, y = -4$, which corresponds to choice C.

Use a Matrix Approach to Solve the System (x−4y=114x+3y=−32)\left( \begin{array} { c c c c } x - 4 y & = & 11 \\4 x + 3 y & = & - 32\end{array} \right)(x−4y4x+3y​==​11−32​)

Use a Matrix Approach to Solve the System $\left( \begin{array} { c c c c } x - 4 y & = & 11 \\4 x + 3 y & = & - 32\end{array} \right)$