Use a matrix approach to solve the system. $$\left( \begin{array} { c c c c }
5 x - 4 y & = & 56 \
x + 3 y & = & - 23
\end{array} \right)$$
A) (4, -9)
B) (6, -9)
C) (4, -8)
D) (6, -8)
E) inconsistent

Question

Use a matrix approach to solve the system. $$\left( \begin{array} { c c c c } 
5 x - 4 y & = & 56 \
x + 3 y & = & - 23
\end{array} \right)$$
A) (4, -9)
B) (6, -9)
C) (4, -8)
D) (6, -8)
E) inconsistent

Quizplus · Accepted Answer

To solve the system using a matrix approach, we first write the system of equations in matrix form as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. For the given system, we have:$$A = \left( \begin{array}{cc} 5 & -4 \ 1 & 3 \end{array} \right), \quad X = \left( \begin{array}{c} x \ y \end{array} \right), \quad B = \left( \begin{array}{c} 56 \ -23 \end{array} \right)$$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:$$A^{-1} = \frac{1}{(5)(3) - (-4)(1)} \left( \begin{array}{cc} 3 & 4 \ -1 & 5 \end{array} \right) = \frac{1}{15 + 4} \left( \begin{array}{cc} 3 & 4 \ -1 & 5 \end{array} \right) = \frac{1}{19} \left( \begin{array}{cc} 3 & 4 \ -1 & 5 \end{array} \right)$$Multiplying $A^{-1}$ by B gives:$$X = \frac{1}{19} \left( \begin{array}{cc} 3 & 4 \ -1 & 5 \end{array} \right) \left( \begin{array}{c} 56 \ -23 \end{array} \right) = \frac{1}{19} \left( \begin{array}{c} 3(56) + 4(-23) \ -1(56) + 5(-23) \end{array} \right) = \frac{1}{19} \left( \begin{array}{c} 168 - 92 \ -56 - 115 \end{array} \right) = \frac{1}{19} \left( \begin{array}{c} 76 \ -171 \end{array} \right) = \left( \begin{array}{c} 4 \ -9 \end{array} \right)$$Therefore, the solution to the system is $x = 4, y = -9$, which corresponds to choice A.

Use a Matrix Approach to Solve the System (5x−4y=56x+3y=−23)\left( \begin{array} { c c c c } 5 x - 4 y & = & 56 \\x + 3 y & = & - 23\end{array} \right)(5x−4yx+3y​==​56−23​)

Use a Matrix Approach to Solve the System $\left( \begin{array} { c c c c } 5 x - 4 y & = & 56 \\x + 3 y & = & - 23\end{array} \right)$