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Question

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Solve the following system of equations using an inverse matrix. You must also indicate the inverse matrix,

A^{-1}

, that was used to solve the system. You may optionally write the inverse matrix with a scalar coefficient.

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\begin{array}{l} 7 x+8 y=3 \\ 4 x+6 y=-2 \end{array}

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Answer anempticet do

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A^{-1}=\square[\square] \quad x=\square=\square

Write System of Equations: First, let's write down the system of equations in matrix form, $A \times X = B$ , where $A$ is the coefficient matrix, $X$ is the vector of variables, and $B$ is the constant vector. $\newline$ $A = \left[\begin{array}{cc} 7 & 8 \ 4 & 6 \end{array}\right]$ $\newline$ $B = \left[\begin{array}{c} 3 \ -2 \end{array}\right]$ $\newline$ $X = \left[\begin{array}{c} x \ y \end{array}\right]$
Calculate Determinant: Next, calculate the determinant of matrix $A$ , $\text{det}(A)$ , to ensure it's invertible (non-zero determinant). $\newline$ $\text{det}(A) = 7 \times 6 - 8 \times 4 = 42 - 32 = 10$
Find Inverse of Matrix: Since the determinant is non-zero, $A$ is invertible. Now, calculate the inverse of matrix $A$ , $A^{-1}$ . $A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A)$ $\text{adj}(A) = \left[\begin{array}{cc}6 & -8 \ -4 & 7\end{array}\right]$ (adjugate of $A$ ) $A^{-1} = \frac{1}{10} \times \left[\begin{array}{cc}6 & -8 \ -4 & 7\end{array}\right]$ $A^{-1} = \left[\begin{array}{cc}0.6 & -0.8 \ -0.4 & 0.7\end{array}\right]$
Multiply Inverse by Constant: Multiply the inverse matrix $A^{-1}$ by the constant vector $B$ to find the values of $x$ and $y$ .
$X = A^{-1} \times B$
$X = \left[\begin{array}{cc} 0.6 & -0.8 \ -0.4 & 0.7 \end{array}\right] \times \left[\begin{array}{c} 3 \ -2 \end{array}\right]$
$X = \left[\begin{array}{c} 0.6\times3 + (-0.8)\times(-2) \ -0.4\times3 + 0.7\times(-2) \end{array}\right]$
$X = \left[\begin{array}{c} 1.8 + 1.6 \ -1.2 - 1.4 \end{array}\right]$
$X = \left[\begin{array}{c} 3.4 \ -2.6 \end{array}\right]$