Question

# Use an inverse matrix to solve each system of equations, if possible.\begin{aligned} &3 x-6 y=9\\ &-5 x-8 y=-6 \end{aligned}

Solution

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$\textbf{\color{#4257b2}Write the system in matrix form AX=B.}$

\begin{align*} \left[ \begin{array}{rr} 3 & -6 \\ -5 & -8 \end{array} \right] \cdot \left[ \begin{array}{r} x \\ y \end{array} \right] = \left[ \begin{array}{r} 9\\ -6 \end{array} \right] \end{align*}

$\textbf{\color{#4257b2}Use the formula for the inverse of a 2\times 2 matrix to find the inverse A^{-1}.}$

\begin{align*} A^{-1} &= \dfrac{1}{ad-cb}\left[\begin{array}{ rr} d & -b\\ -c & a \end{array}\right]\\ &= \dfrac{1}{3(-8)-(-5)(-6)}\left[\begin{array}{ rr} -8 & -(-6)\\ -(-5) & 3 \end{array}\right]\\ &= \dfrac{1}{-54}\left[\begin{array}{ rr} -8 & 6\\ 5 & 3 \end{array}\right]\\ &= \left[\begin{array}{ rr} \frac{4}{27} & -\frac{1}{9}\\ -\frac{5}{54} & -\frac{1}{18} \end{array}\right] \end{align*}

$\textbf{\color{#4257b2}Multiply A^{-1} by B to solve the system.}$

\begin{align*} X&=A^{-1}B\\ &= \left[\begin{array}{ rr} \frac{4}{27} & -\frac{1}{9}\\ -\frac{5}{54} & -\frac{1}{18} \end{array}\right] \left[\begin{array}{r} 9\\ -6 \end{array}\right]\\ &= \left[\begin{array}{r} \frac{4}{27}(9)+(-\frac{1}{9})\left(-6\right) \\ -\frac{5}{54}(9)+(-\frac{1}{18})\left(-6\right) \end{array}\right]\\ &= \left[\begin{array}{r} 2 \\ -\frac{1}{2} \end{array}\right] \end{align*}

So, the solution of the system is:

$\color{#c34632}\left(2,-\dfrac{1}{2}\right)\color{white}\tag{1}$

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