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Question

Write the system of equations represented by the matrix equation below. Then solve matrix equation.

[3254][xy]=[13]\left[ \begin{array} { r r } { - 3 } & { - 2 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l } { x } \\ { y } \end{array} \right] = \left[ \begin{array} { r } { 1 } \\ { - 3 } \end{array} \right]

Solution

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Step 1
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The given matrix equation is in the form:

AX=BAX=B

where AA is the coefficient matrix, XX is the variable matrix, and BB is the constant matrix.

By multiplying AA and XX we have:

AX=[3(x)+(2)(y)5(x)+4(y)]=[3x2y5x+4y]AX= \left[\begin{array}{rr} -3(x)+(-2)(y) \\ 5(x)+4(y) \end{array}\right]= \left[\begin{array}{rr} -3x-2y\\ 5x+4y \end{array}\right]

So, the system of equations is:

{3x2y=15x+4y=3(1)\color{#c34632}{\begin{cases} -3x-2y=1\\ 5x+4y=-3 \end{cases}}\color{white}{\tag{1}}

The solution of the system is:

X=A1BX=A^{-1}B

Using a graphing calculator, we input the coefficient matrix on [A] and the constant matrix on [B]:

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