Solve the Matrix Equation [[-2,-5],[1,3]][[x],[y]]=[[2],[-3]]

$[\begin{array}{c} - 2 & - 5 \\ 1 & 3 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 2 \\ - 3 \end{array}]$

Step 1

Find the inverse matrix.

The inverse of a $2 \times 2$ matrix can be found using the formula $\frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where $| A |$ is the determinant of $A$ .

If $A = [\begin{array}{c} a & b \\ c & d \end{array}]$ then $A^{- 1} = \frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$

Find the determinant of $[\begin{array}{c} - 2 & - 5 \\ 1 & 3 \end{array}]$ .

These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} - 2 & - 5 \\ 1 & 3 \end{array}] = | \begin{array}{c} - 2 & - 5 \\ 1 & 3 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$(- 2) (3) - 1 \cdot - 5$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $3$ .

$- 6 - 1 \cdot - 5$

Multiply $- 1$ by $- 5$ .

$- 6 + 5$

Add $- 6$ and $5$ .

$- 1$

Substitute the known values into the formula for the inverse of a matrix.

$\frac{1}{- 1} [\begin{array}{c} 3 & - (- 5) \\ - (1) & - 2 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (- 5)$ .

$\frac{1}{- 1} [\begin{array}{c} 3 & 5 \\ - (1) & - 2 \end{array}]$

Rearrange $- (1)$ .

$\frac{1}{- 1} [\begin{array}{c} 3 & 5 \\ - 1 & - 2 \end{array}]$

Multiply $\frac{1}{- 1}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{- 1} \cdot 3 & \frac{1}{- 1} \cdot 5 \\ \frac{1}{- 1} \cdot - 1 & \frac{1}{- 1} \cdot - 2 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{- 1} \cdot 3$ .

$[\begin{array}{c} - 3 & \frac{1}{- 1} \cdot 5 \\ \frac{1}{- 1} \cdot - 1 & \frac{1}{- 1} \cdot - 2 \end{array}]$

Rearrange $\frac{1}{- 1} \cdot 5$ .

$[\begin{array}{c} - 3 & - 5 \\ \frac{1}{- 1} \cdot - 1 & \frac{1}{- 1} \cdot - 2 \end{array}]$

Rearrange $\frac{1}{- 1} \cdot - 1$ .

$[\begin{array}{c} - 3 & - 5 \\ 1 & \frac{1}{- 1} \cdot - 2 \end{array}]$

Rearrange $\frac{1}{- 1} \cdot - 2$ .

$[\begin{array}{c} - 3 & - 5 \\ 1 & 2 \end{array}]$

Step 2

Assuming that $X$ is the matrix to solve for, multiply the inverse matrix by both sides of the equation.

$[\begin{array}{c} - 3 & - 5 \\ 1 & 2 \end{array}] \cdot ([\begin{array}{c} - 2 & - 5 \\ 1 & 3 \end{array}] x) = [\begin{array}{c} - 3 & - 5 \\ 1 & 2 \end{array}] \cdot [\begin{array}{c} 2 \\ - 3 \end{array}]$

Step 3

Simplify the left side of the equation.

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} - 3 \cdot - 2 - 5 \cdot 1 & - 3 \cdot - 5 - 5 \cdot 3 \\ 1 \cdot - 2 + 2 \cdot 1 & 1 \cdot - 5 + 2 \cdot 3 \end{array}] x = [\begin{array}{c} - 3 & - 5 \\ 1 & 2 \end{array}] \cdot [\begin{array}{c} 2 \\ - 3 \end{array}]$

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} - 3 & - 5 \\ 1 & 2 \end{array}] \cdot [\begin{array}{c} 2 \\ - 3 \end{array}]$

Multiplying the identity matrix by any matrix $x$ is matrix $x$ .

$x = [\begin{array}{c} - 3 & - 5 \\ 1 & 2 \end{array}] \cdot [\begin{array}{c} 2 \\ - 3 \end{array}]$

Step 4

Simplify the right side of the equation.

Multiply each row in the first matrix by each column in the second matrix.

$x = [\begin{array}{c} - 3 \cdot 2 - 5 \cdot - 3 \\ 1 \cdot 2 + 2 \cdot - 3 \end{array}]$

Simplify each element of the matrix by multiplying out all the expressions.

$x = [\begin{array}{c} 9 \\ - 4 \end{array}]$

The matrix is in the most simplified form.

$x = [\begin{array}{c} 9 \\ - 4 \end{array}]$

Step 5

Solving for the variables in $[\begin{array}{c} x \\ y \end{array}]$ , the answer is $x = 9, y = - 4$ .

$x = 9, y = - 4$

Solve the Matrix Equation [[-2,-5],[1,3]][[x],[y]]=[[2],[-3]]

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