Solve the Matrix Equation [[9,-15],[-21,39]]=[[2,-3],[-7,9]]x

$[\begin{array}{c} 9 & - 15 \\ - 21 & 39 \end{array}] = [\begin{array}{c} 2 & - 3 \\ - 7 & 9 \end{array}] x$

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 $[\begin{array}{c} 2 & - 3 \\ - 7 & 9 \end{array}]$ .

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

$determinant [\begin{array}{c} 2 & - 3 \\ - 7 & 9 \end{array}] = | \begin{array}{c} 2 & - 3 \\ - 7 & 9 \end{array} |$

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

$(2) (9) + 7 \cdot - 3$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $9$ .

$18 + 7 \cdot - 3$

Multiply $7$ by $- 3$ .

$18 - 21$

Subtract $21$ from $18$ .

$- 3$

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

$\frac{1}{- 3} [\begin{array}{c} 9 & - (- 3) \\ - (- 7) & 2 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (- 3)$ .

$\frac{1}{- 3} [\begin{array}{c} 9 & 3 \\ - (- 7) & 2 \end{array}]$

Rearrange $- (- 7)$ .

$\frac{1}{- 3} [\begin{array}{c} 9 & 3 \\ 7 & 2 \end{array}]$

Multiply $\frac{1}{- 3}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{- 3} \cdot 9 & \frac{1}{- 3} \cdot 3 \\ \frac{1}{- 3} \cdot 7 & \frac{1}{- 3} \cdot 2 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{- 3} \cdot 9$ .

$[\begin{array}{c} - 3 & \frac{1}{- 3} \cdot 3 \\ \frac{1}{- 3} \cdot 7 & \frac{1}{- 3} \cdot 2 \end{array}]$

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

$[\begin{array}{c} - 3 & - 1 \\ \frac{1}{- 3} \cdot 7 & \frac{1}{- 3} \cdot 2 \end{array}]$

Rearrange $\frac{1}{- 3} \cdot 7$ .

$[\begin{array}{c} - 3 & - 1 \\ - \frac{7}{3} & \frac{1}{- 3} \cdot 2 \end{array}]$

Rearrange $\frac{1}{- 3} \cdot 2$ .

$[\begin{array}{c} - 3 & - 1 \\ - \frac{7}{3} & - \frac{2}{3} \end{array}]$

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

$[\begin{array}{c} - 3 & - 1 \\ - \frac{7}{3} & - \frac{2}{3} \end{array}] \cdot ([\begin{array}{c} 2 & - 3 \\ - 7 & 9 \end{array}] x) = [\begin{array}{c} - 3 & - 1 \\ - \frac{7}{3} & - \frac{2}{3} \end{array}] \cdot [\begin{array}{c} 9 & - 15 \\ - 21 & 39 \end{array}]$

Simplify the left side of the equation.

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

$[\begin{array}{c} - 3 \cdot 2 - 1 \cdot - 7 & - 3 \cdot - 3 - 1 \cdot 9 \\ - \frac{7}{3} \cdot 2 - \frac{2}{3} \cdot - 7 & - \frac{7}{3} \cdot - 3 - \frac{2}{3} \cdot 9 \end{array}] x = [\begin{array}{c} - 3 & - 1 \\ - \frac{7}{3} & - \frac{2}{3} \end{array}] \cdot [\begin{array}{c} 9 & - 15 \\ - 21 & 39 \end{array}]$

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

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} - 3 & - 1 \\ - \frac{7}{3} & - \frac{2}{3} \end{array}] \cdot [\begin{array}{c} 9 & - 15 \\ - 21 & 39 \end{array}]$

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

$x = [\begin{array}{c} - 3 & - 1 \\ - \frac{7}{3} & - \frac{2}{3} \end{array}] \cdot [\begin{array}{c} 9 & - 15 \\ - 21 & 39 \end{array}]$

Simplify the right side of the equation.

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

$x = [\begin{array}{c} - 3 \cdot 9 - 1 \cdot - 21 & - 3 \cdot - 15 - 1 \cdot 39 \\ - \frac{7}{3} \cdot 9 - \frac{2}{3} \cdot - 21 & - \frac{7}{3} \cdot - 15 - \frac{2}{3} \cdot 39 \end{array}]$

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

$x = [\begin{array}{c} - 6 & 6 \\ - 7 & 9 \end{array}]$

The matrix is in the most simplified form.

$x = [\begin{array}{c} - 6 & 6 \\ - 7 & 9 \end{array}]$

Solve the Matrix Equation [[9,-15],[-21,39]]=[[2,-3],[-7,9]]x

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