Solve the Matrix Equation [[3,7],[5,12]]x=[[3,-1],[4,3]]

$[\begin{array}{c} 3 & 7 \\ 5 & 12 \end{array}] x = [\begin{array}{c} 3 & - 1 \\ 4 & 3 \end{array}]$

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 $[\begin{array}{c} 3 & 7 \\ 5 & 12 \end{array}]$ .

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

$determinant [\begin{array}{c} 3 & 7 \\ 5 & 12 \end{array}] = | \begin{array}{c} 3 & 7 \\ 5 & 12 \end{array} |$

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

$(3) (12) - 5 \cdot 7$

Simplify the determinant.

Simplify each term.

Multiply $3$ by $12$ .

$36 - 5 \cdot 7$

Multiply $- 5$ by $7$ .

$36 - 35$

Subtract $35$ from $36$ .

$1$

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

$\frac{1}{1} [\begin{array}{c} 12 & - (7) \\ - (5) & 3 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (7)$ .

$\frac{1}{1} [\begin{array}{c} 12 & - 7 \\ - (5) & 3 \end{array}]$

Rearrange $- (5)$ .

$\frac{1}{1} [\begin{array}{c} 12 & - 7 \\ - 5 & 3 \end{array}]$

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

$[\begin{array}{c} \frac{1}{1} \cdot 12 & \frac{1}{1} \cdot - 7 \\ \frac{1}{1} \cdot - 5 & \frac{1}{1} \cdot 3 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{1} \cdot 12$ .

$[\begin{array}{c} 12 & \frac{1}{1} \cdot - 7 \\ \frac{1}{1} \cdot - 5 & \frac{1}{1} \cdot 3 \end{array}]$

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

$[\begin{array}{c} 12 & - 7 \\ \frac{1}{1} \cdot - 5 & \frac{1}{1} \cdot 3 \end{array}]$

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

$[\begin{array}{c} 12 & - 7 \\ - 5 & \frac{1}{1} \cdot 3 \end{array}]$

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

$[\begin{array}{c} 12 & - 7 \\ - 5 & 3 \end{array}]$

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

$[\begin{array}{c} 12 & - 7 \\ - 5 & 3 \end{array}] \cdot ([\begin{array}{c} 3 & 7 \\ 5 & 12 \end{array}] x) = [\begin{array}{c} 12 & - 7 \\ - 5 & 3 \end{array}] \cdot [\begin{array}{c} 3 & - 1 \\ 4 & 3 \end{array}]$

Simplify the left side of the equation.

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

$[\begin{array}{c} 12 \cdot 3 - 7 \cdot 5 & 12 \cdot 7 - 7 \cdot 12 \\ - 5 \cdot 3 + 3 \cdot 5 & - 5 \cdot 7 + 3 \cdot 12 \end{array}] x = [\begin{array}{c} 12 & - 7 \\ - 5 & 3 \end{array}] \cdot [\begin{array}{c} 3 & - 1 \\ 4 & 3 \end{array}]$

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

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} 12 & - 7 \\ - 5 & 3 \end{array}] \cdot [\begin{array}{c} 3 & - 1 \\ 4 & 3 \end{array}]$

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

$x = [\begin{array}{c} 12 & - 7 \\ - 5 & 3 \end{array}] \cdot [\begin{array}{c} 3 & - 1 \\ 4 & 3 \end{array}]$

Simplify the right side of the equation.

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

$x = [\begin{array}{c} 12 \cdot 3 - 7 \cdot 4 & 12 \cdot - 1 - 7 \cdot 3 \\ - 5 \cdot 3 + 3 \cdot 4 & - 5 \cdot - 1 + 3 \cdot 3 \end{array}]$

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

$x = [\begin{array}{c} 8 & - 33 \\ - 3 & 14 \end{array}]$

The matrix is in the most simplified form.

$x = [\begin{array}{c} 8 & - 33 \\ - 3 & 14 \end{array}]$

Solve the Matrix Equation [[3,7],[5,12]]x=[[3,-1],[4,3]]

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