Solve Using an Inverse Matrix 5x+4y=8 , 6x-3y=33

$5 x + 4 y = 8$ , $6 x - 3 y = 33$

Step 1

Find the $A X = B$ from the system of equations.

$[\begin{array}{c} 5 & 4 \\ 6 & - 3 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 8 \\ 33 \end{array}]$

Step 2

Find the inverse of the coefficient 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 $[\begin{array}{c} 5 & 4 \\ 6 & - 3 \end{array}]$ .

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

$determinant [\begin{array}{c} 5 & 4 \\ 6 & - 3 \end{array}] = | \begin{array}{c} 5 & 4 \\ 6 & - 3 \end{array} |$

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

$(5) (- 3) - 6 \cdot 4$

Simplify the determinant.

Simplify each term.

Multiply $5$ by $- 3$ .

$- 15 - 6 \cdot 4$

Multiply $- 6$ by $4$ .

$- 15 - 24$

Subtract $24$ from $- 15$ .

$- 39$

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

$\frac{1}{- 39} [\begin{array}{c} - 3 & - (4) \\ - (6) & 5 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (4)$ .

$\frac{1}{- 39} [\begin{array}{c} - 3 & - 4 \\ - (6) & 5 \end{array}]$

Rearrange $- (6)$ .

$\frac{1}{- 39} [\begin{array}{c} - 3 & - 4 \\ - 6 & 5 \end{array}]$

Multiply $\frac{1}{- 39}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{- 39} \cdot - 3 & \frac{1}{- 39} \cdot - 4 \\ \frac{1}{- 39} \cdot - 6 & \frac{1}{- 39} \cdot 5 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{- 39} \cdot - 3$ .

$[\begin{array}{c} \frac{1}{13} & \frac{1}{- 39} \cdot - 4 \\ \frac{1}{- 39} \cdot - 6 & \frac{1}{- 39} \cdot 5 \end{array}]$

Rearrange $\frac{1}{- 39} \cdot - 4$ .

$[\begin{array}{c} \frac{1}{13} & \frac{4}{39} \\ \frac{1}{- 39} \cdot - 6 & \frac{1}{- 39} \cdot 5 \end{array}]$

Rearrange $\frac{1}{- 39} \cdot - 6$ .

$[\begin{array}{c} \frac{1}{13} & \frac{4}{39} \\ \frac{2}{13} & \frac{1}{- 39} \cdot 5 \end{array}]$

Rearrange $\frac{1}{- 39} \cdot 5$ .

$[\begin{array}{c} \frac{1}{13} & \frac{4}{39} \\ \frac{2}{13} & - \frac{5}{39} \end{array}]$

Step 3

Left multiply both sides of the matrix equation by the inverse matrix.

$([\begin{array}{c} \frac{1}{13} & \frac{4}{39} \\ \frac{2}{13} & - \frac{5}{39} \end{array}] \cdot [\begin{array}{c} 5 & 4 \\ 6 & - 3 \end{array}]) \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{13} & \frac{4}{39} \\ \frac{2}{13} & - \frac{5}{39} \end{array}] \cdot [\begin{array}{c} 8 \\ 33 \end{array}]$

Step 4

Any matrix multiplied by its inverse is equal to $1$ all the time. $A \cdot A^{- 1} = 1$ .

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{13} & \frac{4}{39} \\ \frac{2}{13} & - \frac{5}{39} \end{array}] \cdot [\begin{array}{c} 8 \\ 33 \end{array}]$

Step 5

Simplify the right side of the equation.

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

$[\begin{array}{c} \frac{1}{13} \cdot 8 + \frac{4}{39} \cdot 33 \\ \frac{2}{13} \cdot 8 - \frac{5}{39} \cdot 33 \end{array}]$

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

$[\begin{array}{c} 4 \\ - 3 \end{array}]$

Step 6

Simplify the left and right side.

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 4 \\ - 3 \end{array}]$

Step 7

Find the solution.

$x = 4$

$y = - 3$

Solve Using an Inverse Matrix 5x+4y=8 , 6x-3y=33

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