Linear Algebra Examples

$- 3 x - 4 y = 2$ ,

$8 y = - 6 x - 4$

Step 1

Find the

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

$[\begin{array}{c} - 3 & - 4 \\ 6 & 8 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 2 \\ - 4 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

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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

$A$ .

$A = [\begin{array}{c} a & b \\ c & d \end{array}]$ then

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

Find the determinant of

$[\begin{array}{c} - 3 & - 4 \\ 6 & 8 \end{array}]$ .

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

$determinant [\begin{array}{c} - 3 & - 4 \\ 6 & 8 \end{array}] = | \begin{array}{c} - 3 & - 4 \\ 6 & 8 \end{array} |$

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$(- 3) (8) - 6 \cdot - 4$

Simplify the determinant.

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Simplify each term.

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Multiply

$- 3$ by

$8$ .

$- 24 - 6 \cdot - 4$

Multiply

$- 6$ by

$- 4$ .

$- 24 + 24$

Add

$- 24$ and

$24$ .

$0$

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

$\frac{1}{0} [\begin{array}{c} 8 & - (- 4) \\ - (6) & - 3 \end{array}]$

Simplify each element in the matrix.

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Rearrange

$- (- 4)$ .

$\frac{1}{0} [\begin{array}{c} 8 & 4 \\ - (6) & - 3 \end{array}]$

Rearrange

$- (6)$ .

$\frac{1}{0} [\begin{array}{c} 8 & 4 \\ - 6 & - 3 \end{array}]$

Multiply

$\frac{1}{0}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{0} \cdot 8 & \frac{1}{0} \cdot 4 \\ \frac{1}{0} \cdot - 6 & \frac{1}{0} \cdot - 3 \end{array}]$

Rearrange

$\frac{1}{0} \cdot 8$ .

$[\begin{array}{c} Undefined & \frac{1}{0} \cdot 4 \\ \frac{1}{0} \cdot - 6 & \frac{1}{0} \cdot - 3 \end{array}]$

Since the matrix is undefined, it cannot be solved.

$Undefined$

Undefined