Linear Algebra Examples

$y = \frac{1}{3} x + 2$ , $y = \frac{1}{3} x + 3$

Step 1

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

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

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 $A$ .

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

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

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

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

$(- \frac{1}{3}) (1) + \frac{1}{3} \cdot 1$

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $1$ .

$- \frac{1}{3} + \frac{1}{3} \cdot 1$

Multiply $\frac{1}{3}$ by $1$ .

$- \frac{1}{3} + \frac{1}{3}$

Combine fractions.

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Combine the numerators over the common denominator.

$\frac{- 1 + 1}{3}$

Simplify the expression.

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Add $- 1$ and $1$ .

$\frac{0}{3}$

Divide $0$ by $3$ .

$0$

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

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

Simplify each element in the matrix.

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Rearrange $- (1)$ .

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

Rearrange $- (- \frac{1}{3})$ .

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

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

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

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

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

Since the matrix is undefined, it cannot be solved.

$Undefined$

Undefined