Linear Algebra Examples

\frac{1}{3} y - \frac{2}{3} x = 1

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

10 x - 5 y = - 15

$10 x - 5 y = - 15$

Step 1

Find the

A X = B

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

[\begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ 10 & - 5 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 \\ - 15 \end{array}]

$[\begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ 10 & - 5 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 \\ - 15 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

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The inverse of a

2 \times 2

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

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

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

| A |

$| A |$ is the determinant of

A

$A$ .

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

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

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

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

Find the determinant of

[\begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ 10 & - 5 \end{array}]

$[\begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ 10 & - 5 \end{array}]$ .

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

determinant [\begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ 10 & - 5 \end{array}] = | \begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ 10 & - 5 \end{array} |

$determinant [\begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ 10 & - 5 \end{array}] = | \begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ 10 & - 5 \end{array} |$

The determinant of a

2 \times 2

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

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

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

(- \frac{2}{3}) (- 5) - 10 (\frac{1}{3})

$(- \frac{2}{3}) (- 5) - 10 (\frac{1}{3})$

Simplify the determinant.

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Simplify each term.

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Multiply

(- \frac{2}{3}) (- 5)

$(- \frac{2}{3}) (- 5)$ .

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Multiply

- 5

$- 5$ by

- 1

$- 1$ .

5 (\frac{2}{3}) - 10 (\frac{1}{3})

$5 (\frac{2}{3}) - 10 (\frac{1}{3})$

Combine

5

$5$ and

\frac{2}{3}

$\frac{2}{3}$ .

\frac{5 \cdot 2}{3} - 10 (\frac{1}{3})

$\frac{5 \cdot 2}{3} - 10 (\frac{1}{3})$

Multiply

5

$5$ by

2

$2$ .

\frac{10}{3} - 10 (\frac{1}{3})

$\frac{10}{3} - 10 (\frac{1}{3})$

\frac{10}{3} - 10 (\frac{1}{3})

$\frac{10}{3} - 10 (\frac{1}{3})$

Combine

- 10

$- 10$ and

\frac{1}{3}

$\frac{1}{3}$ .

\frac{10}{3} + \frac{- 10}{3}

$\frac{10}{3} + \frac{- 10}{3}$

Move the negative in front of the fraction.

\frac{10}{3} - \frac{10}{3}

$\frac{10}{3} - \frac{10}{3}$

\frac{10}{3} - \frac{10}{3}

$\frac{10}{3} - \frac{10}{3}$

Combine fractions.

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Combine the numerators over the common denominator.

\frac{10 - 10}{3}

$\frac{10 - 10}{3}$

Simplify the expression.

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Subtract

10

$10$ from

10

$10$ .

\frac{0}{3}

$\frac{0}{3}$

Divide

0

$0$ by

3

$3$ .

0

$0$

0

$0$

0

$0$

0

$0$

0

$0$

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

\frac{1}{0} [\begin{array}{c} - 5 & - (\frac{1}{3}) \\ - (10) & - \frac{2}{3} \end{array}]

$\frac{1}{0} [\begin{array}{c} - 5 & - (\frac{1}{3}) \\ - (10) & - \frac{2}{3} \end{array}]$

Simplify each element in the matrix.

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Rearrange

- (\frac{1}{3})

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

\frac{1}{0} [\begin{array}{c} - 5 & - \frac{1}{3} \\ - (10) & - \frac{2}{3} \end{array}]

$\frac{1}{0} [\begin{array}{c} - 5 & - \frac{1}{3} \\ - (10) & - \frac{2}{3} \end{array}]$

Rearrange

- (10)

$- (10)$ .

\frac{1}{0} [\begin{array}{c} - 5 & - \frac{1}{3} \\ - 10 & - \frac{2}{3} \end{array}]

$\frac{1}{0} [\begin{array}{c} - 5 & - \frac{1}{3} \\ - 10 & - \frac{2}{3} \end{array}]$

\frac{1}{0} [\begin{array}{c} - 5 & - \frac{1}{3} \\ - 10 & - \frac{2}{3} \end{array}]

$\frac{1}{0} [\begin{array}{c} - 5 & - \frac{1}{3} \\ - 10 & - \frac{2}{3} \end{array}]$

Multiply

\frac{1}{0}

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

[\begin{array}{c} \frac{1}{0} \cdot - 5 & \frac{1}{0} \cdot (- \frac{1}{3}) \\ \frac{1}{0} \cdot - 10 & \frac{1}{0} \cdot (- \frac{2}{3}) \end{array}]

$[\begin{array}{c} \frac{1}{0} \cdot - 5 & \frac{1}{0} \cdot (- \frac{1}{3}) \\ \frac{1}{0} \cdot - 10 & \frac{1}{0} \cdot (- \frac{2}{3}) \end{array}]$

Rearrange

\frac{1}{0} \cdot - 5

$\frac{1}{0} \cdot - 5$ .

[\begin{array}{c} Undefined & \frac{1}{0} \cdot (- \frac{1}{3}) \\ \frac{1}{0} \cdot - 10 & \frac{1}{0} \cdot (- \frac{2}{3}) \end{array}]

$[\begin{array}{c} Undefined & \frac{1}{0} \cdot (- \frac{1}{3}) \\ \frac{1}{0} \cdot - 10 & \frac{1}{0} \cdot (- \frac{2}{3}) \end{array}]$

Since the matrix is undefined, it cannot be solved.

Undefined

$Undefined$

Undefined