Linear Algebra Examples

[\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}] \cdot B = [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]

$[\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}] \cdot B = [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]$

Step 1

Find the inverse of

[\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}]

$[\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}]$ .

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Step 1.1

The inverse of a

2 \times 2

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

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

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

a d - b c

$a d - b c$ is the determinant.

Step 1.2

Find the determinant.

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Step 1.2.1

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{1}{9} \cdot 27 - \frac{1}{3} \cdot 6

$\frac{1}{9} \cdot 27 - \frac{1}{3} \cdot 6$

Step 1.2.2

Simplify the determinant.

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Step 1.2.2.1

Simplify each term.

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Step 1.2.2.1.1

Cancel the common factor of

9

$9$ .

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Step 1.2.2.1.1.1

Factor

9

$9$ out of

27

$27$ .

\frac{1}{9} \cdot (9 (3)) - \frac{1}{3} \cdot 6

$\frac{1}{9} \cdot (9 (3)) - \frac{1}{3} \cdot 6$

Step 1.2.2.1.1.2

Cancel the common factor.

\frac{1}{9} \cdot (9 \cdot 3) - \frac{1}{3} \cdot 6

$\frac{1}{9} \cdot (9 \cdot 3) - \frac{1}{3} \cdot 6$

Step 1.2.2.1.1.3

Rewrite the expression.

3 - \frac{1}{3} \cdot 6

$3 - \frac{1}{3} \cdot 6$

3 - \frac{1}{3} \cdot 6

$3 - \frac{1}{3} \cdot 6$

Step 1.2.2.1.2

Cancel the common factor of

3

$3$ .

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Step 1.2.2.1.2.1

Move the leading negative in

- \frac{1}{3}

$- \frac{1}{3}$ into the numerator.

3 + \frac{- 1}{3} \cdot 6

$3 + \frac{- 1}{3} \cdot 6$

Step 1.2.2.1.2.2

Factor

3

$3$ out of

6

$6$ .

3 + \frac{- 1}{3} \cdot (3 (2))

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

Step 1.2.2.1.2.3

Cancel the common factor.

3 + \frac{- 1}{3} \cdot (3 \cdot 2)

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

Step 1.2.2.1.2.4

Rewrite the expression.

3 - 1 \cdot 2

$3 - 1 \cdot 2$

3 - 1 \cdot 2

$3 - 1 \cdot 2$

Step 1.2.2.1.3

Multiply

- 1

$- 1$ by

2

$2$ .

3 - 2

$3 - 2$

3 - 2

$3 - 2$

Step 1.2.2.2

Subtract

2

$2$ from

3

$3$ .

1

$1$

1

$1$

1

$1$

Step 1.3

Since the determinant is non-zero, the inverse exists.

Step 1.4

Substitute the known values into the formula for the inverse.

\frac{1}{1} [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]

$\frac{1}{1} [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]$

Step 1.5

Divide

1

$1$ by

1

$1$ .

1 [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]

$1 [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]$

Step 1.6

Multiply

1

$1$ by each element of the matrix.

[\begin{array}{c} 1 \cdot 27 & 1 \cdot - 6 \\ 1 (- \frac{1}{3}) & 1 (\frac{1}{9}) \end{array}]

$[\begin{array}{c} 1 \cdot 27 & 1 \cdot - 6 \\ 1 (- \frac{1}{3}) & 1 (\frac{1}{9}) \end{array}]$

Step 1.7

Simplify each element in the matrix.

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Step 1.7.1

Multiply

27

$27$ by

1

$1$ .

[\begin{array}{c} 27 & 1 \cdot - 6 \\ 1 (- \frac{1}{3}) & 1 (\frac{1}{9}) \end{array}]

$[\begin{array}{c} 27 & 1 \cdot - 6 \\ 1 (- \frac{1}{3}) & 1 (\frac{1}{9}) \end{array}]$

Step 1.7.2

Multiply

- 6

$- 6$ by

1

$1$ .

[\begin{array}{c} 27 & - 6 \\ 1 (- \frac{1}{3}) & 1 (\frac{1}{9}) \end{array}]

$[\begin{array}{c} 27 & - 6 \\ 1 (- \frac{1}{3}) & 1 (\frac{1}{9}) \end{array}]$

Step 1.7.3

Multiply

- \frac{1}{3}

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

1

$1$ .

[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & 1 (\frac{1}{9}) \end{array}]

$[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & 1 (\frac{1}{9}) \end{array}]$

Step 1.7.4

Multiply

\frac{1}{9}

$\frac{1}{9}$ by

1

$1$ .

[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]

$[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]$

[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]

$[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]$

[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]

$[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}]$

Step 2

Multiply both sides by the inverse of

[\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}]

$[\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}]$ .

[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}] B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]

$[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}] B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]$

Step 3

Simplify the equation.

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Step 3.1

Multiply

[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}]

$[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} \frac{1}{9} & 6 \\ \frac{1}{3} & 27 \end{array}]$ .

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Step 3.1.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is

2 \times 2

$2 \times 2$ and the second matrix is

2 \times 2

$2 \times 2$ .

Step 3.1.2

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

[\begin{array}{c} 27 (\frac{1}{9}) - 6 (\frac{1}{3}) & 27 \cdot 6 - 6 \cdot 27 \\ - \frac{1}{3} \cdot \frac{1}{9} + \frac{1}{9} \cdot \frac{1}{3} & - \frac{1}{3} \cdot 6 + \frac{1}{9} \cdot 27 \end{array}] B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]

$[\begin{array}{c} 27 (\frac{1}{9}) - 6 (\frac{1}{3}) & 27 \cdot 6 - 6 \cdot 27 \\ - \frac{1}{3} \cdot \frac{1}{9} + \frac{1}{9} \cdot \frac{1}{3} & - \frac{1}{3} \cdot 6 + \frac{1}{9} \cdot 27 \end{array}] B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]$

Step 3.1.3

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

[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]$

[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]$

Step 3.2

Multiplying the identity matrix by any matrix

A

$A$ is the matrix

A

$A$ itself.

B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]

$B = [\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]$

Step 3.3

Multiply

[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]

$[\begin{array}{c} 27 & - 6 \\ - \frac{1}{3} & \frac{1}{9} \end{array}] [\begin{array}{c} - 10 & 7 \\ - 48 & 30 \end{array}]$ .

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Step 3.3.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is

2 \times 2

$2 \times 2$ and the second matrix is

2 \times 2

$2 \times 2$ .

Step 3.3.2

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

B = [\begin{array}{c} 27 \cdot - 10 - 6 \cdot - 48 & 27 \cdot 7 - 6 \cdot 30 \\ - \frac{1}{3} \cdot - 10 + \frac{1}{9} \cdot - 48 & - \frac{1}{3} \cdot 7 + \frac{1}{9} \cdot 30 \end{array}]

$B = [\begin{array}{c} 27 \cdot - 10 - 6 \cdot - 48 & 27 \cdot 7 - 6 \cdot 30 \\ - \frac{1}{3} \cdot - 10 + \frac{1}{9} \cdot - 48 & - \frac{1}{3} \cdot 7 + \frac{1}{9} \cdot 30 \end{array}]$

Step 3.3.3

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

B = [\begin{array}{c} 18 & 9 \\ - 2 & 1 \end{array}]

$B = [\begin{array}{c} 18 & 9 \\ - 2 & 1 \end{array}]$

B = [\begin{array}{c} 18 & 9 \\ - 2 & 1 \end{array}]

$B = [\begin{array}{c} 18 & 9 \\ - 2 & 1 \end{array}]$

B = [\begin{array}{c} 18 & 9 \\ - 2 & 1 \end{array}]

$B = [\begin{array}{c} 18 & 9 \\ - 2 & 1 \end{array}]$