Linear Algebra Examples

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

Step 1

Multiply $[\begin{array}{c} 3 & 2 \\ - 1 & 0 \\ 0 & - 1 \end{array}] [\begin{array}{c} - 1 & 0 & - 1 \\ 2 & 1 & 0 \end{array}]$ .

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Step 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 $3 \times 2$ and the second matrix is $2 \times 3$ .

Step 1.2

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

$[\begin{array}{c} 3 \cdot - 1 + 2 \cdot 2 & 3 \cdot 0 + 2 \cdot 1 & 3 \cdot - 1 + 2 \cdot 0 \\ - - 1 + 0 \cdot 2 & - 0 + 0 \cdot 1 & - - 1 + 0 \cdot 0 \\ 0 \cdot - 1 - 1 \cdot 2 & 0 \cdot 0 - 1 \cdot 1 & 0 \cdot - 1 - 0 \end{array}] [\begin{array}{c} 1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{array}]$

Step 1.3

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

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

Step 2

Multiply $[\begin{array}{c} 1 & 2 & - 3 \\ 1 & 0 & 1 \\ - 2 & - 1 & 0 \end{array}] [\begin{array}{c} 1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{array}]$ .

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Step 2.1

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 $3 \times 3$ and the second matrix is $3 \times 3$ .

Step 2.2

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

$[\begin{array}{c} 1 \cdot 1 + 2 \cdot 2 - 3 \cdot 3 & 1 \cdot 1 + 2 \cdot 1 - 3 \cdot 0 & 1 \cdot 0 + 2 \cdot 0 - 3 \cdot 1 \\ 1 \cdot 1 + 0 \cdot 2 + 1 \cdot 3 & 1 \cdot 1 + 0 \cdot 1 + 1 \cdot 0 & 1 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 \\ - 2 \cdot 1 - 1 \cdot 2 + 0 \cdot 3 & - 2 \cdot 1 - 1 \cdot 1 + 0 \cdot 0 & - 2 \cdot 0 - 0 + 0 \cdot 1 \end{array}]$

Step 2.3

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

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

Step 3

Find the determinant.

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Step 3.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $3$ by its cofactor and add.

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Step 3.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 3.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 3.1.3

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} 3 & - 3 \\ 1 & 1 \end{array} |$

Step 3.1.4

Multiply element $a_{31}$ by its cofactor.

$- 4 | \begin{array}{c} 3 & - 3 \\ 1 & 1 \end{array} |$

Step 3.1.5

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

$| \begin{array}{c} - 4 & - 3 \\ 4 & 1 \end{array} |$

Step 3.1.6

Multiply element $a_{32}$ by its cofactor.

$3 | \begin{array}{c} - 4 & - 3 \\ 4 & 1 \end{array} |$

Step 3.1.7

The minor for $a_{33}$ is the determinant with row $3$ and column $3$ deleted.

$| \begin{array}{c} - 4 & 3 \\ 4 & 1 \end{array} |$

Step 3.1.8

Multiply element $a_{33}$ by its cofactor.

$0 | \begin{array}{c} - 4 & 3 \\ 4 & 1 \end{array} |$

Step 3.1.9

Add the terms together.

$- 4 | \begin{array}{c} 3 & - 3 \\ 1 & 1 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ 4 & 1 \end{array} | + 0 | \begin{array}{c} - 4 & 3 \\ 4 & 1 \end{array} |$

Step 3.2

Multiply $0$ by $| \begin{array}{c} - 4 & 3 \\ 4 & 1 \end{array} |$ .

$- 4 | \begin{array}{c} 3 & - 3 \\ 1 & 1 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ 4 & 1 \end{array} | + 0$

Step 3.3

Evaluate $| \begin{array}{c} 3 & - 3 \\ 1 & 1 \end{array} |$ .

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Step 3.3.1

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

$- 4 (3 \cdot 1 - 1 \cdot - 3) + 3 | \begin{array}{c} - 4 & - 3 \\ 4 & 1 \end{array} | + 0$

Step 3.3.2

Simplify the determinant.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Multiply $3$ by $1$ .

$- 4 (3 - 1 \cdot - 3) + 3 | \begin{array}{c} - 4 & - 3 \\ 4 & 1 \end{array} | + 0$

Step 3.3.2.1.2

Multiply $- 1$ by $- 3$ .

$- 4 (3 + 3) + 3 | \begin{array}{c} - 4 & - 3 \\ 4 & 1 \end{array} | + 0$

Step 3.3.2.2

Add $3$ and $3$ .

$- 4 \cdot 6 + 3 | \begin{array}{c} - 4 & - 3 \\ 4 & 1 \end{array} | + 0$

Step 3.4

Evaluate $| \begin{array}{c} - 4 & - 3 \\ 4 & 1 \end{array} |$ .

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Step 3.4.1

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

$- 4 \cdot 6 + 3 (- 4 \cdot 1 - 4 \cdot - 3) + 0$

Step 3.4.2

Simplify the determinant.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Multiply $- 4$ by $1$ .

$- 4 \cdot 6 + 3 (- 4 - 4 \cdot - 3) + 0$

Step 3.4.2.1.2

Multiply $- 4$ by $- 3$ .

$- 4 \cdot 6 + 3 (- 4 + 12) + 0$

Step 3.4.2.2

Add $- 4$ and $12$ .

$- 4 \cdot 6 + 3 \cdot 8 + 0$

Step 3.5

Simplify the determinant.

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Step 3.5.1

Simplify each term.

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Step 3.5.1.1

Multiply $- 4$ by $6$ .

$- 24 + 3 \cdot 8 + 0$

Step 3.5.1.2

Multiply $3$ by $8$ .

$- 24 + 24 + 0$

Step 3.5.2

Add $- 24$ and $24$ .

$0 + 0$

Step 3.5.3

Add $0$ and $0$ .

$0$

Step 4

There is no inverse because the determinant is $0$ .