Álgebra linear Exemplos

$[\begin{array}{c} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 0 & 2 & - 1 \end{array}] X = [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Etapa 1

Find the inverse of $[\begin{array}{c} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 0 & 2 & - 1 \end{array}]$ .

Toque para ver mais passagens...

Etapa 1.1

Reescreva.

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

Etapa 1.2

Find the determinant.

Toque para ver mais passagens...

Etapa 1.2.1

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 column $1$ by its cofactor and add.

Toque para ver mais passagens...

Etapa 1.2.1.1

Consider the corresponding sign chart.

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

Etapa 1.2.1.2

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

Etapa 1.2.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

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

Etapa 1.2.1.4

Multiply element $a_{11}$ by its cofactor.

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

Etapa 1.2.1.5

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

$| \begin{array}{c} - 1 & 1 \\ 2 & - 1 \end{array} |$

Etapa 1.2.1.6

Multiply element $a_{21}$ by its cofactor.

$- 2 | \begin{array}{c} - 1 & 1 \\ 2 & - 1 \end{array} |$

Etapa 1.2.1.7

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

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

Etapa 1.2.1.8

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

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

Etapa 1.2.1.9

Add the terms together.

$1 | \begin{array}{c} 3 & 0 \\ 2 & - 1 \end{array} | - 2 | \begin{array}{c} - 1 & 1 \\ 2 & - 1 \end{array} | + 0 | \begin{array}{c} - 1 & 1 \\ 3 & 0 \end{array} |$

Etapa 1.2.2

Multiplique $0$ por $| \begin{array}{c} - 1 & 1 \\ 3 & 0 \end{array} |$ .

$1 | \begin{array}{c} 3 & 0 \\ 2 & - 1 \end{array} | - 2 | \begin{array}{c} - 1 & 1 \\ 2 & - 1 \end{array} | + 0$

Etapa 1.2.3

Avalie $| \begin{array}{c} 3 & 0 \\ 2 & - 1 \end{array} |$ .

Toque para ver mais passagens...

Etapa 1.2.3.1

O determinante de uma matriz $2 \times 2$ pode ser encontrado ao usar a fórmula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$1 (3 \cdot - 1 - 2 \cdot 0) - 2 | \begin{array}{c} - 1 & 1 \\ 2 & - 1 \end{array} | + 0$

Etapa 1.2.3.2

Simplifique o determinante.

Toque para ver mais passagens...

Etapa 1.2.3.2.1

Simplifique cada termo.

Toque para ver mais passagens...

Etapa 1.2.3.2.1.1

Multiplique $3$ por $- 1$ .

$1 (- 3 - 2 \cdot 0) - 2 | \begin{array}{c} - 1 & 1 \\ 2 & - 1 \end{array} | + 0$

Etapa 1.2.3.2.1.2

Multiplique $- 2$ por $0$ .

$1 (- 3 + 0) - 2 | \begin{array}{c} - 1 & 1 \\ 2 & - 1 \end{array} | + 0$

Etapa 1.2.3.2.2

Some $- 3$ e $0$ .

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

Etapa 1.2.4

Avalie $| \begin{array}{c} - 1 & 1 \\ 2 & - 1 \end{array} |$ .

Toque para ver mais passagens...

Etapa 1.2.4.1

O determinante de uma matriz $2 \times 2$ pode ser encontrado ao usar a fórmula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$1 \cdot - 3 - 2 (- - 1 - 2 \cdot 1) + 0$

Etapa 1.2.4.2

Simplifique o determinante.

Toque para ver mais passagens...

Etapa 1.2.4.2.1

Simplifique cada termo.

Toque para ver mais passagens...

Etapa 1.2.4.2.1.1

Multiplique $- 1$ por $- 1$ .

$1 \cdot - 3 - 2 (1 - 2 \cdot 1) + 0$

Etapa 1.2.4.2.1.2

Multiplique $- 2$ por $1$ .

$1 \cdot - 3 - 2 (1 - 2) + 0$

Etapa 1.2.4.2.2

Subtraia $2$ de $1$ .

$1 \cdot - 3 - 2 \cdot - 1 + 0$

Etapa 1.2.5

Simplifique o determinante.

Toque para ver mais passagens...

Etapa 1.2.5.1

Simplifique cada termo.

Toque para ver mais passagens...

Etapa 1.2.5.1.1

Multiplique $- 3$ por $1$ .

$- 3 - 2 \cdot - 1 + 0$

Etapa 1.2.5.1.2

Multiplique $- 2$ por $- 1$ .

$- 3 + 2 + 0$

Etapa 1.2.5.2

Some $- 3$ e $2$ .

$- 1 + 0$

Etapa 1.2.5.3

Some $- 1$ e $0$ .

$- 1$

Etapa 1.3

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

Etapa 1.4

Set up a $3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

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

Etapa 1.5

Encontre a forma escalonada reduzida por linhas.

Toque para ver mais passagens...

Etapa 1.5.1

Perform the row operation $R_{2} = R_{2} - 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

Toque para ver mais passagens...

Etapa 1.5.1.1

Perform the row operation $R_{2} = R_{2} - 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Etapa 1.5.1.2

Simplifique $R_{2}$ .

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

Etapa 1.5.2

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

Toque para ver mais passagens...

Etapa 1.5.2.1

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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

Etapa 1.5.2.2

Simplifique $R_{2}$ .

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

Etapa 1.5.3

Perform the row operation $R_{3} = R_{3} - 2 R_{2}$ to make the entry at $3, 2$ a $0$ .

Toque para ver mais passagens...

Etapa 1.5.3.1

Perform the row operation $R_{3} = R_{3} - 2 R_{2}$ to make the entry at $3, 2$ a $0$ .

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

Etapa 1.5.3.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{2}{5} & \frac{1}{5} & 0 \\ 0 & 0 & - \frac{1}{5} & \frac{4}{5} & - \frac{2}{5} & 1 \end{array}]$

Etapa 1.5.4

Multiply each element of $R_{3}$ by $- 5$ to make the entry at $3, 3$ a $1$ .

Toque para ver mais passagens...

Etapa 1.5.4.1

Multiply each element of $R_{3}$ by $- 5$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{2}{5} & \frac{1}{5} & 0 \\ - 5 \cdot 0 & - 5 \cdot 0 & - 5 (- \frac{1}{5}) & - 5 (\frac{4}{5}) & - 5 (- \frac{2}{5}) & - 5 \cdot 1 \end{array}]$

Etapa 1.5.4.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{2}{5} & \frac{1}{5} & 0 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Etapa 1.5.5

Perform the row operation $R_{2} = R_{2} + \frac{2}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

Toque para ver mais passagens...

Etapa 1.5.5.1

Perform the row operation $R_{2} = R_{2} + \frac{2}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 + \frac{2}{5} \cdot 0 & 1 + \frac{2}{5} \cdot 0 & - \frac{2}{5} + \frac{2}{5} \cdot 1 & - \frac{2}{5} + \frac{2}{5} \cdot - 4 & \frac{1}{5} + \frac{2}{5} \cdot 2 & 0 + \frac{2}{5} \cdot - 5 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Etapa 1.5.5.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Etapa 1.5.6

Perform the row operation $R_{1} = R_{1} - R_{3}$ to make the entry at $1, 3$ a $0$ .

Toque para ver mais passagens...

Etapa 1.5.6.1

Perform the row operation $R_{1} = R_{1} - R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 0 & - 1 - 0 & 1 - 1 & 1 + 4 & 0 - 2 & 0 + 5 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Etapa 1.5.6.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 5 & - 2 & 5 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Etapa 1.5.7

Perform the row operation $R_{1} = R_{1} + R_{2}$ to make the entry at $1, 2$ a $0$ .

Toque para ver mais passagens...

Etapa 1.5.7.1

Perform the row operation $R_{1} = R_{1} + R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 5 - 2 & - 2 + 1 \cdot 1 & 5 - 2 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Etapa 1.5.7.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 3 & - 1 & 3 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Etapa 1.6

The right half of the reduced row echelon form is the inverse.

$[\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}]$

Etapa 2

Multiply both sides by the inverse of $[\begin{array}{c} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 0 & 2 & - 1 \end{array}]$ .

$[\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 0 & 2 & - 1 \end{array}] X = [\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Etapa 3

Simplifique a equação.

Toque para ver mais passagens...

Etapa 3.1

Multiplique $[\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 0 & 2 & - 1 \end{array}]$ .

Toque para ver mais passagens...

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

Etapa 3.1.2

Multiplique cada linha na primeira matriz por cada coluna na segunda matriz.

$[\begin{array}{c} 3 \cdot 1 - 1 \cdot 2 + 3 \cdot 0 & 3 \cdot - 1 - 1 \cdot 3 + 3 \cdot 2 & 3 \cdot 1 - 0 + 3 \cdot - 1 \\ - 2 \cdot 1 + 1 \cdot 2 - 2 \cdot 0 & - 2 \cdot - 1 + 1 \cdot 3 - 2 \cdot 2 & - 2 \cdot 1 + 1 \cdot 0 - 2 \cdot - 1 \\ - 4 \cdot 1 + 2 \cdot 2 - 5 \cdot 0 & - 4 \cdot - 1 + 2 \cdot 3 - 5 \cdot 2 & - 4 \cdot 1 + 2 \cdot 0 - 5 \cdot - 1 \end{array}] X = [\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Etapa 3.1.3

Simplifique cada elemento da matriz multiplicando todas as expressões.

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] X = [\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Etapa 3.2

Multiplying the identity matrix by any matrix $A$ is the matrix $A$ itself.

$X = [\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Etapa 3.3

Multiplique $[\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$ .

Toque para ver mais passagens...

Etapa 3.3.1

Etapa 3.3.2

Multiplique cada linha na primeira matriz por cada coluna na segunda matriz.

$X = [\begin{array}{c} 3 \cdot 6 - 1 \cdot 8 + 3 \cdot 7 & 3 \cdot - 1 - 0 + 3 \cdot 9 & 3 \cdot 9 - - 7 + 3 \cdot - 11 & 3 \cdot 11 - 0 + 3 \cdot 6 & 3 \cdot 12 - 1 \cdot 1 + 3 \cdot 1 \\ - 2 \cdot 6 + 1 \cdot 8 - 2 \cdot 7 & - 2 \cdot - 1 + 1 \cdot 0 - 2 \cdot 9 & - 2 \cdot 9 + 1 \cdot - 7 - 2 \cdot - 11 & - 2 \cdot 11 + 1 \cdot 0 - 2 \cdot 6 & - 2 \cdot 12 + 1 \cdot 1 - 2 \cdot 1 \\ - 4 \cdot 6 + 2 \cdot 8 - 5 \cdot 7 & - 4 \cdot - 1 + 2 \cdot 0 - 5 \cdot 9 & - 4 \cdot 9 + 2 \cdot - 7 - 5 \cdot - 11 & - 4 \cdot 11 + 2 \cdot 0 - 5 \cdot 6 & - 4 \cdot 12 + 2 \cdot 1 - 5 \cdot 1 \end{array}]$

Etapa 3.3.3

Simplifique cada elemento da matriz multiplicando todas as expressões.

$X = [\begin{array}{c} 31 & 24 & 1 & 51 & 38 \\ - 18 & - 16 & - 3 & - 34 & - 25 \\ - 43 & - 41 & 5 & - 74 & - 51 \end{array}]$