Matematica discreta Esempi

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

Passaggio 1

Find the determinant.

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Passaggio 1.1

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 $2$ by its cofactor and add.

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Passaggio 1.1.1

Consider the corresponding sign chart.

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

Passaggio 1.1.2

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

Passaggio 1.1.3

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

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

Passaggio 1.1.4

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

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

Passaggio 1.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

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

Passaggio 1.1.6

Multiply element $a_{22}$ by its cofactor.

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

Passaggio 1.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

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

Passaggio 1.1.8

Multiply element $a_{23}$ by its cofactor.

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

Passaggio 1.1.9

Add the terms together.

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

Passaggio 1.2

Moltiplica $0$ per $| \begin{array}{c} 0 & - 1 \\ - 2 & - 2 \end{array} |$ .

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

Passaggio 1.3

Moltiplica $0$ per $| \begin{array}{c} 2 & 0 \\ 3 & - 2 \end{array} |$ .

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

Passaggio 1.4

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

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Passaggio 1.4.1

È possibile trovare il determinante di una matrice $2 \times 2$ usando la formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

Passaggio 1.4.2

Semplifica il determinante.

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Passaggio 1.4.2.1

Semplifica ciascun termine.

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Passaggio 1.4.2.1.1

Moltiplica $2$ per $- 2$ .

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

Passaggio 1.4.2.1.2

Moltiplica $- 3$ per $- 1$ .

$0 - 1 (- 4 + 3) + 0$

Passaggio 1.4.2.2

Somma $- 4$ e $3$ .

$0 - 1 \cdot - 1 + 0$

Passaggio 1.5

Semplifica il determinante.

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Passaggio 1.5.1

Moltiplica $- 1$ per $- 1$ .

$0 + 1 + 0$

Passaggio 1.5.2

Somma $0$ e $1$ .

$1 + 0$

Passaggio 1.5.3

Somma $1$ e $0$ .

$1$

Passaggio 2

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

Passaggio 3

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

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

Passaggio 4

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 4.1

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

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Passaggio 4.1.1

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

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

Passaggio 4.1.2

Semplifica $R_{1}$ .

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

Passaggio 4.2

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

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Passaggio 4.2.1

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

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

Passaggio 4.2.2

Semplifica $R_{3}$ .

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

Passaggio 4.3

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

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Passaggio 4.3.1

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

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

Passaggio 4.3.2

Semplifica $R_{2}$ .

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

Passaggio 4.4

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

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Passaggio 4.4.1

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

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

Passaggio 4.4.2

Semplifica $R_{3}$ .

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

Passaggio 4.5

Multiply each element of $R_{3}$ by $- 2$ to make the entry at $3, 3$ a $1$ .

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Passaggio 4.5.1

Multiply each element of $R_{3}$ by $- 2$ to make the entry at $3, 3$ a $1$ .

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

Passaggio 4.5.2

Semplifica $R_{3}$ .

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

Passaggio 4.6

Perform the row operation $R_{1} = R_{1} + \frac{1}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Passaggio 4.6.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

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

Passaggio 4.6.2

Semplifica $R_{1}$ .

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

Passaggio 5

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

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