Matematica discreta Esempi

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

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 column $2$ 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 $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

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

Passaggio 1.1.4

Multiply element $a_{12}$ by its cofactor.

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

Passaggio 1.1.5

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

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

Passaggio 1.1.6

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

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

Passaggio 1.1.7

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

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

Passaggio 1.1.8

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

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

Passaggio 1.1.9

Add the terms together.

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

Passaggio 1.2

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

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

Passaggio 1.3

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

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

Passaggio 1.4

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

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Passaggio 1.4.1

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

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

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 $0$ per $1$ .

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

Passaggio 1.4.2.1.2

Moltiplica $- 3$ per $1$ .

$0 - 2 (0 - 3) + 0$

Passaggio 1.4.2.2

Sottrai $3$ da $0$ .

$0 - 2 \cdot - 3 + 0$

Passaggio 1.5

Semplifica il determinante.

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Passaggio 1.5.1

Moltiplica $- 2$ per $- 3$ .

$0 + 6 + 0$

Passaggio 1.5.2

Somma $0$ e $6$ .

$6 + 0$

Passaggio 1.5.3

Somma $6$ e $0$ .

$6$

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.

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

Passaggio 4

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 4.1

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

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Passaggio 4.1.1

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

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

Passaggio 4.1.2

Semplifica $R_{2}$ .

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

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 $0$ .

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

Passaggio 4.2.2

Semplifica $R_{3}$ .

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

Passaggio 4.3

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

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Passaggio 4.3.1

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

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

Passaggio 4.3.2

Semplifica $R_{2}$ .

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

Passaggio 4.4

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

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Passaggio 4.4.1

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

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

Passaggio 4.4.2

Semplifica $R_{3}$ .

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

Passaggio 4.5

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

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Passaggio 4.5.1

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

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

Passaggio 4.5.2

Semplifica $R_{2}$ .

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

Passaggio 4.6

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

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Passaggio 4.6.1

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

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

Passaggio 4.6.2

Semplifica $R_{1}$ .

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

Passaggio 5

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

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