Algebra lineare Esempi

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 3 & 3 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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

Passaggio 1

Find the inverse of

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array}]$ .

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Passaggio 1.1

Riscrivi.

∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣

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

Passaggio 1.2

Find the determinant.

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Passaggio 1.2.1

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in row

1

$1$ by its cofactor and add.

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Passaggio 1.2.1.1

Consider the corresponding sign chart.

∣ ∣ ∣ ∣ \begin{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

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

Passaggio 1.2.1.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

Passaggio 1.2.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

Passaggio 1.2.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

Passaggio 1.2.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

Passaggio 1.2.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

Passaggio 1.2.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

Passaggio 1.2.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

Passaggio 1.2.1.9

Add the terms together.

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

Passaggio 1.2.2

Moltiplica

0

$0$ per

∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 + 0 ∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

Passaggio 1.2.3

Moltiplica

0

$0$ per

∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 + 0

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

Passaggio 1.2.4

Calcola

∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

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Passaggio 1.2.4.1

È possibile trovare il determinante di una matrice

2 \times 2

$2 \times 2$ usando la formula

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

1 (1 \cdot 1 - 1 \cdot 0) + 0 + 0

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

Passaggio 1.2.4.2

Semplifica il determinante.

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Passaggio 1.2.4.2.1

Moltiplica

1

$1$ per

1

$1$ .

1 (1 - 1 \cdot 0) + 0 + 0

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

Passaggio 1.2.4.2.2

Sottrai

0

$0$ da

1

$1$ .

1 \cdot 1 + 0 + 0

$1 \cdot 1 + 0 + 0$

1 \cdot 1 + 0 + 0

$1 \cdot 1 + 0 + 0$

1 \cdot 1 + 0 + 0

$1 \cdot 1 + 0 + 0$

Passaggio 1.2.5

Semplifica il determinante.

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Passaggio 1.2.5.1

Moltiplica

1

$1$ per

1

$1$ .

1 + 0 + 0

$1 + 0 + 0$

Passaggio 1.2.5.2

Somma

1

$1$ e

0

$0$ .

1 + 0

$1 + 0$

Passaggio 1.2.5.3

Somma

1

$1$ e

0

$0$ .

1

$1$

1

$1$

1

$1$

Passaggio 1.3

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

Passaggio 1.4

Set up a

3 \times 6

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 1 & 1 & 0 & 0 & 1 & 0 1 & 1 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

Passaggio 1.5

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 1.5.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Passaggio 1.5.1.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 1 - 1 & 1 - 0 & 0 - 0 & 0 - 1 & 1 - 0 & 0 - 0 1 & 1 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

Passaggio 1.5.1.2

Semplifica

R_{2}

$R_{2}$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 1 & 1 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 1 & 1 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

Passaggio 1.5.2

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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Passaggio 1.5.2.1

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 1 - 1 & 1 - 0 & 1 - 0 & 0 - 1 & 0 - 0 & 1 - 0 \end{matrix} ⎤ ⎥ ⎦

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

Passaggio 1.5.2.2

Semplifica

R_{3}

$R_{3}$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 0 & 1 & 1 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 0 & 1 & 1 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

Passaggio 1.5.3

Perform the row operation

R_{3} = R_{3} - R_{2}

$R_{3} = R_{3} - R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

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Passaggio 1.5.3.1

Perform the row operation

$R_{3} = R_{3} - R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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

Passaggio 1.5.3.2

Semplifica

$R_{3}$ .

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

Passaggio 1.6

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

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

Passaggio 2

Multiply both sides by the inverse of

$[\begin{array}{c} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array}]$ .

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

Passaggio 3

Semplifica l'equazione.

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Passaggio 3.1

Moltiplica

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

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Passaggio 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

$3 \times 3$ .

Passaggio 3.1.2

Moltiplica ogni riga nella prima matrice per ogni colonna nella seconda matrice.

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

Passaggio 3.1.3

Semplifica ogni elemento della matrice moltiplicando tutte le espressioni.

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

Passaggio 3.2

Multiplying the identity matrix by any matrix

$A$ is the matrix

$A$ itself.

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

Passaggio 3.3

Moltiplica

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

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Passaggio 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

$3 \times 3$ and the second matrix is

$3 \times 2$ .

Passaggio 3.3.2

Moltiplica ogni riga nella prima matrice per ogni colonna nella seconda matrice.

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

Passaggio 3.3.3

Semplifica ogni elemento della matrice moltiplicando tutte le espressioni.

$y = [\begin{array}{c} 1 & 2 \\ 2 & 1 \\ - 1 & - 2 \end{array}]$