Linear Algebra Examples

$[\begin{array}{c} x & y & z \\ 2 x & - y & - z \\ x & 2 y & - z \end{array}]$

Step 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 $1$ by its cofactor and add.

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Step 1.1

Consider the corresponding sign chart.

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

Step 1.2

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

Step 1.3

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

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

Step 1.4

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

$x | \begin{array}{c} - y & - z \\ 2 y & - z \end{array} |$

Step 1.5

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

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

Step 1.6

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

$- y | \begin{array}{c} 2 x & - z \\ x & - z \end{array} |$

Step 1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

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

Step 1.8

Multiply element $a_{13}$ by its cofactor.

$z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 1.9

Add the terms together.

$x | \begin{array}{c} - y & - z \\ 2 y & - z \end{array} | - y | \begin{array}{c} 2 x & - z \\ x & - z \end{array} | + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 2

Evaluate $| \begin{array}{c} - y & - z \\ 2 y & - z \end{array} |$ .

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Step 2.1

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

$x (- y (- z) - 2 y (- z)) - y | \begin{array}{c} 2 x & - z \\ x & - z \end{array} | + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 2.2

Simplify the determinant.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Rewrite using the commutative property of multiplication.

$x (- 1 \cdot - 1 y z - 2 y (- z)) - y | \begin{array}{c} 2 x & - z \\ x & - z \end{array} | + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 2.2.1.2

Multiply $- 1$ by $- 1$ .

$x (1 y z - 2 y (- z)) - y | \begin{array}{c} 2 x & - z \\ x & - z \end{array} | + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 2.2.1.3

Multiply $y$ by $1$ .

$x (y z - 2 y (- z)) - y | \begin{array}{c} 2 x & - z \\ x & - z \end{array} | + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 2.2.1.4

Rewrite using the commutative property of multiplication.

$x (y z - 2 \cdot - 1 y z) - y | \begin{array}{c} 2 x & - z \\ x & - z \end{array} | + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 2.2.1.5

Multiply $- 2$ by $- 1$ .

$x (y z + 2 y z) - y | \begin{array}{c} 2 x & - z \\ x & - z \end{array} | + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 2.2.2

Add $y z$ and $2 y z$ .

$x (3 y z) - y | \begin{array}{c} 2 x & - z \\ x & - z \end{array} | + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 3

Evaluate $| \begin{array}{c} 2 x & - z \\ x & - z \end{array} |$ .

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Step 3.1

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

$x (3 y z) - y (2 x (- z) - x (- z)) + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 3.2

Simplify the determinant.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Rewrite using the commutative property of multiplication.

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

Step 3.2.1.2

Multiply $2$ by $- 1$ .

$x (3 y z) - y (- 2 x z - x (- z)) + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 3.2.1.3

Rewrite using the commutative property of multiplication.

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

Step 3.2.1.4

Multiply $- 1$ by $- 1$ .

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

Step 3.2.1.5

Multiply $x$ by $1$ .

$x (3 y z) - y (- 2 x z + x z) + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 3.2.2

Add $- 2 x z$ and $x z$ .

$x (3 y z) - y (- x z) + z | \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$

Step 4

Evaluate $| \begin{array}{c} 2 x & - y \\ x & 2 y \end{array} |$ .

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Step 4.1

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

$x (3 y z) - y (- x z) + z (2 x (2 y) - x (- y))$

Step 4.2

Simplify the determinant.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Rewrite using the commutative property of multiplication.

$x (3 y z) - y (- x z) + z (2 \cdot 2 x y - x (- y))$

Step 4.2.1.2

Multiply $2$ by $2$ .

$x (3 y z) - y (- x z) + z (4 x y - x (- y))$

Step 4.2.1.3

Rewrite using the commutative property of multiplication.

$x (3 y z) - y (- x z) + z (4 x y - 1 \cdot - 1 x y)$

Step 4.2.1.4

Multiply $- 1$ by $- 1$ .

$x (3 y z) - y (- x z) + z (4 x y + 1 x y)$

Step 4.2.1.5

Multiply $x$ by $1$ .

$x (3 y z) - y (- x z) + z (4 x y + x y)$

Step 4.2.2

Add $4 x y$ and $x y$ .

$x (3 y z) - y (- x z) + z (5 x y)$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Rewrite using the commutative property of multiplication.

$3 x y z - y (- x z) + z (5 x y)$

Step 5.1.2

Multiply $- y (- x z)$ .

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Step 5.1.2.1

Multiply $- 1$ by $- 1$ .

$3 x y z + 1 y (x z) + z (5 x y)$

Step 5.1.2.2

Multiply $y$ by $1$ .

$3 x y z + y (x z) + z (5 x y)$

Step 5.1.3

Rewrite using the commutative property of multiplication.

$3 x y z + y x z + 5 z x y$

Step 5.2

Add $3 x y z$ and $y x z$ .

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Step 5.2.1

Reorder $y$ and $x$ .

$3 x y z + x y z + 5 z x y$

Step 5.2.2

Add $3 x y z$ and $x y z$ .

$4 x y z + 5 z x y$

Step 5.3

Add $4 x y z$ and $5 z x y$ .

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Step 5.3.1

Move $z$ .

$4 x y z + 5 x y z$

Step 5.3.2

Add $4 x y z$ and $5 x y z$ .

$9 x y z$