Linear Algebra Examples

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

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 column $1$ 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.

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

Step 1.4

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

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

Step 1.5

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

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

Step 1.6

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

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

Step 1.7

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

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

Step 1.8

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

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

Step 1.9

Add the terms together.

$p | \begin{array}{c} 1 & q \\ 1 & p \end{array} | - q | \begin{array}{c} 1 & 1 \\ 1 & p \end{array} | + 1 | \begin{array}{c} 1 & 1 \\ 1 & q \end{array} |$

Step 2

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

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Step 2.1

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

$p (1 p - q) - q | \begin{array}{c} 1 & 1 \\ 1 & p \end{array} | + 1 | \begin{array}{c} 1 & 1 \\ 1 & q \end{array} |$

Step 2.2

Multiply $p$ by $1$ .

$p (p - q) - q | \begin{array}{c} 1 & 1 \\ 1 & p \end{array} | + 1 | \begin{array}{c} 1 & 1 \\ 1 & q \end{array} |$

Step 3

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

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Step 3.1

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

$p (p - q) - q (1 p - 1 \cdot 1) + 1 | \begin{array}{c} 1 & 1 \\ 1 & q \end{array} |$

Step 3.2

Simplify each term.

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Step 3.2.1

Multiply $p$ by $1$ .

$p (p - q) - q (p - 1 \cdot 1) + 1 | \begin{array}{c} 1 & 1 \\ 1 & q \end{array} |$

Step 3.2.2

Multiply $- 1$ by $1$ .

$p (p - q) - q (p - 1) + 1 | \begin{array}{c} 1 & 1 \\ 1 & q \end{array} |$

Step 4

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

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Step 4.1

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

$p (p - q) - q (p - 1) + 1 (1 q - 1 \cdot 1)$

Step 4.2

Simplify each term.

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Step 4.2.1

Multiply $q$ by $1$ .

$p (p - q) - q (p - 1) + 1 (q - 1 \cdot 1)$

Step 4.2.2

Multiply $- 1$ by $1$ .

$p (p - q) - q (p - 1) + 1 (q - 1)$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Apply the distributive property.

$p \cdot p + p (- q) - q (p - 1) + 1 (q - 1)$

Step 5.1.2

Multiply $p$ by $p$ .

$p^{2} + p (- q) - q (p - 1) + 1 (q - 1)$

Step 5.1.3

Rewrite using the commutative property of multiplication.

$p^{2} - p q - q (p - 1) + 1 (q - 1)$

Step 5.1.4

Apply the distributive property.

$p^{2} - p q - q p - q \cdot - 1 + 1 (q - 1)$

Step 5.1.5

Multiply $- q \cdot - 1$ .

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Step 5.1.5.1

Multiply $- 1$ by $- 1$ .

$p^{2} - p q - q p + 1 q + 1 (q - 1)$

Step 5.1.5.2

Multiply $q$ by $1$ .

$p^{2} - p q - q p + q + 1 (q - 1)$

Step 5.1.6

Multiply $q - 1$ by $1$ .

$p^{2} - p q - q p + q + q - 1$

Step 5.2

Subtract $q p$ from $- p q$ .

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Step 5.2.1

Move $q$ .

$p^{2} - p q - 1 p q + q + q - 1$

Step 5.2.2

Subtract $p q$ from $- p q$ .

$p^{2} - 2 p q + q + q - 1$

Step 5.3

Add $q$ and $q$ .

$p^{2} - 2 p q + 2 q - 1$