Algebra Examples

$[\begin{array}{c} x & 0 & 0 \\ 2 & x & 1 \\ 5 & 2 & 1 \end{array}]$

Step 1

Multiply every element in the $1 s t$ row by its cofactor and add.

$x | \begin{array}{c} x & 1 \\ 2 & 1 \end{array} | + 0 | \begin{array}{c} 2 & 1 \\ 5 & 1 \end{array} | + 0 | \begin{array}{c} 2 & x \\ 5 & 2 \end{array} |$

Step 2

Multiply $0$ by $| \begin{array}{c} 2 & 1 \\ 5 & 1 \end{array} |$ .

$x | \begin{array}{c} x & 1 \\ 2 & 1 \end{array} | + 0 + 0 | \begin{array}{c} 2 & x \\ 5 & 2 \end{array} |$

Step 3

Multiply $0$ by $| \begin{array}{c} 2 & x \\ 5 & 2 \end{array} |$ .

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

Step 4

Evaluate $| \begin{array}{c} x & 1 \\ 2 & 1 \end{array} |$ .

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

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

Simplify each term.

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Multiply $x$ by $1$ .

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

Multiply $- 2$ by $1$ .

$x (x - 2) + 0 + 0$

Step 5

Simplify the determinant.

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Combine the opposite terms in $x (x - 2) + 0 + 0$ .

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Add $x (x - 2)$ and $0$ .

$x (x - 2) + 0$

Add $x (x - 2)$ and $0$ .

$x (x - 2)$

Apply the distributive property.

$x \cdot x + x \cdot - 2$

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 2$

Move $- 2$ to the left of $x$ .

$x^{2} - 2 x$