Find the Determinant [[a,1,0],[1,0,b],[c,0,1]]

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

Set up the determinant by breaking it into smaller components.

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

Find the determinant of $- 1 | \begin{array}{c} 1 & b \\ c & 1 \end{array} |$ .

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

$- 1 ((1) (1) - c b) + 0 | \begin{array}{c} a & 0 \\ c & 1 \end{array} | + 0 | \begin{array}{c} a & 0 \\ 1 & b \end{array} |$

Simplify the determinant.

Multiply $1$ by $1$ .

$- 1 (1 - c b) + 0 | \begin{array}{c} a & 0 \\ c & 1 \end{array} | + 0 | \begin{array}{c} a & 0 \\ 1 & b \end{array} |$

Apply the distributive property.

$- 1 \cdot 1 - 1 (- c b) + 0 | \begin{array}{c} a & 0 \\ c & 1 \end{array} | + 0 | \begin{array}{c} a & 0 \\ 1 & b \end{array} |$

Multiply $- 1$ by $1$ .

$- 1 - 1 (- c b) + 0 | \begin{array}{c} a & 0 \\ c & 1 \end{array} | + 0 | \begin{array}{c} a & 0 \\ 1 & b \end{array} |$

Multiply $- 1 (- c b)$ .

Multiply $- 1$ by $- 1$ .

$- 1 + 1 (c b) + 0 | \begin{array}{c} a & 0 \\ c & 1 \end{array} | + 0 | \begin{array}{c} a & 0 \\ 1 & b \end{array} |$

Multiply $c$ by $1$ .

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

Since the matrix is multiplied by $0$ , the determinant is $0$ .

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

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$- 1 + c b + 0 + 0$

Combine the opposite terms in $- 1 + c b + 0 + 0$ .

Add $- 1 + c b$ and $0$ .

$- 1 + c b + 0$

Add $- 1 + c b$ and $0$ .

$- 1 + c b$

Find the Determinant [[a,1,0],[1,0,b],[c,0,1]]

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