Find the Determinant [[4,1,-6],[5,-1,1],[3,4,4]]

$[\begin{array}{c} 4 & 1 & - 6 \\ 5 & - 1 & 1 \\ 3 & 4 & 4 \end{array}]$

Step 1

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

$4 | \begin{array}{c} - 1 & 1 \\ 4 & 4 \end{array} | - | \begin{array}{c} 5 & 1 \\ 3 & 4 \end{array} | - 6 | \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$

Step 2

Evaluate $| \begin{array}{c} - 1 & 1 \\ 4 & 4 \end{array} |$ .

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

$4 (- 1 \cdot 4 - 4 \cdot 1) - | \begin{array}{c} 5 & 1 \\ 3 & 4 \end{array} | - 6 | \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 1$ by $4$ .

$4 (- 4 - 4 \cdot 1) - | \begin{array}{c} 5 & 1 \\ 3 & 4 \end{array} | - 6 | \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$

Multiply $- 4$ by $1$ .

$4 (- 4 - 4) - | \begin{array}{c} 5 & 1 \\ 3 & 4 \end{array} | - 6 | \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$

Subtract $4$ from $- 4$ .

$4 \cdot - 8 - | \begin{array}{c} 5 & 1 \\ 3 & 4 \end{array} | - 6 | \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$

Step 3

Evaluate $| \begin{array}{c} 5 & 1 \\ 3 & 4 \end{array} |$ .

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

$4 \cdot - 8 - (5 \cdot 4 - 3 \cdot 1) - 6 | \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $5$ by $4$ .

$4 \cdot - 8 - (20 - 3 \cdot 1) - 6 | \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$

Multiply $- 3$ by $1$ .

$4 \cdot - 8 - (20 - 3) - 6 | \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$

Subtract $3$ from $20$ .

$4 \cdot - 8 - 1 \cdot 17 - 6 | \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$

Step 4

Evaluate $| \begin{array}{c} 5 & - 1 \\ 3 & 4 \end{array} |$ .

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

$4 \cdot - 8 - 1 \cdot 17 - 6 (5 \cdot 4 - 3 \cdot - 1)$

Simplify the determinant.

Simplify each term.

Multiply $5$ by $4$ .

$4 \cdot - 8 - 1 \cdot 17 - 6 (20 - 3 \cdot - 1)$

Multiply $- 3$ by $- 1$ .

$4 \cdot - 8 - 1 \cdot 17 - 6 (20 + 3)$

Add $20$ and $3$ .

$4 \cdot - 8 - 1 \cdot 17 - 6 \cdot 23$

Step 5

Simplify the determinant.

Simplify each term.

Multiply $4$ by $- 8$ .

$- 32 - 1 \cdot 17 - 6 \cdot 23$

Multiply $- 1$ by $17$ .

$- 32 - 17 - 6 \cdot 23$

Multiply $- 6$ by $23$ .

$- 32 - 17 - 138$

Subtract $17$ from $- 32$ .

$- 49 - 138$

Subtract $138$ from $- 49$ .

$- 187$

Find the Determinant [[4,1,-6],[5,-1,1],[3,4,4]]

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