Solve Matrix

$$\left( \begin{array} { l l } { 1 } & { 3 } & { 2 } \\ { 1 } & { 1 } & { - 1 } \\ { 7 } & { 7 } & { 5 } \end{array} \right)$$

Calculate Determinant

$$-24$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}1&3&2\\1&1&-1\\7&7&5\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}1&3&2&1&3\\1&1&-1&1&1\\7&7&5&7&7\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$5+3\left(-1\right)\times 7+2\times 7=-2$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$7\times 2+7\left(-1\right)+5\times 3=22$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$-2-22$$

Subtract 22 from -2.

$$-24$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}1&3&2\\1&1&-1\\7&7&5\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$det(\left(\begin{matrix}1&-1\\7&5\end{matrix}\right))-3det(\left(\begin{matrix}1&-1\\7&5\end{matrix}\right))+2det(\left(\begin{matrix}1&1\\7&7\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$5-7\left(-1\right)-3\left(5-7\left(-1\right)\right)+2\left(7-7\right)$$

Simplify.

$$12-3\times 12$$

Add the terms to obtain the final result.

$$-24$$

Evaluate

$$\left(\begin{matrix}1&3&2\\1&1&-1\\7&7&5\end{matrix}\right)$$