Solve {lll}{3}&{-3}&{4}{2}&{-3}&{4}{0}&{-1}

$$\left( \begin{array} { l l l } { 3 } & { - 3 } & { 4 } \\ { 2 } & { - 3 } & { 4 } \\ { 0 } & { - 1 } \end{array} \right)$$

Calculate Determinant

$$4$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&-3&4\\2&-3&4\\0&-1&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&-3&4&3&-3\\2&-3&4&2&-3\\0&-1&0&0&-1\end{matrix}\right)$$

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

$$4\times 2\left(-1\right)=-8$$

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

$$-4\times 3=-12$$

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

$$-8-\left(-12\right)$$

Subtract -12 from -8.

$$4$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&-3&4\\2&-3&4\\0&-1&0\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}-3&4\\-1&0\end{matrix}\right))-\left(-3det(\left(\begin{matrix}2&4\\0&0\end{matrix}\right))\right)+4det(\left(\begin{matrix}2&-3\\0&-1\end{matrix}\right))$$

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

$$3\left(-\left(-4\right)\right)+4\times 2\left(-1\right)$$

Simplify.

$$3\times 4+4\left(-2\right)$$

Add the terms to obtain the final result.

$$4$$

Evaluate

$$\left(\begin{matrix}3&-3&4\\2&-3&4\\0&-1&0\end{matrix}\right)$$