Solve {ccc}{1}&{3}&{4}{5}&{78}{4}&{21}

$$\left| \begin{array} { c c c } { 1 } & { 3 } & { 4 } \\ { 5 } & { 78 } \\ { 4 } & { 21 } \end{array} \right|$$

Evaluate

$$-828$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&4\\5&78&0\\4&21&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&4&1&3\\5&78&0&5&78\\4&21&0&4&21\end{matrix}\right)$$

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

$$4\times 5\times 21=420$$

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

$$4\times 78\times 4=1248$$

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

$$420-1248$$

Subtract 1248 from 420.

$$-828$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&4\\5&78&0\\4&21&0\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}78&0\\21&0\end{matrix}\right))-3det(\left(\begin{matrix}5&0\\4&0\end{matrix}\right))+4det(\left(\begin{matrix}5&78\\4&21\end{matrix}\right))$$

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

$$4\left(5\times 21-4\times 78\right)$$

Simplify.

$$4\left(-207\right)$$

Add the terms to obtain the final result.

$$-828$$

Factor

$$-828$$