Solve {ccc}{5}&{-1}&{3}{0}&{2}&{-1}{5}&{3}&{1}

$$\left| \begin{array} { c c c } { 5 } & { - 1 } & { 3 } \\ { 0 } & { 2 } & { - 1 } \\ { 5 } & { 3 } & { 1 } \end{array} \right|$$

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&-1&3\\0&2&-1\\5&3&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&-1&3&5&-1\\0&2&-1&0&2\\5&3&1&5&3\end{matrix}\right)$$

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

$$5\times 2-\left(-5\right)=15$$

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

$$5\times 2\times 3+3\left(-1\right)\times 5=15$$

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

$$15-15$$

Subtract 15 from 15.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&-1&3\\0&2&-1\\5&3&1\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}2&-1\\3&1\end{matrix}\right))-\left(-det(\left(\begin{matrix}0&-1\\5&1\end{matrix}\right))\right)+3det(\left(\begin{matrix}0&2\\5&3\end{matrix}\right))$$

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

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

Simplify.

$$5\times 5-\left(-5\right)+3\left(-10\right)$$

Add the terms to obtain the final result.

$$0$$

Factor

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