Solve {rrr}{1}&{-3}&{5}{2}&{-1}&{4}{1}&{1}&{0}

$$\left( \begin{array} { r r r } { 1 } & { - 3 } & { 5 } \\ { 2 } & { - 1 } & { 4 } \\ { 1 } & { 1 } & { 0 } \end{array} \right)$$

Calculate Determinant

$$-1$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&-3&5\\2&-1&4\\1&1&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&-3&5&1&-3\\2&-1&4&2&-1\\1&1&0&1&1\end{matrix}\right)$$

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

$$-3\times 4+5\times 2=-2$$

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

$$-5+4=-1$$

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

$$-2-\left(-1\right)$$

Subtract -1 from -2.

$$-1$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&-3&5\\2&-1&4\\1&1&0\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-1&4\\1&0\end{matrix}\right))-\left(-3det(\left(\begin{matrix}2&4\\1&0\end{matrix}\right))\right)+5det(\left(\begin{matrix}2&-1\\1&1\end{matrix}\right))$$

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

$$-4-\left(-3\left(-4\right)\right)+5\left(2-\left(-1\right)\right)$$

Simplify.

$$-4-\left(-3\left(-4\right)\right)+5\times 3$$

Add the terms to obtain the final result.

$$-1$$

Evaluate

$$\left(\begin{matrix}1&-3&5\\2&-1&4\\1&1&0\end{matrix}\right)$$