Solve {ccc}{i}&{j}&{k}{4}&{-1}&{0}{5}&{0}&{1}

$$\left| \begin{array} { c c c } { i } & { j } & { k } \\ { 4 } & { - 1 } & { 0 } \\ { 5 } & { 0 } & { 1 } \end{array} \right|$$

Evaluate

$$-i+5k-4j$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&j&k\\4&-1&0\\5&0&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&j&k&i&j\\4&-1&0&4&-1\\5&0&1&5&0\end{matrix}\right)$$

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

$$-i=-i$$

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

$$5\left(-1\right)k+4j=4j-5k$$

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

$$-i-\left(4j-5k\right)$$

Subtract -5k+4j from -i.

$$-i+5k-4j$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&j&k\\4&-1&0\\5&0&1\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}-1&0\\0&1\end{matrix}\right))-jdet(\left(\begin{matrix}4&0\\5&1\end{matrix}\right))+kdet(\left(\begin{matrix}4&-1\\5&0\end{matrix}\right))$$

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

$$-i-j\times 4+k\left(-5\left(-1\right)\right)$$

Simplify.

$$-i-j\times 4+k\times 5$$

Add the terms to obtain the final result.

$$-i+5k-4j$$