Solve {ccc}{i}&{j}&{k}{2}&{1}&{-3}{5}&{5}&{-4}

$$\left| \begin{array} { c c c } { i } & { j } & { k } \\ { 2 } & { 1 } & { - 3 } \\ { 5 } & { 5 } & { - 4 } \end{array} \right|$$

Evaluate

$$11i+5k-7j$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-4i+j\left(-3\right)\times 5+k\times 2\times 5=-4i+10k-15j$$

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

$$5k+5\times \left(-3i\right)-4\times 2j=-15i+5k-8j$$

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

$$-4i+10k-15j-\left(-15i+5k-8j\right)$$

Subtract 5k-15i-8j from -4i-15j+10k.

$$11i+5k-7j$$

Steps Using Expansion by Minors

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

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

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.

$$idet(\left(\begin{matrix}1&-3\\5&-4\end{matrix}\right))-jdet(\left(\begin{matrix}2&-3\\5&-4\end{matrix}\right))+kdet(\left(\begin{matrix}2&1\\5&5\end{matrix}\right))$$

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

$$i\left(-4-5\left(-3\right)\right)-j\left(2\left(-4\right)-5\left(-3\right)\right)+k\left(2\times 5-5\right)$$

Simplify.

$$11i-j\times 7+k\times 5$$

Add the terms to obtain the final result.

$$11i+5k-7j$$