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

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

Evaluate

$$4j-k+3i$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-3i+j\times 3\times 3+k\times 5\left(-2\right)=9j-10k-3i$$

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

$$3\left(-3\right)k-2\times \left(3i\right)+5j=5j-9k-6i$$

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

$$9j-10k-3i-\left(5j-9k-6i\right)$$

Subtract -9k-6i+5j from -3i+9j-10k.

$$4j-k+3i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3i-j\left(-4\right)+k\left(-1\right)$$

Add the terms to obtain the final result.

$$4j-k+3i$$