Solve {ccc}{i}&{j}&{k}{1}&{-2}&{-3}{0}&{-1}&{1}

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

Evaluate

$$-5i-k-j$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2i+k\left(-1\right)=-2i-k$$

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

$$-\left(-3i\right)+j=j+3i$$

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

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

Subtract 3i+j from -2i-k.

$$-5i-k-j$$

Steps Using Expansion by Minors

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

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

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.

$$idet(\left(\begin{matrix}-2&-3\\-1&1\end{matrix}\right))-jdet(\left(\begin{matrix}1&-3\\0&1\end{matrix}\right))+kdet(\left(\begin{matrix}1&-2\\0&-1\end{matrix}\right))$$

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

$$i\left(-2-\left(-\left(-3\right)\right)\right)-j+k\left(-1\right)$$

Simplify.

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

Add the terms to obtain the final result.

$$-5i-k-j$$