Solve {lll}{1}&{2}&{3}{x}&{y}&{z}{5}&{7}&{8}

$$\left( \begin{array} { l l l } { 1 } & { 2 } & { 3 } \\ { x } & { y } & { z } \\ { 5 } & { 7 } & { 8 } \end{array} \right)$$

Calculate Determinant

$$5x-7y+3z$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\x&y&z\\5&7&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\x&y&z&x&y\\5&7&8&5&7\end{matrix}\right)$$

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

$$y\times 8+2z\times 5+3x\times 7=21x+8y+10z$$

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

$$5y\times 3+7z+8x\times 2=16x+15y+7z$$

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

$$21x+8y+10z-\left(16x+15y+7z\right)$$

Subtract 15y+7z+16x from 8y+10z+21x.

$$5x-7y+3z$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\x&y&z\\5&7&8\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}y&z\\7&8\end{matrix}\right))-2det(\left(\begin{matrix}x&z\\5&8\end{matrix}\right))+3det(\left(\begin{matrix}x&y\\5&7\end{matrix}\right))$$

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

$$y\times 8-7z-2\left(x\times 8-5z\right)+3\left(x\times 7-5y\right)$$

Simplify.

$$8y-7z-2\left(8x-5z\right)+3\left(7x-5y\right)$$

Add the terms to obtain the final result.

$$5x-7y+3z$$

Evaluate

$$\left(\begin{matrix}1&2&3\\x&y&z\\5&7&8\end{matrix}\right)$$