Solve {lll}{a}&{b}&{c}{x}&{y}&{z}{a}&{b}&{c}

$$\left| \begin{array} { l l l } { a } & { b } & { c } \\ { x } & { y } & { z } \\ { a } & { b } & { c } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}a&b&c\\x&y&z\\a&b&c\end{matrix}\right))$$

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

$$\left(\begin{matrix}a&b&c&a&b\\x&y&z&x&y\\a&b&c&a&b\end{matrix}\right)$$

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

$$ayc+bza+cxb=bcx+acy+abz$$

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

$$bcx+acy+abz-\left(bcx+acy+abz\right)$$

Subtract ayc+bza+cxb from ayc+bza+cxb.

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Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}a&b&c\\x&y&z\\a&b&c\end{matrix}\right))$$

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.

$$adet(\left(\begin{matrix}y&z\\b&c\end{matrix}\right))-bdet(\left(\begin{matrix}x&z\\a&c\end{matrix}\right))+cdet(\left(\begin{matrix}x&y\\a&b\end{matrix}\right))$$

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

$$a\left(yc-bz\right)-b\left(xc-az\right)+c\left(xb-ay\right)$$

Simplify.

$$a\left(cy-bz\right)-b\left(cx-az\right)+c\left(bx-ay\right)$$

Add the terms to obtain the final result.

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Factor

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