Solve {c|c|c}{a}&{b}&{b}hline+15&{18}&{}hline-28&{-12}&{}hline

$$\left. \begin{array} { c | c | c } { a } & { b } & { b } \\ \hline + 15 & { 18 } & { } \\ \hline - 28 & { - 12 } & { } \\ \hline \end{array} \right.$$

Calculate Determinant

$$324b$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}a&b&b\\15&18&0\\-28&-12&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}a&b&b&a&b\\15&18&0&15&18\\-28&-12&0&-28&-12\end{matrix}\right)$$

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

$$b\times 15\left(-12\right)=-180b$$

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

$$-28\times 18b=-504b$$

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

$$-180b-\left(-504b\right)$$

Subtract -504b from -180b.

$$324b$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}a&b&b\\15&18&0\\-28&-12&0\end{matrix}\right))$$

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.

$$adet(\left(\begin{matrix}18&0\\-12&0\end{matrix}\right))-bdet(\left(\begin{matrix}15&0\\-28&0\end{matrix}\right))+bdet(\left(\begin{matrix}15&18\\-28&-12\end{matrix}\right))$$

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

$$b\left(15\left(-12\right)-\left(-28\times 18\right)\right)$$

Simplify.

$$b\times 324$$

Add the terms to obtain the final result.

$$324b$$

Evaluate

$$\left(\begin{matrix}a&b&b\\15&18&0\\-28&-12&0\end{matrix}\right)$$