Solve operatorname{det}{ccc}{0.8}&{0.4}&{1}{1}&{-0.7}&{0}{0}&{1}&{2}

$$\operatorname { det } \left( \begin{array} { c c c } { 0.8 } & { 0.4 } & { 1 } \\ { 1 } & { - 0.7 } & { 0 } \\ { 0 } & { 1 } & { 2 } \end{array} \right)$$

Evaluate

$$-0.92$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0.8&0.4&1\\1&-0.7&0\\0&1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}0.8&0.4&1&0.8&0.4\\1&-0.7&0&1&-0.7\\0&1&2&0&1\end{matrix}\right)$$

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

$$0.8\left(-0.7\right)\times 2+1=-0.12$$

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

$$2\times 0.4=0.8$$

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

$$-0.12-0.8$$

Subtract 0.8 from -0.12 by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-0.92$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0.8&0.4&1\\1&-0.7&0\\0&1&2\end{matrix}\right))$$

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.

$$0.8det(\left(\begin{matrix}-0.7&0\\1&2\end{matrix}\right))-0.4det(\left(\begin{matrix}1&0\\0&2\end{matrix}\right))+det(\left(\begin{matrix}1&-0.7\\0&1\end{matrix}\right))$$

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

$$0.8\left(-0.7\right)\times 2-0.4\times 2+1$$

Simplify.

$$0.8\left(-1.4\right)-0.4\times 2+1$$

Add the terms to obtain the final result.

$$-0.92$$

Factor

$$-0.92$$