Solve {ccc}{0}&{1,5}&{2}{0,4}&{0}&{0}{0}&{0,5}&{0,6}

$$\left( \begin{array} { c c c } { 0 } & { 1,5 } & { 2 } \\ { 0,4 } & { 0 } & { 0 } \\ { 0 } & { 0,5 } & { 0,6 } \end{array} \right)$$

Calculate Determinant

$$0,04$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0&1,5&2\\0,4&0&0\\0&0,5&0,6\end{matrix}\right))$$

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

$$\left(\begin{matrix}0&1,5&2&0&1,5\\0,4&0&0&0,4&0\\0&0,5&0,6&0&0,5\end{matrix}\right)$$

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

$$2\times 0,4\times 0,5=0,4$$

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

$$0,6\times 0,4\times 1,5=0,36$$

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

$$0,4-0,36$$

Subtract 0,36 from 0,4 by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$0,04$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0&1,5&2\\0,4&0&0\\0&0,5&0,6\end{matrix}\right))$$

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.

$$-1,5det(\left(\begin{matrix}0,4&0\\0&0,6\end{matrix}\right))+2det(\left(\begin{matrix}0,4&0\\0&0,5\end{matrix}\right))$$

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

$$-1,5\times 0,4\times 0,6+2\times 0,4\times 0,5$$

Simplify.

$$-1,5\times 0,24+2\times 0,2$$

Add the terms to obtain the final result.

$$0,04$$

Evaluate

$$\left(\begin{matrix}0&1,5&2\\0,4&0&0\\0&0,5&0,6\end{matrix}\right)$$