Solve {ccc}{1}&{3}&{1}{1}&{-1}&{m}{2}&{0}&{1}

$$\left| \begin{array} { c c c } { 1 } & { 3 } & { 1 } \\ { 1 } & { - 1 } & { m } \\ { 2 } & { 0 } & { 1 } \end{array} \right|$$

Evaluate

$$6m-2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&1\\1&-1&m\\2&0&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&1&1&3\\1&-1&m&1&-1\\2&0&1&2&0\end{matrix}\right)$$

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

$$-1+3m\times 2=6m-1$$

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

$$2\left(-1\right)+3=1$$

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

$$6m-1-1$$

Subtract 1 from -1+6m.

$$6m-2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&1\\1&-1&m\\2&0&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-1&m\\0&1\end{matrix}\right))-3det(\left(\begin{matrix}1&m\\2&1\end{matrix}\right))+det(\left(\begin{matrix}1&-1\\2&0\end{matrix}\right))$$

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

$$-1-3\left(1-2m\right)-2\left(-1\right)$$

Simplify.

$$-1-3\left(1-2m\right)+2$$

Add the terms to obtain the final result.

$$6m-2$$

Integrate w.r.t. m

$$3m^{2}-2m+С$$