Solve {ccc}{overline{T}}&{j}&{overline{k}}{-12}&{2}&{-4}{-4}&{2}&{3}

$$\left| \begin{array} { c c c } { \overline { T } } & { j } & { \overline { k } } \\ { - 12 } & { 2 } & { - 4 } \\ { - 4 } & { 2 } & { 3 } \end{array} \right|$$

Evaluate

$$52j$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0&j&0\\-12&2&-4\\-4&2&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}0&j&0&0&j\\-12&2&-4&-12&2\\-4&2&3&-4&2\end{matrix}\right)$$

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

$$j\left(-4\right)\left(-4\right)=16j$$

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

$$3\left(-12\right)j=-36j$$

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

$$16j-\left(-36j\right)$$

Subtract -36j from 16j.

$$52j$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0&j&0\\-12&2&-4\\-4&2&3\end{matrix}\right))$$

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.

$$-jdet(\left(\begin{matrix}-12&-4\\-4&3\end{matrix}\right))$$

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

$$-j\left(-12\times 3-\left(-4\left(-4\right)\right)\right)$$

Simplify.

$$-j\left(-52\right)$$

Add the terms to obtain the final result.

$$52j$$

Integrate w.r.t. j

$$С+26j^{2}$$