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Efficient Way to Calculate Determinants

Before we do this, we have to define what a determinant is. 

Determinant

MathJax TeX Test Page The determinant of a $2\times2$ matrix $\begin{bmatrix}a & b\\c & d\end{bmatrix}$ is $ad - bc$. We can define the determinant of an $n \times n$ as the cofactor expression across any row or down any column. Row i: $$\det A = a_{i1}(-1)^{i + 1}\det(A_{i1}) + ... + a_{in}(-1)^{i + n}\det(A_{in})$$ Column j $$\det A = a_{1j}(-1)^{1 + j}\det(A_{1j}) + ... + a_{nj}(-1)^{n + j}\det(A_{nj})$$

Important Facts

MathJax TeX Test Page $$\det(A^T) = \det(A)$$ $$\text{Row reduction maintains the determinant, meaning } \begin{vmatrix}a & b\\c & d\end{vmatrix} = \begin{vmatrix}a & b\\ka + c & kb + d\end{vmatrix}$$ $$\text{Row swapping negates the determinant, meaning } \begin{vmatrix}a & b\\c & d\end{vmatrix} = -\begin{vmatrix}c & d\\a & b\end{vmatrix}$$ $$\text{Row scaling scales the determinant, meaning } k\cdot\begin{vmatrix}a & b\\c & d\end{vmatrix} = \begin{vmatrix}a & b\\kc & kd\end{vmatrix}$$ $$\det(AB) = \det(A)\cdot\det(B)$$

Example One

MathJax TeX Test Page $$\det\left(\begin{vmatrix}4 & 3 & 9\\0 & 2 & 7\\-5 & 10 & 10\end{vmatrix}\right)$$ $$=-5 \cdot \begin{vmatrix}4 & 3 & 9\\0 & 2 & 7\\1 & -2 & -2\end{vmatrix}$$ $$ \text{ Remember that if you scale inside, you must scale outside}$$ $$-5 \cdot \begin{vmatrix}0 & 11 & 17\\0 & 2 & 7\\1 & -2 & -2\end{vmatrix} \text{$r_1 = r_1 + -4r_2$}$$ $$-5 \cdot \begin{vmatrix}0 & 7 & 3\\0 & 2 & 7\\1 & -2 & -2\end{vmatrix} \text{$r_1 = r_1 - 2r_2$}$$ $$\text{Note- I didn't eliminate the first row, I just reduced the size of the numbers, making the calculation easier}$$ $$-5 \cdot \begin{vmatrix}7 & 3\\2 & 7\end{vmatrix} = -5 \cdot (49-6) = -215$$

Example Two

MathJax TeX Test Page $$\det\left(\begin{vmatrix}5 & -1 & 8\\2 & -2 & -8\\-7 & -2 & -2\end{vmatrix}\right)$$ Remember, the determinant of the transpose is the same as the determinant of the regular matrix, so we can do column operations!!! (basically, you transpose, then do a row operation, the transpose back) $$\det\left(\begin{vmatrix}0 & -1 & 8\\-8 & -2 & -8\\-17 & -2 & -2\end{vmatrix}\right) \text{$C_1 = C_1 + 5C_2$}$$ $$\det\left(\begin{vmatrix}0 & -1 & 0\\-8 & -2 & -24\\-17 & -2 & -18\end{vmatrix}\right) \text{$C_3 = C_3 + 8C_2$}$$ $$-6\det\left(\begin{vmatrix}0 & -1 & 0\\-8 & -2 & 4\\-17 & -2 & 3\end{vmatrix}\right) \text{$C_3 = C_3/6$}$$ $$-6\det\left(\begin{vmatrix}0 & -1 & 0\\0 & -2 & 4\\-11 & -2 & 3\end{vmatrix}\right) \text{$C_1 = C_1 + 2C_3$}$$ $$-6\det\left(\begin{vmatrix}0 & 4\\-11 & 3\end{vmatrix}\right) = -6 \cdot 44 = \boxed{-264}$$

Vector Spaces and Subspaces

TFAE