Determinants

A=(abcd),det(A)=ad−bc (can be generalized to any SQUARE matrix and is also notated as |A|)Theorem: A is invertible if and only if det(A)≠0A = \begin{pmatrix} a & b \\ c & d\end{pmatrix}, det(A) = ad - bc \text{ (can be generalized to any SQUARE matrix and is also notated as |A|)}\\ \text{Theorem: A is invertible if and only if } det(A) \neq 0 \\

Examples

If A is a 1 x 1 matrix, then det(A)=A, such asdet([−7])=−7det(11−11)=1∗1−1∗(−1)=2\text{If A is a 1 x 1 matrix, then } det(A) = A \text{, such as} det([-7]) = -7 \\ det\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} = 1 * 1 - 1 * (-1) = 2 \\

Finding the Determinant of a 3 x 3 matrix

det(230203−111)det\begin{pmatrix} 2 & 3 & 0 \\ 2 & 0 & 3 \\ -1 & 1 & 1 \end{pmatrix}

Step 1: Choose a row OR column (with many zeroes)

  • In this example, choose row 1 because it has a 0

Step 2: Construct a sign table, where each sign alternates from the last

[+−+−+−+−+]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}

This signifies what we should be doing with our chosen row; we should do +2 * something - 3 * something + 0 * something

  • This is why we choose a row with zeroes; makes it easier because we don’t need to compute as many terms

Step 3: Cancelling time

  • Find the determinants of the elements of the rows/columns that the numbers aren’t in
2∗det(0311)−3∗det(23−11)+0∗det(20−11)=2(−3)−3(5)=−212 * det\begin{pmatrix}0 & 3 \\ 1 & 1\end{pmatrix} - 3 * det\begin{pmatrix}2 & 3 \\ -1 & 1\end{pmatrix} + 0 * det\begin{pmatrix}2 & 0 \\ -1 & 1\end{pmatrix} \\ = 2(-3) - 3(5) = -21

More Examples

det(10−1305697)=0∗det(3667)−0∗det(1−167)+9∗det(1−135)=−9(5+3)=−72det\begin{pmatrix} 1 & 0 & -1 \\ 3 & 0 & 5 \\ 6 & 9 & 7 \end{pmatrix} = 0 * det\begin{pmatrix} 3 & 6 \\ 6 & 7 \end{pmatrix} - 0 * det\begin{pmatrix} 1 & -1 \\ 6 & 7 \end{pmatrix} + 9 * det\begin{pmatrix} 1 & -1 \\ 3 & 5 \end{pmatrix} \\ = -9(5 + 3) = -72 det(100005401010γπe342π+eλ0023πe0057)=1∗det(401010γe342π+e00230057)=1∗3∗det(41010γ023057)=3∗4∗det(2357)=12∗(14−15)=−12det\begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ \sqrt{5} & 4 & 0 & 10^{10} & \gamma \\ \pi & e & 3 & 42 & \pi + e \\ \lambda & 0 & 0 & 2 & 3 \\ \pi^e & 0 & 0 & 5 & 7 \end{pmatrix} = 1 * det\begin{pmatrix} 4 & 0 & 10^{10} & \gamma \\ e & 3 & 42 & \pi + e \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 5 & 7 \end{pmatrix} \\ = 1 * 3 * det\begin{pmatrix} 4 & 10^{10} & \gamma \\ 0 & 2 & 3 \\ 0 & 5 & 7 \end{pmatrix} = 3 * 4 * det\begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \\ = 12 * (14 - 15) = -12