Keywords
[1] The determinant of \(A = \begin{bmatrix} a & b \\\ c & d \end{bmatrix}\) is \(ad-bc\). Singular matrix \(A = \begin{bmatrix} a & xa \\\ c & xc \end{bmatrix}\) has det = 0.
[2] Row change reverses signs: \(PA = \begin{bmatrix} 0 & 1 \\\ 1 & 0\end{bmatrix} \begin{bmatrix} c & d \\\ a & b \end{bmatrix}\) has \(det(PA) = bc-ad=-det(A)\)
[3] The determinant of \(\begin{bmatrix} xa & yA & xb + yB \\\ c & d\end{bmatrix}\) is \(x(ad-bc) + y(Ad-Bc)\): Determinant is linear in row 1 by itself.
[4] Elimination \(EA = \begin{bmatrix} a & b \\\ 0 & d-\frac{c}{a}b \end{bmatrix}\), \(det(EA)=a(d-\frac{c}{a}b)\) = product of pivots = \(det(A)\)
[5] If \(A\) is \(n\) by \(n\) then [1], [2], [3], [4] remain true and always \(det(BA) = det(B)det(A)\) and \(det(A^T)=det(A)\)
Determinant
The determinant of a square matrix is a single number.
The determinant has zero when the matrix has no inverse. When \(A\) is invertible, \(det(A^{-1}) = 1\ /\ (det(A))\)
\[A = \begin{bmatrix} a & b \\\ c & d \end{bmatrix}\] \[det(A) = \lvert A \rvert = ad - bc\] \[A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\\ -c & a \end{bmatrix}\]Product of Pivots
The product of the pivots is the determinant:
\[a\left(d - \frac{c}{a}b\right) = ad - bc = det(A)\]Determinant Rules
[1] The determinant of \(n\) by \(n\) identity matrix if \(1\).
\[\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1\][2] The determinant changes sign when two rows are exchanged (sign reversal)
\[\begin{vmatrix} c & d \\\ a & b \end{vmatrix} = -\begin{vmatrix} a & b \\\ c & d\end{vmatrix}\]From this property, we can find \(det(P)\) for any permutation matrix. If there was an even number of row changes, then \(det(P)=1\), when there’s an odd number of row exchanges, \(det(P)=-1\).
[3] The determinant is a linear function of each row separately (other rows stay fixed)
If the first row is multiplied by \(t\), the determinant is multiplied by \(t\). If the first rows are added, the determinants are added. Remember that this rule only applies when the other rows don’t change!
\[\begin{vmatrix} ta & tb \\\ c d \end{vmatrix} = t\begin{vmatrix} a & b \\\ c & d \end{vmatrix}\] \[\begin{vmatrix} a+a' & b+b' \\\ c & d\end{vmatrix} = \begin{vmatrix} a & b \\\ c & d\end{vmatrix} + \begin{vmatrix} a' & b' \\\ c & d \end{vmatrix}\]This rule also applies to \(n\) by \(n\) matrix \(A\) when only one row changes
\[A = \begin{vmatrix} \textbf{4} & \textbf{8} & \textbf{8} \\\ 0 & 1 & 1 \\\ 0 & 0 & 1\end{vmatrix} = \textbf{4}\begin{vmatrix} \textbf{1} & \textbf{2} & \textbf{2} \\\ 0 & 1 & 1 \\\ 0 & 0 & 1\end{vmatrix}\] \[\begin{vmatrix} \textbf{4} & \textbf{8} & \textbf{8} \\\ 0 & 1 & 1 \\\ 0 & 0 & 1 \end{vmatrix} = \begin{vmatrix} \textbf{4} & \textbf{0} & \textbf{0} \\\ 0 & 1 & 1 \\\ 0 & 0 & 1\end{vmatrix} + \begin{vmatrix} \textbf{0} & \textbf{8} & \textbf{8} \\\ 0 & 1 & 1 \\\ 0 & 0 & 1\end{vmatrix}\][WARNING!] This linearity DOES NOT mean \(det(A+B) = det(A) + det(B)\)! This is false
For example, \(det(2I) \neq 2det(I)\). To obtain \(2I\), we have to multiply both rows by 2. Hence,
\[\begin{vmatrix} t & 0 \\\ 0 & t\end{vmatrix} = t^2\][4] If two rows of \(A\) are equal, then \(det(A)=0\)
Intuitively, a matrix with two equals rows is singular so its determinant must be zero. Another reasoning is from rule [2]. If we exchange the two identical rows, the sign changes but since it’s still the same matrix, the only number which satisfies \(-D=D\) is \(0\).
[5] Subtracting a multiple of one row from another row leaves \(det(A)\) unchanged
\[\begin{vmatrix} a & b \\\ c-la & d-lb\end{vmatrix} = \begin{vmatrix} a & b \\\ c & d \end{vmatrix}\]Rule [3] splits the left side into the right side plus \(-l\begin{vmatrix} a & b \\\ a & b\end{vmatrix}\) which is equal to \(0\) by rule [4].
The determinant is not changed by the usual elimination from \(A\) to \(U\) (except possibly sign): \(det(A)=\pm(U)\).
[6] A matrix with a row of zeros has \(det(A)=0\)
Add a non-zero row to a zero row then the matrix has two equal rows so \(det(A)=0\).
[7] If \(A\) is triangular then \(det(A)=a_{11}a_{22}\cdots a_{nn}\) = product of diagonal entries
Suppose all diagonal entries are nonzero. Remove the off-diagonal entries by elimination. By rule [5], the determinant is not changed. Using rule [3], factor each diagonal entry sequentially. Then the determinant becomes the product of pivots.
[8] If \(A\) is singular, then \(det(A)=0\). If \(A\) is invertible then \(det(A) \neq 0\)
If \(A\) is signular then the eliminated matrix \(U\) has a zero row so its determinant is \(0\). If \(A\) is invertible, then \(U\) has the pivots along its diagonal.
[9] \(det(AB) = det(A)det(B)\)
When \(B=A^{-1}\), the formula becomes \(det(I) = det(A)det(A^{-1})\) where \(det(I) = 1\).
\[det(A) det(A^{-1}) = 1\][10] \(det(A^T)=det(A)\)
For example,
\[\begin{vmatrix} a & b \\\ c & d\end{vmatrix} = \begin{vmatrix} a & c \\\ b & d\end{vmatrix}\]Every rule also works for columns!
Every rule for the rows we talked about also apply to the columns since \(\lvert A \rvert = \lvert A^T \rvert\).
References
[1] Introduction to Linear Algebra, 5th edition