24.3 Determinants
Determinant of a matrix written as det\(A=|A|\).
Let \(A= \begin {pmatrix} a & b\\ c & d\\ \end {pmatrix} \). Then det\(A=ad-cb\).
We can only find a determinant of a matrix if it is a square matrix.
Determinant of a \(3\times 3\) matrix
Let \(A= \begin {pmatrix} a & b & c\\ d & e & f\\ g & h & i\\ \end {pmatrix} \)
\begin {align*} \text {Then}\quad \det A &= |A|= \begin {vmatrix} a & b & c\\ d & e & f\\ g & h & i\\ \end {vmatrix}\\\\ & = a \begin {vmatrix} e & f\\ h & i\\ \end {vmatrix} - b \begin {vmatrix} d & f\\ g & i\\ \end {vmatrix} + c \begin {vmatrix} d & e\\ g & h\\ \end {vmatrix}\\\\ & = a(ei-fh)-b(di-fg)+c(dh-eg)\\ \end {align*}
Let \(A= \begin {pmatrix} 1 & -3 & 1\\ 3 & 0 & 4\\ -2 & 1 & 0\\ \end {pmatrix} \)
\begin {align*} |A| & = \begin {vmatrix} 1 & -3 & 1\\ 3 & 0 & 4\\ -2 & 1 & 0\\ \end {vmatrix}\\\\ & = 1 \begin {vmatrix} 0 & 4\\ 1 & 0\\ \end {vmatrix} +3 \begin {vmatrix} 3 & 4\\ -2 & 0\\ \end {vmatrix} +1 \begin {vmatrix} 3 & 0\\ -2 & 1\\ \end {vmatrix}\\\\ & = 1 (0-4)+3(0+8)+1(3-0)\\ & = -4+24+3\\ & = 23\\ \end {align*}
Note 24.3. Watch the signs when expanding. The middle term of the expansion carries a minus, and here the entry itself is \(-3\), so the two minuses give \(+3\) — which is why the second term appears added rather than subtracted. Losing track of that sign is the commonest error in evaluating a \(3\times 3\) determinant.
The signs alternate in the pattern \[\begin {pmatrix}+&-&+\\-&+&-\\+&-&+\end {pmatrix},\] starting with \(+\) in the top-left corner, and the same chessboard pattern extends to larger determinants.
A useful check: expanding along a different row or column must give the same answer. Expanding this one down the first column gives \(1(0\cdot 0-4\cdot 1)-3\left ((-3)(0)-1(1)\right )+(-2)\left ((-3)(4)-1(0)\right ) =-4+3+24=23\), agreeing. Choosing the row or column with the most zeros makes the work shortest.
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