The determinant of a 2 x 2 matrix \(A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) is \(\det(A)=ad-bc.\)
For example, \(\det\left(\begin{bmatrix}1&-2\\3&-4\end{bmatrix}\right)=\)\((1)(-4)-(3)(-2)=-4+6=2.\)
In general, for an n x n matrix A with n ≥ 2, we use the cofactor expansion:
- Choose a column or row, preferably the one with the most 1s and 0s.
- For the entry aij, let Aij denote the matrix obtained by deleting row i and column j.
- Calculate \(c_{ij}(A)=(-1)^{i+j}\det(A_{ij}),\) this is called the (i, j)- cofactor. You can use this diagram to find (-1)i + j for each entry:
- Repeat steps 2 and 3 for each entry in the row/column you chose.
- Multiply each cofactor by its corresponding entry, then add them all together. Here is the determinant of A using the first row: \(\det(A)=a_{11}c_{11}(A)+a_{12}c_{12}(A)+a_{13}c_{13}(A)+...+a_{1n}c_{1n}(A)\)
Keep in mind if any of the a11, ........., a1n is 0, then the corresponding term aijcij(A) disappears, so there is no need to calculate the cofactor!
For example, the determinant of a 3 x 3 matrix \(B=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\) using Row 1:
\(\left.\det(B)=a\det\left(\begin{bmatrix}&e&f&\\&h&i&\end{bmatrix}\right.\right)-b\det\left(\begin{bmatrix}&d&f&\\&g&i&\end{bmatrix}\right)+c\det\left(\begin{bmatrix}&d&e&\\&g&h&\end{bmatrix}\right)\)\(=a(ei-fh)-b(di-fg)+c(dh-eg)\)
Example. Find the determinant of the following matrices
\(\mathrm{(a)}\quad M=\begin{bmatrix}&1&4&1\\&2&7&1\\&3&1&4\end{bmatrix}\quad\mathrm{(b)}\quad M=\begin{bmatrix}9&9&4\\\\0&2&0\\\\1&3&2\end{bmatrix}\)
Solution Part (a).
We will find the determinant using Column 3, since it has two 1s. Here are the calculations for the cofactors:
\(c_{13}(M)=(-1)^{1+3}\det\left(\begin{bmatrix}&2&7\\&3&1\end{bmatrix}\right)=(1)(2-21)=-19\)
\(c_{23}(M)=(-1)^{2+3}\det\left(\begin{bmatrix}&1&4\\&3&1\end{bmatrix}\right)=(-1)(1-12)=11\)
\(\left.c_{33}(M)=(-1)^{3+3}\det\left(\left\lfloor\begin{array}{cc}1&4\\2&7\end{array}\right.\right\rfloor\right)=(1)(7-8)=-1\)
Now, we can calculate the determinant:
\(\begin{aligned}
\mathrm{det}(M)& =a_{13}c_{13}(M)+a_{23}c_{23}(M)+a_{33}c_{33}(M) \\
&=(1)(-19)+(1)(11)+(4)(-1) \\
&=-9
\end{aligned}\)
Solution Part (b).
Since Row 2 has two 0s we will use that row, it will make the calculation much easier. We will only calculate the (2, 2)-cofactor since the other two aren’t needed:
\(c_{22}(M)=(-1)^{2+2}\det\left(\begin{bmatrix}&9&4\\&1&2\end{bmatrix}\right)=(1)(18-4)=14\)
Calculating the determinant, we get:
\(\det(M)=(0)c_{21}(M)+(2)c_{22}(M)+(0)c_{23}(M)\\=(0)(0)+(2)(14)+(0)(0)=28\)