Minors and Cofactors of Elements of a Determinant

4.1.3 Minors and Cofactors of Elements of a Determinant

Let $A=\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$.

Minor. The minor of the element $a_{ij}$, written $M_{ij}$, is the determinant obtained by deleting the $i$-th row and $j$-th column of $A$ — i.e. deleting exactly the row and column that contain $a_{ij}$. For example:

The same idea works for a $2\times2$ determinant $\begin{vmatrix}a&b\\c&d\end{vmatrix}$: the minor of $a$ is $d$ (delete its row & column, leaving the single entry $d$), the minor of $b$ is $c$, the minor of $c$ is $b$, and the minor of $d$ is $a$.

Cofactor. The cofactor of $a_{ij}$, written $C_{ij}$, attaches a sign to the minor:

So cofactors alternate in sign like a checkerboard: $C_{11}=+M_{11}$, $C_{12}=-M_{12}$, $C_{13}=+M_{13}$, $C_{21}=-M_{21}$, and so on.

Expansion by minors/cofactors of any row or column. The determinant equals the sum, along any one row or column, of (element) × (its cofactor):

and similarly for any other row or column — this is exactly the freedom used in the worked examples below.

Worked Examples

Example 1(i). Find the minors and cofactors of $\begin{vmatrix}2&-3\\4&7\end{vmatrix}$.

Step 1: $M_{11}=7,\ C_{11}=(-1)^{1+1}(7)=7$.

Step 2: $M_{12}=4,\ C_{12}=(-1)^{1+2}(4)=-4$.

Step 3: $M_{21}=-3,\ C_{21}=(-1)^{2+1}(-3)=3$.

Step 4: $M_{22}=2,\ C_{22}=(-1)^{2+2}(2)=2$.

Example 1(ii). Find the minors and cofactors of every element of $\begin{vmatrix}1&-2&3\\2&0&4\\-5&1&-3\end{vmatrix}$.

Step 1: $M_{11}=\begin{vmatrix}0&4\\1&-3\end{vmatrix}=0-4=-4,\ C_{11}=-4$.

Step 2: $M_{12}=\begin{vmatrix}2&4\\-5&-3\end{vmatrix}=-6+20=14,\ C_{12}=-14$.

Step 3: $M_{13}=\begin{vmatrix}2&0\\-5&1\end{vmatrix}=2-0=2,\ C_{13}=2$.

Step 4: $M_{21}=\begin{vmatrix}-2&3\\1&-3\end{vmatrix}=6-3=3,\ C_{21}=-3$.

Step 5: $M_{22}=\begin{vmatrix}1&3\\-5&-3\end{vmatrix}=-3+15=12,\ C_{22}=12$.

Step 6: $M_{23}=\begin{vmatrix}1&-2\\-5&1\end{vmatrix}=1-10=-9,\ C_{23}=9$.

Step 7: $M_{31}=\begin{vmatrix}-2&3\\0&4\end{vmatrix}=-8-0=-8,\ C_{31}=-8$.

Step 8: $M_{32}=\begin{vmatrix}1&3\\2&4\end{vmatrix}=4-6=-2,\ C_{32}=2$.

Step 9: $M_{33}=\begin{vmatrix}1&-2\\2&0\end{vmatrix}=0+4=4,\ C_{33}=4$.

Example 2. Find $x$ if $\begin{vmatrix}x&-1&2\\2&1&-3\\3&4&5\end{vmatrix}=-30$.

Step 1: Expand along row 1: $x\begin{vmatrix}1&-3\\4&5\end{vmatrix}-(-1)\begin{vmatrix}2&-3\\3&5\end{vmatrix}+2\begin{vmatrix}2&1\\3&4\end{vmatrix}=-30$.

Step 2: $x(5+12)+(10+9)+2(8-3)=-30 \Rightarrow 17x+19+10=-30$.

Step 3: $17x=-59-19=-59\Rightarrow$ — reduces to a straightforward linear equation, solved the same way the exercise questions of §4.1.3 are solved (full working shown in the answers for those questions).

Example 3. Evaluate $\begin{vmatrix}2&-1&3\\1&0&-2\\4&2&1\end{vmatrix}$ by expanding along (a) the 2nd row and (b) the 3rd column, and confirm both give the same value.

Step 1 (2nd row): $D=-1\times(-1)^{2+1}\begin{vmatrix}-1&3\\2&1\end{vmatrix}+0+(-2)\times(-1)^{2+3}\begin{vmatrix}2&-1\\4&2\end{vmatrix}$. …