We came across the determinant of a \(2\times 2\) matrix in a previous lesson. That determinant can be useful when to find determinants of larger square matrices. This is known as the cofactor method.
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The determinant of a \(2\times 2\) matrix \(A=\left[\begin{array}{cc} a_{11} & a_{12}\\a_{21}&a_{22}\end{array}\right]\) is defined by \[detA=\left|\begin{array}{cc} a_{11} & a_{12}\\a_{21}&a_{22}\end{array}\right|=det\left[\begin{array}{cc} a_{11} & a_{12}\\a_{21}&a_{22}\end{array}\right]=a_{11}a_{22}-a_{12}a_{21}\] |
Example: Find the determinant of \(\left[\begin{array}{cc} 10&-3\\4&-8\end{array}\right]\)
Solution: \(\left|\begin{array}{cc} 10&-3\\4&-8\end{array}\right|=10(-8)-4(-3)=-68\)
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Cofactors of a \(3\times 3\) matrix Let \(A\in M_{3\times 3}(\mathbb{R})\) and let \(A(i,j)\) denote the \(2\times 2\) submatrix obtained from \(A\) by deleting the i-th row and the j-th column. Define the \((i,j)-cofactor of A to be \[C_{ij}=(-1)^{i+j}|A(i,j)|\] |
Example: Let \(A=\left[\begin{array}{ccc} 1&0&3\\ 0&-1&3\\1&-2&3\end{array}\right]\)
Cofactor \(C_{11}\) is defined by \(C_{11}=(-1)^{1+1}|A(1,1)|\)
This means to remove row 1 and column 1 from the matrix and calculate the determinant of the remaining submatrix \(A(1,1)\).
\[A(1,1)=\left[\begin{array}{cc} -1&3\\-2&3\end{array}\right]\]
The determinant \(|A(1,1)|=\left|\begin{array}{cc} -1&3\\-2&3\end{array}\right|=3\), therefore, the cofactor \(C_{11}=+3\)
Let's calculate \(C_{12}\), which requires us to remove row 1 and column 2, with submatrix
\[A(1,2)=\left[\begin{array}{cc} 0&3\\1&3\end{array}\right]\]
The determinant \(|A(1,2)|=\left|\begin{array}{cc} 0&3\\1&3\end{array}\right|=-3\), therefore, the cofactor \(C_{12}=(-1)^{1+2}(-3)=-(-3)=3\)
Continuing with this process, the determinant \(|A(1,3)|=\left|\begin{array}{cc} 0&-1\\1&-2\end{array}\right|=1\), therefore, the cofactor \(C_{13}=(-1)^{1+3}(-3)=+(1)=1\).
There are 9 possible cofactors of matrix \(A\). Can you find them all?
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Determinant of \(n\times n\) matrix using Cofactors Let \(A\in M_{n\times n}(\mathbb{R})\) with \(n>2\). Let \(A(i,j)\) denote the \((n-1)\times (n-1)\) submatrix obtained from \(A\) by deleting the i-th row and j-th column. The determinant of \(A\in M_{n\times n}(\mathbb{R})\) is defined by \[|A|=a_{k1}C_{k1}+a_{k2}C_{k2}+\ldots+a_{kn}C_{kn}\] where \(k\) is any row and the \((i,j)\)-cofactor of \(A\) is defined to be \[C_{ij}=(-1)^{i+j}|A(i,j)|\] The determinant can also be defined along any column \(l\) \[|A|=a_{1l}C_{1l}+a_{2l}C_{2l}+\ldots+a_{nl}C_{nl}\] |
Example: Find the determinant of \(A=\left[\begin{array}{ccc} 1&0&3\\ 0&-1&3\\1&-2&3\end{array}\right]\)
Solution
We can find the determinant using cofactors of row 1.
\begin{align} |A|&=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}\\ &=1(3)+0(3)+3(1)\\&=6\end{align}
Tips:
