Definition of \(A(i\mid j)\)

Let \(A\) be an \(n \times n\) matrix. Let \(i, j \in \{1,\ldots,n \}\). We define \(A(i \mid j)\) to be the matrix obtained from \(A\) by removing row \(i\) and column \(j\) from \(A\).

Example

Let \(A = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix}\). Then \(A(1 \mid 1) = \begin{bmatrix} 5 & 6\\ 8 & 9 \end{bmatrix},\) \(A(2 \mid 2) = \begin{bmatrix} 1 & 3\\ 7 & 9 \end{bmatrix},\) \(A(3 \mid 1) = \begin{bmatrix} 2 & 3\\ 5 & 6 \end{bmatrix}.\)

Cofactor expansion

One way of computing the determinant of an \(n \times n\) matrix \(A\) is to use the following formula called the cofactor formula.

Pick any \(i \in \{1,\ldots, n\}\). Then \[ \det(A) = (-1)^{i+1}A_{i,1}\det(A(i \mid 1)) + (-1)^{i+2}A_{i,2}\det(A(i \mid 2)) + \cdots + (-1)^{i+n}A_{i,n}\det(A(i \mid n)).\]

We often say the right-hand side is the cofactor expansion of the determinant along row \(i\). (This formula can be proved directly from the definition of the determinant.)

There is also a formula for expanding along column \(j\): \[ \det(A) = (-1)^{1+j}A_{1,j}\det(A(1 \mid j)) + (-1)^{2+j}A_{2,j}\det(A(2 \mid j)) + \cdots + (-1)^{n+j}A_{n,j}\det(A(n \mid j)).\]

Sometimes, out of convenience, one uses \(C_{i,j}\) to denote \((-1)^{i+j}\det(A(i \mid j))\). We call \(C_{i,j}\) a cofactor of \(A\).

Hence, the cofactor expansion along row \(i\) could be written as \(\displaystyle\det(A) = \sum_{j = 1}^n A_{i,j}C_{i,j},\) and the cofactor expansion along column \(j\) could be written as \(\displaystyle\det(A) = \sum_{i = 1}^n A_{i,j}C_{i,j}.\)

Examples

1. Let \(A = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix}\). We compute \(\det(A)\) by expanding along row \(2\). \begin{eqnarray} \det(A) & = & (-1)^{2+1}A_{2,1}\det(A(2 \mid 1)) + (-1)^{2+2}A_{2,2}\det(A(2 \mid 2)) + (-1)^{2+3}A_{2,3}\det(A(2 \mid 3)) \\ & = & -4\left|\begin{matrix} 2 & 3 \\ 8 & 9\end{matrix}\right| + 5\left|\begin{matrix} 1 & 3 \\ 7 & 9\end{matrix}\right| - 6\left|\begin{matrix} 1 & 2 \\ 7 & 8\end{matrix}\right| \\ & = & -4(2\cdot 9 - 3\cdot 8) + 5(1 \cdot 9- 3\cdot 7) - 6(1\cdot 8 - 2 \cdot 7) \\ & = & 24 - 60 + 36 \\ & = & 0 \end{eqnarray}

2. Cofactor expansion can be very handy when the matrix has many \(0\)'s. Let \(A = \begin{bmatrix} 1 & a \\ 0_{n-1} & B \end{bmatrix}\) where \(a\) is \(1 \times (n-1)\), \(B\) is \( (n-1)\times (n-1)\), and \(0_{n-1}\) is an \((n-1)\)-tuple of \(0\)'s. Using the formula for expanding along column 1, we obtain just one term since \(A_{i,1} = 0\) for all \(i \geq 2\). Hence, \(\det(A) = (-1)^{1+1}A_{1,1}\det(A(1 \mid 1)) = 1 \det(B) = \det(B).\)

Quick Quiz

Exercises

1. Derive the cofactor expansion formulas for computing the determinant of a \(3 \times 3\) matrix directly from the definition of the determinant.

2. Compute the determinant of \(\begin{bmatrix} 1 & 0 & 2\\ 3 & 4 & 5\\ 2 & -1 & -4\end{bmatrix}\) by expanding along the second column.  

   \begin{eqnarray*} \begin{vmatrix} 1 & 0 & 2\\ 3 & 4 & 5\\ 2 & -1 & -4\end{vmatrix} & = & (-1)^{1+2} (0) \begin{vmatrix} 3 & 5 \\ 2 & - 4\end{vmatrix} +(-1)^{2+2} (4) \begin{vmatrix} 1 & 2 \\ 2 & - 4\end{vmatrix} +(-1)^{3+2} (-1) \begin{vmatrix} 1 & 2 \\ 3 & 5\end{vmatrix} \\ & = & -32 + (-1) = -33 \end{eqnarray*}

3. Prove the cofactor formula for expanding along row \(i\).