Definition C.1.1. Determinant of \(1 \times 1\) and a \(2 \times 2\) matrices.
- If \(A\) is a \(1 \times 1\) matrix, then \(\lvert A \rvert = A_{1,1}\)
- If \(A\) is a \(2 \times 2\) matrix, then \(\lvert A \rvert = A_{1,1} A_{2,2} - A_{1,2} A_{2,1}\)
Section C.1 Definition
Associated with every square matrix is a number called its determinant. The most important information it provides us with is whether the matrix is invertible. A matrix has an inverse if and only if its determinant is nonzero. If \(A\) is a square matrix, then the determinant of \(A\) is commonly denoted either \(\det(A)\) or \(\lvert A \rvert\text{.}\) Strictly speaking, we only need to define the determinant of a \(1 \times 1\) matrix here and then define the higher ordered ones recursively, but for convenience we also recall the definition of the determinant of a \(2 \times 2\) matrix.
We now proceed to define the determinant of an \(n \times n\) matrix where \(n > 2\text{.}\) This definition requires two preliminary definitions those of minors and cofactors.
Definition C.1.2. Matrix Minor.
Let \(A\) be an \(n \times n\) matrix, \(n \geq 2\text{.}\) The determinant of the \((n-1) \times (n-1)\) matrix formed by removing the \(i^{th}\) row and \(j^{th}\) column of \(A\) is the minor denoted by \(M(A)_{i,j}\text{.}\)
Example C.1.3.
Let \(A = \begin{pmatrix}
3 & 4 & 1 \\
1 & 3 & 4 \\
4 & 1 & 3
\end{pmatrix} \) then \(A\) has nine minors, one of which is
\begin{equation*}
M(A)_{1,1} = \begin{vmatrix}
3 & 4 \\
1 & 3
\end{vmatrix} = 3 \cdot 3 - 4 \cdot 1 =5
\end{equation*}
For our purposes in computing \(\lvert A \rvert\text{,}\) we only need minors corresponding to any one row or column. Completing the minors in the first row we have \(M(A)_{1,2} = -13 \) and \(M(A)_{1,3} = -11 \)
Definition C.1.4. Cofactor.
Let \(A\) be an \(n \times n\) matrix, \(n \geq 2\text{.}\) The \(i^{th}\) row, \(j^{th}\) column cofactor of \(A\text{,}\) denoted \(C(A)_{i,j}\text{,}\) is defined by
\begin{equation*}
C(A)_{i,j} = (-1)^{i+j} M(A)_{i,j}
\end{equation*}
Example C.1.5.
Using the values of minors computed in Example 3, we have \(C(A)_{1,1} = (-1)^2 M(A)_{1,1} = 5\text{,}\) \(C(A)_{1,2} = (-1)^3 M(A)_{1,2} = 13\text{,}\) and \(C(A)_{1,3} = (-1)^4 M(A)_{1,3} = -11\text{.}\)
Finally, we will define the determinant of a square matrix. Our definition is practical in that you can apply it easily to any matrix. It isn’t the most general, nor is it the best definiton for the purposes of proving properties of determinants. The more general definition is beyond our current scope, but can be easily stated with background in permutation groups.
Definition C.1.6. Determinant of a Square Matrix.
Let \(A\) be an \(n \times n\) matrix, \(n \geq 2\text{.}\) The determinant of \(A\) is equal to
\begin{equation*}
\sum_{j=1}^{n} A_{1,j}\cdot C(A)_{1,j}
\end{equation*}
Our definition of a determinant involves what is called expansion along the first row of the matrix A. It is certainly not obvious, but it is true, that the determinant of a matrix can be found by expanding along any row or any column.
Example C.1.7.
We have computed the cofactors for row 1 of \(A = \begin{pmatrix}
3 & 4 & 1 \\
1 & 3 & 4 \\
4 & 1 & 3
\end{pmatrix} \) above and so the determinant is only a few operations away.
\begin{equation*}
\begin{split}
\lvert A \rvert &= A_{1,1}\cdot C(A)_{1,1} +A_{1,2}\cdot C(A)_{1,2}+A_{1,3}\cdot C(A)_{1,3}\\
&= 3\cdot 5 + 4\cdot 13 + 1 \cdot (-11)\\
&= 56
\end{split}
\end{equation*}
Example C.1.8.
Associated with any square matrix, \(A\text{,}\) is a characteristic polynomial which is defined to be the \(\lvert A - \lambda I \rvert\text{.}\) The roots of this polynomial are the eigenvalues of the matrix. Here, we compute the characteristic polynomial of \(A = \begin{pmatrix}
3 & 4 & 1 \\
1 & 3 & 4 \\
4 & 1 & 3
\end{pmatrix}\text{.}\)
To compute the determinant we expand along the first row.
\begin{equation*}
\begin{split}
\det{(A - \lambda I)} &= \begin{vmatrix}
3 - \lambda & 4 & 1 \\
1 & 3 - \lambda & 4 \\
4 & 1 & 3 - \lambda
\end{vmatrix}\\
&= (3-\lambda)\cdot \begin{vmatrix}
3 - \lambda & 4 \\
1 & 3 - \lambda
\end{vmatrix}
+ 4 \cdot (-1)\cdot \begin{vmatrix}
1 & 4 \\
4 & 3 - \lambda
\end{vmatrix}
+ 1\cdot \begin{vmatrix}
1 & 3 - \lambda \\
4 & 1
\end{vmatrix} \\
&=(3-\lambda)((3-\lambda)^2-4) - 4((3-\lambda)-16) + (1-4(3-\lambda))\\
&=-\lambda ^3+9 \lambda ^2-15 \lambda +56
\end{split}
\end{equation*}
You have attempted of activities on this page.
