Computing determinants

Matrices: Matrices

Computing determinants

We describe a number of properties of the determinant of a matrix that help in calculations.

Let $A$ be a square matrix. Then:

if $A$ contains a n null row or null column, then $\det(A)=0$ ;
if $A$ contains two identical rows or two identical columns, then $\det(A)=0$ ;
if $A$ is an upper or lower triangular matrix, then $\det(A)$ is equal to the product of the diagonal elements;
$\det(A^{\top})=\det(A)$ .

Let $B$ be a matrix that is obtained from the square matrix $A$ by

multiplying a row or column with a scalar $c$ t, then $\det(B)=c\cdot\det(A)$ ;
interchanging two rows or two columns of $A$ , then $\det(B)=-\det(A)$ ;
adding a scalar multiple of a row (column) of $A$ to another row (column), then $\det(B)=\det(A)$ .

The rules above describe how the determinant changes for elementary row and column operations. If you keep the accounts consistently during reduction and have arrived, for example, at a lower or upper triangular form, then you can easily calculate a determinant. When you arrive during this reduction process in a situation of a row or column in which 1 appears at index $(i,j)$ and for the rest in the horizontal or vertical direction only zeros, then you can calculate the determinant obtained from the matrix by omitting the $i$ th row and $j$ th column, and by multiplying this result by $(-1)^{i+j}$ .

We give two examples of calculations of determinants.

Calculate $\det\matrix{3 & -2 & -5 & 4\\ -5 & 2 & 8 & -5\\ -2 & 4 & 7 & -3\\ 2 & -3 &- 5 & 8}$

Here, it is more convenient to use the notation $|A|$ rather than $\det(A)$ .
References to columns go through $C_1,\ldots, C_4$ .
$\begin{array}{rlll} \left|\begin{array}{rrrr}3 & -2 & -5 & 4\\ -5 & 2 & 8 & -5\\ -2 & 4 & 7 & -3\\ 2 & -3 &- 5 & 8\end{array}\right| &=& \left|\begin{array}{rrrr}1 & 2 & 2 & 1\\ -5 & 2 & 8 & -5\\ -2 & 4 & 7 & -3\\ 0 & 1 & 2 & 5\end{array}\right| & \blue{\begin{array}{l}R_1+R_3\\ \\ \\ R_3+R_4\end{array}}\\ \\ &=& \left|\begin{array}{rrrr}1 & 2 & 2 & 1\\ 0 & 12 & 18 & 0\\ 0 & 8 & 11 & -1\\ 0 & 1 & 2 & 5\end{array}\right| & \blue{\begin{array}{l} \\ R_2+5R_1\\ R_3+2R_1\\ \\ \end{array}}\\ \\ &=& \left|\begin{array}{rrr} 12 & 18 & 0\\ 8 & 11 & -1\\ 1 & 2 & 5\end{array}\right| & \\ \\ &=& 6\cdot \left|\begin{array}{rrr} 2 & 3 & 0\\ 8 & 11 & -1\\ 1 & 2 & 5\end{array}\right| & \\ \\ &=& 6\cdot \left|\begin{array}{rrr} 2 & 0 & 0\\ 8 & -1 & -1\\ 1 & \frac{1}{2} & 5\end{array}\right| & \blue{\begin{array}{l}\\ C_2\rightarrow C_2-\frac{3}{2}C_1 \\ \\ \end{array}}\\ \\ &=& 6\cdot 2\cdot \left|\begin{array}{rr} -1 & -1\\ \frac{1}{2} & 5\end{array}\right| & \\ \\ &=& 6\cdot 2\cdot (-1\cdot 5- ((-1)\cdot \dfrac{1}{2}) & \\ &=& 6\cdot 2\cdot (-\dfrac{9}{2}) & \\ &=& -54 & \end{array}$

New example

The following theorems are two of the most important theorems about determinants.

Let $A$ be an $n\times n$ matrix. Then the following statements are equivalent:

$A$ is invertible.
$\text{rank}(A)=n$ , i.e., the equation $A\vec{x}=\vec{0}$ has only $\vec{0}$ as a solution.
$\det(A)\neq 0$ .

Let $A$ and $B$ be square matrices of the same size. Then: $\det(A\,B)=\det(A)\cdot \det(B)\tiny.$

If $A$ is invertible then it follows from this theorem that $\det(A^{-1})=\bigl(\det(A)\bigr)^{-1}= \dfrac{1}{\det(A)}\tiny.$

The following theorem shows that some properties of the determinant can be generalized to matrices whose structure has been defined by submatrices.

The determinant of a square matrix of the form $M = \matrix{A&C\\ 0&B}$ where $A$ and $B$ are square submatrices, and $C$ is an arbitrary submatrix of appropriate size, is equal to the product of the determinants of the two submatrices along the diagonal: $\det(M) = \det(A)\cdot\det(B)$