18. Properties of Determinant
Dated: 22-04-2025
Theorem
- If a multiple of one row is added to another row of \(A\) then the
determinant remains the same.
- If two rows of \(A\) are interchanged to produce \(B\) then \(\det (B) = - \det (A)\).
- If one of the rows of \(A\) is multiplied by a constant \(k\) to produce \(B\) then \(\det (B) = k \times \det (A)\).
Example
\[
A = \begin{vmatrix}
2 & 3 & 1 & 0 & 1 \\
1 & 1 & 3 & 1 & 2 \\
2 & 1 & 2 & 3 & 4 \\
3 & 2 & 1 & 1 & 2 \\
4 & 1 & 1 & 0 & 0
\end{vmatrix}
\]
Interchanging \(R_1\) and \(R_2\).
\[
A = - \begin{vmatrix}
1 & 1 & 3 & 1 & 2 \\
2 & 3 & 1 & 0 & 1 \\
2 & 1 & 2 & 3 & 4 \\
3 & 2 & 1 & 1 & 2 \\
4 & 1 & 1 & 0 & 0
\end{vmatrix}
\]
\[
\Big\downarrow R_2 = R_2 - 2R_1
\]
\[
\Big\downarrow R_3 = R_3 - 2R_1
\]
\[
\Big\downarrow R_4 = R_4 - 3R_1
\]
\[
\Big\downarrow R_5 = R_5 - 4R_1
\]
\[
= - \begin{vmatrix}
1 & 1 & 3 & 1 & 2 \\
0 & 1 & -5 & -2 & -3 \\
0 & -1 & -4 & 1 & 0 \\
0 & -1 & -8 & -2 & -4 \\
0 & -3 & -11 & -4 & -8
\end{vmatrix}
\]
Expand from \(C_1\)
\[
= - \begin{vmatrix}
1 & -5 & -2 & -3 \\
-1 & -4 & 1 & 0 \\
-1 & -8 & -2 & -4 \\
-3 & -11 & -4 & -8
\end{vmatrix}
\]
\[
\Big\downarrow R_2 = R_2 + R_1
\]
\[
\Big\downarrow R_3 = R_3 + R_1
\]
\[
\Big\downarrow R_4 = R_4 + 3R_1
\]
\[
= - \begin{vmatrix}
1 & -5 & -2 & -3 \\
0 & -9 & -1 & -3 \\
0 & -13 & -4 & -7 \\
0 & -26 & -10 & -17
\end{vmatrix}
\]
Expand from \(C_1\)
\[
= - \begin{vmatrix}
-9 & -1 & -3 \\
-13 & -4 & -7 \\
-26 & -10 & -17
\end{vmatrix}
\]
Taking \(-1\) common out from all rows
\[
= \begin{vmatrix}
9 & 1 & 3 \\
13 & 4 & 7 \\
26 & 10 & 17
\end{vmatrix}
\]
Interchange \(C_1\) and \(C_2\)
\[
= - \begin{vmatrix}
1 & 9 & 3 \\
4 & 13 & 7 \\
10 & 26 & 17
\end{vmatrix}
\]
\[
\Big\downarrow C_2 = C_2 - 9C_1
\]
\[
\Big\downarrow C_3 = C_3 - 3C_1
\]
\[
= - \begin{vmatrix}
1 & 0 & 0 \\
4 & -23 & -5 \\
10 & -64 & -13
\end{vmatrix}
\]
Expand from \(R_1\)
\[
= - \begin{vmatrix}
-23 & -5 \\
-64 & -13
\end{vmatrix} = -((-23)(-13) - (-5)(-64)) = -(299 - 320) = 21
\]
An Algorithm to Evaluate the Determinant
Step 1
Interchange rows of \(A\) to bring a non zero entry to \((1, 1)\) position unless rest of the entries in that column are \(0\).
Step 2
Perform row operations on \(A\) to reduce \(C_1\) to \(\begin{bmatrix}a_{ij}\\ 0 \\ \vdots \\ 0\end{bmatrix}\).
Expand \(\det (A)\) by \(C_1\).
Step 3
Convert \(A\) to a upper triangular form.
Step 4
\(\det(A)\) is the product of all diagonal entries.
Example
\[
A = \begin{bmatrix}
2 & -8 & 6 & 8 \\
3 & -9 & 5 & 10 \\
-3 & 0 & 1 & -2 \\
1 & -4 & 0 & 6
\end{bmatrix}
\]
Taking \(2\) common from \(R_1\)
\[
\det (A) = 2 \begin{vmatrix}
1 & -4 & 3 & 4 \\
3 & -9 & 5 & 10 \\
-3 & 0 & 1 & -2 \\
1 & -4 & 0 & 6
\end{vmatrix}
\]
\[
\Big\downarrow R_2 = R_2 - 3R_1
\]
\[
\Big\downarrow R_3 = R_3 + 3R_1
\]
\[
\Big\downarrow R_4 = R_4 - R_1
\]
\[
\det (A) = 2 \begin{vmatrix}
1 & -4 & 3 & 4 \\
0 & 3 & -4 & -2 \\
0 & 0 & -6 & 2 \\
0 & 0 & -3 & 2
\end{vmatrix}
\]
\[
\Big\downarrow R_4 = R_4 - \frac 1 2 R_3
\]
\[
\det (A) = 2 \begin{vmatrix}
1 & -4 & 3 & 4 \\
0 & 3 & -4 & -2 \\
0 & 0 & -6 & 2 \\
0 & 0 & 0 & 1
\end{vmatrix}
\]
\[= 2 \times (1)(3)(-6)(1) = -36\]
If \(U\) is the echelon form of \(A\) then.
If \(r\) is the number of row interchanges to convert \(A\) to \(U\) then.
\[
\det(A)
=
\begin{cases}
(-1)^r \cdot (\prod_{i = 1}^n [a_{ii}] \in U) & \text{if } A^{-1} \text{ exists}\\
0 & \text{if } A^{-1} \text{ doesn't exists}\\
\end{cases}
\]
Theorem
\[\det (A_{n \times n}) = \det(A_{n \times n}^T)\]
Theorem
If \(A\) and \(B\) are square matrices.
\[\det (AB) = \det (A) \times \det(B)\]
In general, \(\det(A + B) \neq \det(A) + \det(B)\)
References