A determinant is a scalar value associated with a square matrix.
It serves as a useful tool in solving systems of linear equations and finding the inverse of matrices.
The determinant of a square matrix $[A_{ij}]$ is given |A|
iF $A=\begin{bmatrix}
1 & 2 \\
3 & -1 \\
\end{bmatrix} $
then Determinant is given by
$|A|=\begin{vmatrix}
1 & 2 \\
3 & -1 \\
\end{vmatrix} $
Determinant of a 2x2 Matrix
For a 2x2 matrix \( A \) given by
\[
A = \begin{vmatrix}
a & b \\
c & d
\end{vmatrix}
\]
The determinant \( |A| \) is calculated as:
\[
|A| = ad - bc
\]
Determinant of a 3x3 Matrix
For a 3x3 matrix \( B \) given by
\[
B = \begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}
\]
Determinant of a matrix of order three can be determined by expressing it in terms of second order determinants. This is known as expansion of a determinant along a row (or a column). There are six ways of expanding a determinant of order 3 corresponding to each of three rows (R1, R2 and R3 ) and three columns (C1, C2 and C3) giving the same value
Expansion Along R1
The determinant \( |B| \) is calculated as
\[
|B| = (-1)^{1 +1} a(ei - fh) + (-1)^{1 +1} b(di - fg) + (-1)^{1 +3}c(dh - eg)
\]
This is obtained by
1. Multiply first element a of R1 by $(–1)^{1 + 1} [(–1)^{\text{sum of suffixes in a}}]$ and with the second order determinant obtained by deleting the elements of first row (R1) and first column (C1 ) of | A | as a lies in R1 and C1
$(-1)^{1 + 1} a \begin{vmatrix}
e & f \\
h & i \\
\end{vmatrix} $
$=(-1)^{1 +1} a(ei - fh)$
2.Multiply 2nd element b of R1 by $(–1)^{1 + 2} [(–1)^{\text{sum of suffixes in b}}]$ and the second order determinant obtained by deleting elements of first row (R1) and 2nd column (C2) of | A | as b lies in R1 and C2
$(-1)^{1 + 2} b \begin{vmatrix}
d & f \\
g & i \\
\end{vmatrix} $
$=(-1)^{1 +2} b(di - fg)$
3. Multiply third element c of R1 by $(–1)^{1 + 3} [(–1)^{\text{sum of suffixes in c}}]$ and the second order determinant obtained by deleting elements of first row (R1) and third column (C3) of | A | as c lies in R1
and C3
$(-1)^{1 + 3} c \begin{vmatrix}
d & e \\
g & h \\
\end{vmatrix} $
$=(-1)^{1 +3} c(dh - eg)$
Therefore
|B|=aei -afh -bdi + bfg + cdh -ceg
Expansion Along R2
we can write as
\[
|B| = (-1)^{2 +1} d(bi - ch) + (-1)^{2 +2} e(ai - cg) + (-1)^{2 + 3} f(ah - bg)=-bdi+ cdh + aei -acg -fah +fbg=aei -afh -bdi + bfg + cdh -ceg
\]
So it is the same value
Expansion Along R3
we can write as
\[
|B| = (-1)^{3 +1} g(bf - ec) + (-1)^{3 +2} h(af - cd) + (-1)^{3 + 3} i(ae - bd)=bgf - gec - haf +hcd +aei -bdi=aei -afh -bdi + bfg + cdh -ceg
\]
So it is the same value
Expansion Along C1
we can write as
\[
|B| = (-1)^{1 +1} a(ei - fh) + (-1)^{2 +1} d(bi - ch) + (-1)^{3 + 1} g(bf - ec)=aei - afh - bdi +hcd +gbf -gec=aei -afh -bdi + bfg + cdh -ceg
\]
So it is the same value
Expansion Along C2
we can write as
\[
|B| = (-1)^{1+ 2} b(di - fg) + (-1)^{2 +2} e(ai - cg) + (-1)^{3 + 2} h(af - cd)=-bdi + bfg +aei -ecg -haf +hcd=aei -afh -bdi + bfg + cdh -ceg
\]
So it is the same value
Expansion Along C2
we can write as
\[
|B| = (-1)^{1 +3} c(dh - eg) + (-1)^{2 +3} f(ah - bg) + (-1)^{3 + 3} i(ae - bd)=cdh ceg -afh +bfg +aei -bdi=aei -afh -bdi + bfg + cdh -ceg
\]
So it is the same value
