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What is Determinants




What is Determinants

  • 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

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