Determinant

Given a 2×2 matrix

\(\scriptsize \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\)

then the determinant of matrix A is denoted by

\( \scriptsize \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = \scriptsize a_{11}.a_{22} \; – \; a_{21}. a_{12} \)

Example 1

det A = |A| = \( \scriptsize \begin{vmatrix} 6 & 5 \\ 1 & 2 \end{vmatrix} \)

\(\scriptsize (6 \times 2) \; – \; (1 \times 5)\)

= 12 – 5

= 7

For a 3 × 3 matrix

A = \(\scriptsize \begin{pmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\\a_{3} & b_{3} & c_{3} \end{pmatrix} \)

Thus the determinant,

|A| = \(\scriptsize \begin{vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\\a_{3} & b_{3} & c_{3} \end{vmatrix} \)

Each element in the determinant is associated with minor which is formed by omitting the row and column containing the element concerned. 

The minor of a₁ = \(\scriptsize \begin{vmatrix} b_{2} & c_{2}\\b_{3} & c_{3} \end{vmatrix} \)

The minor of b₁ = \(\scriptsize \begin{vmatrix} a_{2} & c_{2}\\a_{3} & c_{3} \end{vmatrix} \)

The minor of c₁ = \(\scriptsize \begin{vmatrix} a_{2} & b_{2}\\a_{3} & b_{3} \end{vmatrix} \)

To expand a determinant of the third order, we can write down each element along the top row, multiply it by its minor and give the terms a plus or minus sign alternatively.

\(\scriptsize \begin{vmatrix} + & – & + \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\\a_{3} & b_{3} & c_{3} \end{vmatrix} \)

= \(\scriptsize a_1 \begin{vmatrix} b_{2} & c_{2}\\b_{3} & c_{3} \end{vmatrix}\; – b_1 \begin{vmatrix} a_{2} & c_{2}\\a_{3} & c_{3} \end{vmatrix} \; + c_1 \scriptsize \begin{vmatrix} a_{2} & b_{2}\\a_{3} & b_{3} \end{vmatrix} \)

Example 2

= \(\scriptsize 1 \begin{vmatrix} 5 & 7 \\4 & 8 \end{vmatrix}\; – 3\begin{vmatrix} 4 & 7\\2 & 8 \end{vmatrix} \; + 2 \begin{vmatrix} 4 & 5\\2 & 4 \end{vmatrix} \)

= 1( 5 x 8 – 4 x 7) – 3(4 x 8 – 2 x 7) + 2(4 x 4 – 2 x 5)

= 1(12) -3(18) + 2(6)

= 12 – 54 + 12

= -30

Solving pairs of simultaneous linear equations in two variables using  2×2 determinant.

Example 3

The key to the method is:

\( \frac {x}{\Delta_1} = \frac {-y}{\Delta_2} = \frac {1}{\Delta_0} \)

To find, \(\scriptsize \Delta_0 \)pick the coefficients of unknowns and omit the constant terms;

= \(\scriptsize \Delta_0 = \scriptsize \begin{vmatrix} 5 & 2 \\ 3 & 4 \end{vmatrix} \)

= (5 x 4) – (3 x 2)

= 20 – 6

= 14

Now, to find \( \scriptsize \Delta_1 \) , pick the coefficients of y terms and the constant terms and omit the x terms.

= \(\scriptsize \Delta_1 = \scriptsize \begin{vmatrix} 2 & 19 \\ 4 & 17 \end{vmatrix} \)

= (2 x 17) – (4 x 19)

= 34 – 76

= – 42

Similarly, to find \( \scriptsize \Delta_2 \), pick the coefficients of the terms and the constants and omit the terms:

:- \(\scriptsize \Delta_2 = \scriptsize \begin{vmatrix} 5 & 19 \\ 3 & 17 \end{vmatrix} \)

= (5 x 17) – (3 x 19)

= 85 – 57

= 28

Substituting the values of \(\scriptsize \Delta_0, \scriptsize \Delta_1, \scriptsize \Delta_2 \) in the key,

\( \frac {x}{\Delta_1} = \frac {-y}{\Delta_2} = \frac {1}{\Delta_0} \)

=\( \frac {x}{-42} = \frac {-y}{28} = \frac {1}{14} \)

=\( \therefore \frac {x}{-42} = \frac {1}{14} \)

14x = -42

x = -3

=\( \frac {-y}{8} = \frac {1}{14} \)

-14y = 28

y = -2

Simultaneous equations in three unknowns

Example 4

using the key

\( \frac {x}{\Delta_1} = \frac {-y}{\Delta_2} = \frac {z}{\Delta_3}=\frac {1}{\Delta_0} \)

To find\(\scriptsize \Delta_0 \), omit the constant terms and evaluate the determinant of the coefficients of unknown terms:

= \(\scriptsize \Delta_0 = \scriptsize \begin{vmatrix} 2 & 3 & -1 \\ 3 & 1 & 2\\ 1 & 2 & -5\end{vmatrix} \)

= \(\scriptsize 2 \begin{vmatrix} 1 & 2 \\2 & -5 \end{vmatrix}\; – -3\begin{vmatrix} 3 & 2\\1 & -5 \end{vmatrix} \; – 1 \begin{vmatrix} 3 & 1\\1 & 2 \end{vmatrix} \)

= 2(-5 – 4) -3(-15 – 2) -1(6 – 1)

= 2(-9) – 3(-17) -1(5)

= -18 + 51 -5

= 28

Now, to find \(\scriptsize \Delta_1 \), omit the coefficients of x terms and evaluate the determinant of coefficients of y, z and constant terms;

= \(\scriptsize \Delta_1 = \scriptsize \begin{vmatrix} 3 & -1 & -4 \\ 1 & 2 & -13\\ 2 & -5 & 11\end{vmatrix} \)

= \(\scriptsize 3 \begin{vmatrix} 2 & -13 \\-5 & 11 \end{vmatrix}\; – 1\begin{vmatrix} 1 & -13\\2 & 11 \end{vmatrix} \; – 4 \begin{vmatrix} 1 & 2\\2& -5 \end{vmatrix} \)

= 3(22 – 65) + 1(11 + 26) -4(-5 – 4)

= 3(-43) + 1(37) -4(-9)

= -129 + 37 + 36

= -56

In the same manner we evaluate for \(\scriptsize \Delta_2 \), omitting the coefficients of y terms,

= \(\scriptsize \Delta_2 = \scriptsize \begin{vmatrix} 2 & -1 & -4 \\ 3 & 2 & -13\\ 1 & -5 & 11\end{vmatrix} \)

= \(\scriptsize 2 \begin{vmatrix} 2 & -13 \\-5 & 11 \end{vmatrix}\; – 1\begin{vmatrix} 3 & -13\\1 & 11 \end{vmatrix} \; – 4 \begin{vmatrix} 3 & 2\\1& -5 \end{vmatrix} \)

= 2(22 – 65) + 1(33 + 13) -4(-15 – 2)

= 2(43) + 1(46) -4(-17)

= -86 + 46 + 68

= 28

Omit the coefficients of z terms and find the value of using the same process as above;

\(\scriptsize \Delta_2 = -84\)

Substituting the values of \(\scriptsize \Delta_0, \scriptsize \Delta_1, \scriptsize \Delta_2, \; and \; \Delta_3 \) into the key,

\( \frac {x}{\Delta_1} = \frac {-y}{\Delta_2} = \frac {z}{\Delta_3}=\frac {1}{\Delta_0} \)

=\( \frac {x}{-56} = \frac {-y}{28} = \frac {z}{-84}=\frac {1}{28} \)

\( \frac {x}{-56} =\frac {1}{28} \)

28x = -56

x = -2

\( \frac {-y}{28} =\frac {1}{28} \)

-28y = 28

y = -1

\( \frac {z}{-84}=\frac {1}{28} \)

28z = -84

z = – 3