Determinants
Solving Simultaneous Equations
Solving simultaneous equations can be done easily by elimination. But what happens if substitution is used?
Consider the system:
\(ax + by = e\)
\(cx + dy = f\)
\[
\qquad ax + by = e \qquad (1)
\]
\[
\qquad cx + dy = f \qquad (2)
\]
\[ \text{From (2)} \]
\[
y = \frac{\,f - cx\,}{d}
\]
\[
x = \frac{\,f - dy\,}{c}
\]
\[
\text{Substitution for } y \text{ in (1)}
\]
\[
\; ax + b\left(\frac{f - cx}{d}\right) = e
\]
\[
ad\,x + b(f - cx) = de
\]
\[
ad\,x - bcx = de - bf
\]
\[
x(ad - bc) = de - bf
\]
\[
x = \frac{de - bf}{\,ad - bc\,}
\]
\[
\; = \frac{bf - de}{\,bc - ad\,}
\]
\[
\text{Substitution for } x \text{ in (1)}
\]
\[
\; a\left(\frac{f - dy}{c}\right) + by = e
\]
\[
\Rightarrow\; a(f - dy) + bcy = ce
\]
\[
\Rightarrow\; bcy - ady = ce - af
\]
\[
\Rightarrow\; y(bc - ad) = ce - af
\]
\[
\Rightarrow\; y = \frac{\,ce - af\,}{\,bc - ad\,}
\]
\[
\; = \frac{\,af - ce\,}{\,ad - bc\,}
\]
For \( x = \frac{bf - de}{bc - ad} \)
The numerator bf - de is found by cross‑multiplying and subtracting using the y‑coefficients:
The denominator bc- ad is found by reverse cross‑multiplying and subtracting the coefficients of x and y:
Similarly for \( y = \frac{af - ce}{ad - bc} \)
The numerator af - ce is found by cross‑multiplying using the x‑coefficients:
This is the vedic way of solving simultaneous equations.
This method only works when \(ad - bc \neq 0\).
Example
Solve the system:
\(2x + 3y = 7\)
\(3x + 4y = 6\)
\[
x = \frac{bf - de}{\,bc - ad\,}
\]
\[
= \frac{3 \times 6 \;-\; 4 \times 7}{\,3 \times 3 \;-\; 4 \times 2\,}
\]
\[
= \frac{18 - 28}{\,9 - 8\,}
\]
\[
= -10
\]
\[
y = \frac{af - ce}{\,ae - bd\,}
\]
\[
= \frac{2 \times 6 \;-\; 3 \times 7}{\,2 \times 4 \;-\; 3 \times 3\,}
\]
\[
= \frac{12 - 21}{\,8 - 9\,}
\]
\[
= \frac{-9}{-1}
\]
\[
= 9
\]
Cramer's Rule
\[
a_{1}x + b_{1}y = c_{1}
\]
\[
a_{2}x + b_{2}y = c_{2}
\]
\[
x
=
\frac{
\left|
\begin{matrix}
c_{1} & b_{1} \\
c_{2} & b_{2}
\end{matrix}
\right|
}{
\left|
\begin{matrix}
a_{1} & b_{1} \\
a_{2} & b_{2}
\end{matrix}
\right|
}
\]
\[
=
\frac{
c_{1}b_{2} - c_{2}b_{1}
}{
a_{1}b_{2} - a_{2}b_{1}
}
\]
\[
y
=
\frac{
\left|
\begin{matrix}
a_{1} & c_{1} \\
a_{2} & c_{2}
\end{matrix}
\right|
}{
\left|
\begin{matrix}
a_{1} & b_{1} \\
a_{2} & b_{2}
\end{matrix}
\right|
}
\]
\[
=
\frac{
a_{1}c_{2} - a_{2}c_{1}
}{
a_{1}b_{2} - a_{2}b_{1}
}
\]
Example
\(2x + 3y = 7\)
\(3x + 4y = 6\)
\[
x
=
\frac{
\left|
\begin{matrix}
7 & 3 \\
6 & 4
\end{matrix}
\right|
}{
\left|
\begin{matrix}
2 & 3 \\
3 & 4
\end{matrix}
\right|
}
\]
\[
=
\frac{
(7 \times 4) - (6 \times 3)
}{
(2 \times 4) - (3 \times 3)
}
\]
\[
=
\frac{28 - 18}{8 - 9}
\]
\[
=
\frac{10}{-1}
\]
\[
= -10
\]
\[
y
=
\frac{
\left|
\begin{matrix}
2 & 7 \\
3 & 6
\end{matrix}
\right|
}{
\left|
\begin{matrix}
2 & 3 \\
3 & 4
\end{matrix}
\right|
}
\]
\[
=
\frac{
(2 \times 6) - (3 \times 7)
}{
(2 \times 4) - (3 \times 3)
}
\]
\[
=
\frac{12 - 21}{8 - 9}
\]
\[
=
\frac{-9}{-1}
\]
\[
= 9
\]
Determinants
The system \(ax + by = e,\; cx + dy = f\) can be written in matrix form as:
\[ \mathbf{A}\,x = b \]
\[
\mathbf{A}
=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\qquad
\mathbf{x}
=
\begin{bmatrix}
x \\
y
\end{bmatrix}
\qquad
\text{and}\quad
\mathbf{b}
=
\begin{bmatrix}
e \\
f
\end{bmatrix}
\]
The denominator \(ad - bc\) is the determinant of A.
It is written det(A) or: \( \left| \mathbf{A} \right| \)
\[
\mathbf{A}
=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\]
\[
\text{has determinant}
\]
\[
\left|
\begin{matrix}
a & b \\
c & d
\end{matrix}
\right|
= ad - bc
\]
If det(A) = 0, the system has no solution.
Example
Does the system have a solution?
\(3x + 2y = 7\)
\(9x + 6y = 6\)
\[
\left|
\begin{matrix}
3 & 2 \\
9 & 6
\end{matrix}
\right|
= 3 \times 6 \;-\; 9 \times 2
\]
\[
= 18 - 18
\]
\[
= 0
\]
No solution, since determinant = 0.
Determinant of a 3×3 Matrix
\[
\mathbf{A}
=
\begin{bmatrix}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{bmatrix}
\]
\[
\mathbf{det}(\mathbf{A})
=
\left|
\begin{matrix}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{matrix}
\right|
\]
Each element in A is associated with its minor:
To find the determinant:
- Write down each element from the top row.
- Multiply it by its minor.
- Add or subtract alternately according to position.
Place signs:
Example
Find the determinant of:
\[
\mathbf{A}
=
\begin{bmatrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{bmatrix}
\]
\[
\mathbf{det}(\mathbf{A})
=
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\]
\[
= \;
2
\left|
\begin{matrix}
1 & 6 \\
3 & 0
\end{matrix}
\right|
\;-\;
3
\left|
\begin{matrix}
4 & 6 \\
1 & 0
\end{matrix}
\right|
\;+\;
5
\left|
\begin{matrix}
4 & 1 \\
1 & 3
\end{matrix}
\right|
\]
\[
= \; 2(1 \times 0 \;-\; 3 \times 6)
\;-\;
3(4 \times 0 \;-\; 1 \times 6)
\;+\;
5(4 \times 3 \;-\; 1 \times 1)
\]
\[
= \; 2(-18)
\;-\;
3(-6)
\;+\;
5(11)
\]
\[
= -36 + 18 + 55
\]
\[
= 37
\]
Note:
\[
\det(\mathbf{A}\mathbf{B})
=
\det(\mathbf{A})\,\det(\mathbf{B})
\]
Cofactors
Cofactors are minors with their place sign.
Example
Find the cofactors of:
\[
\mathbf{A}
=
\begin{bmatrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{bmatrix}
\]
\[
\text{sign}
\begin{bmatrix}
+ & - & + \\
- & + & - \\
+ & - & +
\end{bmatrix}
\]
First row:
\[
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\qquad
\text{The minor of } a_{1}
\text{ is }
\left|
\begin{matrix}
1 & 6 \\
3 & 0
\end{matrix}
\right|
\]
\[
= (0 - 18) = -18
\]
\[
\text{The cofactor of } a_{1}
\text{ is }
+\;(-18)
= -18
\]
\[
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\qquad
\text{The minor of } b_{1}
\text{ is }
\left|
\begin{matrix}
4 & 6 \\
1 & 0
\end{matrix}
\right|
\]
\[
= 0 - 6 = -6
\]
\[
\text{The cofactor of } b_{1}
\text{ is }
-\;(-6)
= 6
\]
\[
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\qquad
\text{The minor of } c_{1}
\text{ is }
\left|
\begin{matrix}
4 & 1 \\
1 & 3
\end{matrix}
\right|
\]
\[
= 12 - 1 = 11
\]
\[
\text{The cofactor of } c_{1}
\text{ is }
+\,(11)
= 11
\]
Second row:
\[
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\qquad
\text{The minor of } a_{2}
\text{ is }
\left|
\begin{matrix}
3 & 5 \\
3 & 0
\end{matrix}
\right|
\]
\[
= 0 - 15 = -15
\]
\[
\text{The cofactor of } a_{2}
\text{ is }
-\;(-15)
= 15
\]
\[
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\qquad
\text{The minor of } b_{2}
\text{ is }
\left|
\begin{matrix}
2 & 5 \\
1 & 0
\end{matrix}
\right|
\]
\[
= 0 - 5 = -5
\]
\[
\text{The cofactor of } b_{2}
\text{ is }
+\;(-5)
= -5
\]
\[
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\qquad
\text{The minor of } c_{2}
\text{ is }
\left|
\begin{matrix}
2 & 3 \\
1 & 3
\end{matrix}
\right|
\]
\[
= 6 - 3 = 3
\]
\[
\text{The cofactor of } c_{2}
\text{ is }
-\,(3)
= -3
\]
Third row:
\[
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\qquad
\text{The minor of } a_{3}
\text{ is }
\left|
\begin{matrix}
3 & 5 \\
1 & 6
\end{matrix}
\right|
\]
\[
= 18 - 5 = 13
\]
\[
\text{The cofactor of } a_{3}
\text{ is }
+\,(13)
= 13
\]
\[
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\qquad
\text{The minor of } b_{3}
\text{ is }
\left|
\begin{matrix}
2 & 5 \\
4 & 6
\end{matrix}
\right|
\]
\[
= 12 - 20 = -8
\]
\[
\text{The cofactor of } b_{3}
\text{ is }
-\;(-8)
= 8
\]
\[
\left|
\begin{matrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{matrix}
\right|
\qquad
\text{The minor of } c_{3}
\text{ is }
\left|
\begin{matrix}
2 & 3 \\
4 & 1
\end{matrix}
\right|
\]
\[
= 2 - 12 = -10
\]
\[
\text{The cofactor of } c_{3}
\text{ is }
+\;(-10)
= -10
\]
Combining these gives the cofactor matrix C:
\[
C
=
\begin{bmatrix}
-18 & 6 & 11 \\
15 & -5 & -3 \\
13 & 8 & -10
\end{bmatrix}
\]
The Adjoint of A
The adjoint is the transpose of the cofactor matrix:
\[
\operatorname{adj}(A) = C^{T}
\]
Example
Find Adj(A) for:
\[ \mathbf{A} = \begin{bmatrix} 2 & 3 & 5 \\ 4 & 1 & 6 \\ 1 & 3 & 0 \end{bmatrix} \]From above:
\[
\mathbf{A}
=
\begin{bmatrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{bmatrix}
\]
\[
\text{has cofactor matrix}
\]
\[
C
=
\begin{bmatrix}
-18 & 6 & 11 \\
15 & -5 & -3 \\
13 & 8 & -10
\end{bmatrix}
\]
Transposed:
\[
C^{T}
=
\begin{bmatrix}
-18 & 15 & 13 \\
6 & -5 & 8 \\
11 & -3 & -10
\end{bmatrix}
\]
\[
\therefore\;
\operatorname{Adj}(A)
=
C^{T}
=
\begin{bmatrix}
-18 & 15 & 13 \\
6 & -5 & 8 \\
11 & -3 & -10
\end{bmatrix}
\]
Inverse of a Square Matrix
\[
A^{-1}
=
\frac{\operatorname{adj}(A)}{\det(A)}
\]
Example
Find the inverse of:
\[ \mathbf{A} = \begin{bmatrix} 2 & 3 & 5 \\ 4 & 1 & 6 \\ 1 & 3 & 0 \end{bmatrix} \]From above:
\[
\operatorname{Adj}(A)
=
\begin{bmatrix}
-18 & 15 & 13 \\
6 & -5 & 8 \\
11 & -3 & -10
\end{bmatrix}
\]
\[
\det(A) = 37
\]
\[
\therefore\;
A^{-1}
=
\frac{\operatorname{adj}(A)}{\det(A)}
\]
\[
=
\frac{1}{37}
\begin{bmatrix}
-18 & 15 & 13 \\
6 & -5 & 8 \\
11 & -3 & -10
\end{bmatrix}
\]
Note:
\[
\left(\mathbf{AB}\right)^{-1}
=
\mathbf{B}^{-1}\,\mathbf{A}^{-1}
\]
Product of a Square Matrix and its Inverse
\[
\mathbf{A}\,\mathbf{A}^{-1}
=
\mathbf{A}^{-1}\,\mathbf{A}
=
\mathbf{I}
\]
Example
\[
\mathbf{A}\,\mathbf{A}^{-1}
=
\begin{bmatrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{bmatrix}
\cdot
\left[
\frac{1}{37}
\begin{bmatrix}
-18 & 15 & 13 \\
6 & -5 & 8 \\
11 & -3 & -10
\end{bmatrix}
\right]
\]
\[
=
\frac{1}{37}
\begin{bmatrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{bmatrix}
\cdot
\begin{bmatrix}
-18 & 15 & 13 \\
6 & -5 & 8 \\
11 & -3 & -10
\end{bmatrix}
\]
\[
=
\frac{1}{37}
\begin{bmatrix}
-36 + 18 + 55 & 30 - 15 - 15 & 26 + 24 - 50 \\
-72 + 6 + 66 & 60 - 5 - 18 & 52 + 8 - 60 \\
-18 + 18 + 0 & 15 - 15 + 0 & 13 + 24 - 0
\end{bmatrix}
\]
\[
=
\frac{1}{37}
\begin{bmatrix}
37 & 0 & 0 \\
0 & 37 & 0 \\
0 & 0 & 37
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
This means that Gaussian elimination can be used to find the inverse:
Write the matrix with the Identity beside it.
Perform EROs to reduce the left matrix to I,
while applying the same EROs to the Identity.
The result is the inverse.
Example:
\[
\mathbf{A}
=
\begin{bmatrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
2 & 3 & 5 \\
4 & 1 & 6 \\
1 & 3 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
\[
\begin{array}{c}
\text{R1} \;\to\; \tfrac12\,\text{R1} \\[6pt]
\text{R2} \;\to\; 2\text{R1} - \text{R2} \\[6pt]
\text{R3} \;\to\; \text{R1} - 2\text{R3}
\end{array}
\qquad
\left[
\begin{matrix}
1 & \tfrac{3}{2} & \tfrac{5}{2} \\
0 & 5 & 4 \\
0 & -3 & 5
\end{matrix}
\right]
\left[
\begin{matrix}
\tfrac12 & 0 & 0 \\
2 & -1 & 0 \\
1 & 0 & -2
\end{matrix}
\right]
\]
\[
\begin{array}{c}
\text{R2} \;\to\; \tfrac15\,\text{R2} \\[6pt]
\text{R3} \;\to\; \text{R3} + \tfrac35\,\text{R2}
\end{array}
\qquad
\left[
\begin{matrix}
1 & \tfrac{3}{2} & \tfrac{5}{2} \\
0 & 1 & \tfrac{4}{5} \\
0 & 0 & \tfrac{37}{5}
\end{matrix}
\right]
\left[
\begin{matrix}
\tfrac12 & 0 & 0 \\
\tfrac{2}{5} & -\tfrac15 & 0 \\
\tfrac{11}{5} & -\tfrac35 & -2
\end{matrix}
\right]
\]
\[
\text{R3} \;\to\; \tfrac{5}{37}\,\text{R3}
\qquad
\left[
\begin{matrix}
1 & \tfrac{3}{2} & \tfrac{5}{2} \\
0 & 1 & \tfrac{4}{5} \\
0 & 0 & 1
\end{matrix}
\right]
\left[
\begin{matrix}
\tfrac12 & 0 & 0 \\
\tfrac{2}{5} & -\tfrac15 & 0 \\
\tfrac{11}{37} & -\tfrac{3}{37} & -\tfrac{10}{37}
\end{matrix}
\right]
\]
\[
\begin{array}{l}
\text{R1} \;\to\; \text{R1} - \tfrac{5}{2}\,\text{R3} \\[6pt]
\text{R2} \;\to\; \text{R2} - \tfrac{4}{5}\,\text{R3}
\end{array}
\qquad
\left[
\begin{matrix}
1 & \tfrac{3}{2} & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{matrix}
\right]
\left[
\begin{matrix}
-\tfrac{9}{37} & \tfrac{15}{74} & \tfrac{25}{37} \\
\tfrac{6}{37} & -\tfrac{5}{37} & \tfrac{8}{37} \\
\tfrac{11}{37} & -\tfrac{3}{37} & -\tfrac{10}{37}
\end{matrix}
\right]
\]
\[
\text{R1} \;\to\; \text{R1} - \tfrac{3}{2}\,\text{R2}
\qquad
\left[
\begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{matrix}
\right]
\left[
\begin{matrix}
-\tfrac{18}{37} & \tfrac{15}{37} & \tfrac{13}{37} \\
\tfrac{6}{37} & -\tfrac{5}{37} & \tfrac{8}{37} \\
\tfrac{11}{37} & -\tfrac{3}{37} & -\tfrac{10}{37}
\end{matrix}
\right]
\]
\[
\text{so }
A^{-1}
=
\begin{bmatrix}
-\tfrac{18}{37} & \tfrac{15}{37} & \tfrac{13}{37} \\
\tfrac{6}{37} & -\tfrac{5}{37} & \tfrac{8}{37} \\
\tfrac{11}{37} & -\tfrac{3}{37} & -\tfrac{10}{37}
\end{bmatrix}
\]
\[
\text{or }\;
A^{-1}
=
\frac{1}{37}
\begin{bmatrix}
-18 & 15 & 13 \\
6 & -5 & 8 \\
11 & -3 & -10
\end{bmatrix}
\]
