Determinants of the second order
Solving a pair of simultaneous equations by elimination. e.g.
$$ \begin{cases} a_1 x + b_1 y + d_1 = 0 \\ a_2 x + b_2 y + d_2 = 0 \end{cases} $$
We eliminate \(y\) from original equations and find an expression for \(x\), we obtain: \(x = \frac{b_1 d_2 - b_2 d_1} {a_1 b_2 - a_2 b_1}\)
In just the same way, eliminate \(x\) and find an expression for \(y\), we get: \(y = -\left( \frac{a_1 d_2 - a_2 d_1}{a_1 b_2 - a_2 b_1} \right)\)
So, the solutions for \(x\) and \(y\) are: $$x = \frac{b_1 d_2 - b_2 d_1} {a_1 b_2 - a_2 b_1}$$
$$y = -\left(\frac{a_1 d_2 - a_2 d_1}{a_1 b_2 - a_2 b_1} \right)$$
For the solutions to be finite, let’s assume \(a_1 b_2 - a_2 b_1 \neq 0\).
Use a shorthand notation for the above solutions:
e.g.
$$a_1 b_2 - a_2 b_1 = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}$$
We multiple \(\begin{vmatrix} a_1 & \\ & b_2 \end{vmatrix}\) and subtract the product \(\begin{vmatrix} & b_1 \\ a_2 & \end{vmatrix}\) for representing \(a_1 b_2 - a_2 b_1\) .
\(\begin{vmatrix} a1 & b_1 \\ a_2 & b_2 \end{vmatrix}\) is called a _determinant of the second order (since it has two rows and two columns) and represents \(a_1 b_2 - a_2 b_1\). And we can easily remember this as \(+\searrow \quad \text{and} \quad -\nearrow\).
Thus, the solutions \(x = \frac{b_1 d_2 - b_2 d_1} {a_1 b_2 - a_2 b_1}\) and \(y = -\left( \frac{a_1 d_2 - a_2 d_1}{a_1 b_2 - a_2 b_1} \right)\) can each have their numerators and denominators expressed as determinants:
$$ x = \frac{\begin{vmatrix} b_1 & d_1 \\ b_2 & d_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}} \quad \text{and} \quad y = -\frac{\begin{vmatrix} a_1 & d_1 \\ a_2 & d_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}} $$
$$ \therefore \frac{x}{\begin{vmatrix} b_1 & d_1 \\ b_2 & d_2 \end{vmatrix}} = \frac{1}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}} \quad \text{and} \quad \frac{y}{\begin{vmatrix} a_1 & d_1 \\ a_2 & d_2 \end{vmatrix}} = -\frac{1}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}} $$
Combine these results, thus:
$$ \frac{x}{\begin{vmatrix} b_1 & d_1 \\ b_2 & d_2 \end{vmatrix}} = -\frac{y}{\begin{vmatrix} a_1 & d_1 \\ a_2 & d_2 \end{vmatrix}} = \frac{1}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}} $$
So if
$$ \begin{cases} a_1 x + b_1 y + d_1 = 0 \\ a_2 x + b_2 y + d_2 = 0 \end{cases} $$
Then
$$ \frac{x}{\begin{vmatrix} b_1 & d_1 \\ b_2 & d_2 \end{vmatrix}} = -\frac{y}{\begin{vmatrix} a_1 & d_1 \\ a_2 & d_2 \end{vmatrix}} = \frac{1}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}} $$
Denote the determinant in three denominators:
\(\Delta_1 = \begin{vmatrix} b_1 & d_1 \\ b_2 & d_2 \end{vmatrix}\) , \(\Delta_2 = \begin{vmatrix} a_1 & d_1 \\ a_2 & d_2 \end{vmatrix}\), \(\Delta_0 = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}\)
Determinants of the third order
A determinant of the third order will contains 3 rows and 3 columns.
e.g. \(\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c2 \\ a_3 & b_3 & c3 \end{vmatrix}\)
Each element in the determinant is associated with its minor, which is found by omitting the row and column containing the element concerned.
Minor of \(a_1\) is \(\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix}\) ; Minor of \(b_1\) is \(\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix}\) ; Minor of \(c_1\) is \(\begin{vmatrix} a_2 & b_2 \\ a_3 & c_3 \end{vmatrix}\)
Evaluation of a third-order determinant:
$$ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1 \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - b_1 \begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1 \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} $$
Expand along any row or column: multiplying each element by its minor and assigning to each product the appropriate \(+\) or \(-\) sign.
The appropriate place signs are given by:
$$ \begin{array}{cccccc} + & - & + & - & \cdots & \cdots \\ - & + & - & + & \cdots & \cdots \\ + & - & + & - & + & \cdots \\ - & + & - & + & - & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \end{array} $$
Simultaneous equations in three unknowns
Consider the equations:
$$ \begin{cases} a_1 x + b_1 y + c_1 z + d_1 = 0 \\ a_2 x + b_2 y + c_2 z + d_2 = 0 \\ a_3 x + b_3 y + c_3 z + d_3 = 0 \end{cases} $$
Find \(x,y,\) and \(z\) by elimination method, we obtain results that can be expressed in determinant form:
$$ \frac{x}{\begin{vmatrix} b_1 & c_1 & d_1 \\ b_2 & c_2 & d_2 \\ b_3 & c_3 & d_3 \end{vmatrix}} = -\frac{y}{\begin{vmatrix} a_1 & c_1 & d_1 \\ a_2 & c_2 & d_2 \\ a_3 & c_3 & d_3 \end{vmatrix}} = \frac{z}{\begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}} = -\frac{1}{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}} $$
$$ \frac{x}{\Delta_1} = -\frac{y}{\Delta_2} = \frac{z}{\Delta_3} = -\frac{1}{\Delta_0} $$
- \(\Delta_1\) = the determinant of the coefficients omitting the \(x-terms\)
- \(\Delta_2\) = the determinant of the coefficients omitting the \(y-terms\)
- \(\Delta_3\) = the determinant of the coefficients omitting the \(z-terms\)
- \(\Delta_0\) = the determinant of the coefficients omitting the constant terms
Properties of determinants
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The value of a determinant remains unchanged if rows are changed to columns and columns to rows. $$ \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} $$
-
If two rows (or two columns) are interchanged, the sign of the determinant is changed. $$ \begin{vmatrix} a_2 & b_2 \\ a_1 & b_1 \end{vmatrix} = - \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} $$
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If two rows (or two columns) are identical, the value of the determinant is zero. $$ \begin{vmatrix} a_1 & a_1 \\ a_2 & a_2 \end{vmatrix} = 0 $$
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If the elements of any one row (or column) are all multiplied by a common factor, the determinant is multiplied by that factor. $$ \begin{vmatrix} ka_1 & kb_1 \\ a_2 & b_2 \end{vmatrix} = k \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} $$
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If the elements of any row (or column) are increased (or decreased) by equal multiples of the corresponding elements of any other row (or column), the value of the determinant is unchanged. $$ \begin{vmatrix} a_1 + k b_1 & b_1 \\ a_2 + k b_2 & b_2 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} \quad \text{and} \quad \begin{vmatrix} a_1 & b_1 \\ a_2 + k a_1 & b_2 + k b_1 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} $$
Reference
- Engineering Mathematics — K.A.STROUD
- Determinants — Khanacdemy
- Determinant — Wikipedia
- Wolfran MathWorld Determinant
- Determinant of a Matrix
- Matrix Determinant — ncl.ac.uk