Inverse of a matrix

Inverse of a Matrix

The Determinant of a 2 X 2 Matrix

Determinant of a matrix

$$A = \begin{pmatrix} \stackrel{\curvearrowleft}{a} & \stackrel{\curvearrowright}{c} \\ \stackrel{\curvearrowleft}{b} & \stackrel{\curvearrowright}{d} \end{pmatrix}$$ $$\begin{aligned} &\stackrel{\curvearrowleft}{a}\rightarrow \stackrel{\curvearrowleft}{d} = \text{Leading diagonal} \\ &\stackrel{\curvearrowleft}{b}\rightarrow \stackrel{\curvearrowleft}{c} = \text{Main diagonal} \end{aligned}$$

Now the determinant of matrix A is then defined as the difference of the product of elements in the leading diagonal and the product of the elements in the main diagonal.

Example 1

Find

$$\text{If } A = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}, \text{ then:}$$

$|A| = 1 \times 4 - 2 \times 3 = 4 - 6 = -2$

$$\therefore |A| = -2$$

$|A|$ if $A = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$

Solution: $|A| = 1 \times 4 - 2 \times 3 = 4 - 6 = -2$

$\therefore |A| = -2$

Example 2

Considering

If $B = \begin{pmatrix} 1 & 0 \\ 3 & 5 \end{pmatrix}$, find $|A|$

Solution: $|B| = 1 \times 5 - 0 \times 3 = 5 - 0 = 5$

$\therefore |B| = 5$

Example 3

Find the value of x

If $C = \begin{pmatrix} 2x & 3 \\ -2 & 10 \end{pmatrix}$ having determinant 46.

Solution: $C = \begin{pmatrix} 2x & 3 \\ -2 & 10 \end{pmatrix}$

$|C| = 2x \cdot 10 - (-2) \cdot 3 = 46$

$20x + 6 = 46$

$20x = 46 - 6$

$20x = 40$

$x = \frac{40}{20} = 2$

$\therefore$ The value of x is 2.

Singular and non-singular matrices

Definition:

A singular matrix is a matrix whose determinant is zero, while non – singular matrix is the one with a non zero determinant.

Example 4

Find the value of y

The Inverse of a 2 X 2 Matrix

If

$$B = \begin{pmatrix} 4 & 5 \\ 3y & 6 \end{pmatrix}$$

is a singular matrix, then:

$|B| = 4 \times 6 - 5 \times (3y)$

$|B| = 24 - 15y$

Since $B$ is a singular matrix, its determinant must be zero:

$$24 - 15y = 0$$ $$24 = 15y$$ $$y = \frac{24}{15} = \frac{8}{5}$$ $$\therefore y = \frac{8}{5}$$

Inverse of matrices

Definition: If A is a square matrix and B is another matrix with the same order as A, then B is the inverse of A if AB = BA = I where I is the identity matrix.

Thus AB = BA = I means either A is the inverse of B or B is the inverse of A.

Where $B = A^{-1}$, that is B is the inverse of matrix A

$$AA^{-1} = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

So,

$$\begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} p & r \\ q & s \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ $$\begin{pmatrix} ap + cq & ar + cs \\ bp + dq & br + ds \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

From which we obtain:

$$\begin{aligned} &ap + cq = 1 \quad \text{(i)} \\ &ar + cs = 0 \quad \text{(ii)} \\ &bp + dq = 0 \quad \text{(iii)} \\ &br + ds = 1 \quad \text{(iv)} \end{aligned}$$

Since we need the unknown matrix B, we can solve for p and q by using equations (i) and (iii) and we solve for r and s using equations (ii) and (iv)

$$\text{Now: } \begin{cases} ap + cq = 1 \\ bp + dq = 0 \end{cases} \quad \text{and} \quad \begin{cases} ar + cs = 0 \\ br + ds = 1 \end{cases}$$ $$\text{By elimination method}$$ $$\begin{aligned} & \text{From } \begin{cases} ap + cq = 1 \quad \text{(1)} \\ bp + dq = 0 \quad \text{(2)} \end{cases} \\ \text{Multiply (1) by } b: &\quad abp + bcq = b \\ \text{Multiply (2) by } a: &\quad abp + adq = 0 \\ \text{Subtract:} &\quad (bc - ad)q = b \\ \Rightarrow \quad & q = \frac{b}{bc - ad} \end{aligned}$$ $$\text{To find } p:$$ $$\begin{aligned} & ap + cq = 1 \\ & \Rightarrow p = \frac{1 - cq}{a} \end{aligned}$$ $$\text{Since } q = \frac{b}{bc - ad}, \text{ substitute into the equation for } p:$$ $$\begin{aligned} p &= \frac{1 - c\left(\frac{b}{bc - ad}\right)}{a} \\ &= \frac{(bc - ad) - bc}{a(bc - ad)} \\ &= \frac{-ad}{a(bc - ad)} \\ &= \frac{d}{ad - bc} \end{aligned}$$

Also to get r and s, the same procedure must be followed:

And

$$\begin{aligned} & \begin{cases} ar + cs = 0 \\ br + ds = 1 \end{cases} \\[10pt] & \begin{cases} adr + cds = 0 \\ bcr + cds = c \end{cases} \\[10pt] & (ad - bc)r = -c \\ & \text{or } (bc - ad)r = c \\ & r = \frac{c}{bc - ad} \\[10pt] & \text{Therefore } p = \frac{d}{ad - bc}, \quad q = \frac{-b}{ad - bc}, \quad r = \frac{c}{bc - ad}, \quad s = \frac{a}{ad - bc}. \end{aligned}$$

Remember $B = \begin{pmatrix} p & r \\ q & s \end{pmatrix}$ which is the inverse of $A$ where $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

So

$$B = \begin{pmatrix} \dfrac{d}{ad-bc} & \dfrac{c}{bc-ad} \\ \dfrac{b}{bc-ad} & \dfrac{a}{ad-bc} \end{pmatrix}$$

But $bc-ad = -(ad-bc)$.

Now

$$B = \begin{pmatrix} \dfrac{d}{ad-bc} & \dfrac{-c}{ad-bc} \\ \dfrac{-b}{ad-bc} & \dfrac{a}{ad-bc} \end{pmatrix}$$

Example 5

Given that,

If $B = \begin{pmatrix} 4 & 5 \\ 3y & 6 \end{pmatrix}$ is a singular matrix,

Solution:

Then,

$|B| = 4 \times 6 - 5 \times (3y) = 24 - 15y$

Since $B$ is a singular matrix, its determinant is zero:

$|B| = 0 \Rightarrow 24 - 15y = 0$

Solving for $y$:

$$24 = 15y \Rightarrow y = \frac{24}{15} = \frac{8}{5}$$ $$\therefore y = \frac{8}{5}$$

Example 6

(a) $B = \begin{pmatrix} -1 & 3 \\ 2 & 6 \end{pmatrix}$, $|B| = -1 \times 6 - 2 \times 3$

$|B| = -12 \neq 0$

$\therefore B^{-1}$ exists

(b) $C = \begin{pmatrix} 3 & 6 \\ 4 & 8 \end{pmatrix}$, $|C| = 3 \times 8 - 4 \times 6 = 24 - 24 = 0$

$|C| = 0, \text{ so } C^{-1} \text{ does not exist}$

(c) $D = \begin{pmatrix} 2 & 9 \\ 4 & 18 \end{pmatrix}$, $|D| = 2 \times 18 - 9 \times 4 = 36 - 36 = 0$

$\therefore D^{-1}$ does not exist.

Which of the following matrices have inverses?

2 X 2 Matrix to Solve Simultaneous Equations

Solving simultaneous equations by matrix method:

If $A$ and $B$ are two matrices such that

$$A = \begin{pmatrix} a & c \\ b & d \end{pmatrix}, \quad B = \begin{pmatrix} x \\ y \end{pmatrix}$$

Then

$$AB = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + cy \\ bx + dy \end{pmatrix}$$

where $a, b, c, d, x$, and $y$ are any real numbers.

Let $AB = C$ where

$$C = \begin{pmatrix} p \\ q \end{pmatrix}$$

Then,

$$AB = C \Rightarrow \begin{pmatrix} ax + cy \\ bx + dy \end{pmatrix} = \begin{pmatrix} p \\ q \end{pmatrix}$$ $$\begin{aligned} ax + cy &= p \quad \ldots \ldots \ldots (1) \\ bx + dy &= q \quad \ldots \ldots \ldots (2) \end{aligned}$$

Therefore, for any system of simultaneous equations, a matrix method can be used if and only if $ad - bc \neq 0$.

Since the system:

$$ax + cy = p \quad \text{and} \quad bx + dy = q$$

can be written in matrix form as:

$$\begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} p \\ q \end{pmatrix}$$

Let

$$A = \begin{pmatrix} a & c \\ b & d \end{pmatrix}, \quad B = \begin{pmatrix} x \\ y \end{pmatrix}, \quad C = \begin{pmatrix} p \\ q \end{pmatrix}$$

Then the matrix equation becomes:

$$AB = C$$

If $A^{-1}$ is multiplied on each side of the equation, we get:

$$A^{-1} \cdot AB = A^{-1} \cdot C$$ $$(A^{-1}A) \cdot B = A^{-1} \cdot C$$

But $A^{-1}A = I$, where $I$ is the identity matrix, and for any matrix $K$, $KI = IK = K$.

Therefore:

$$I \cdot B = A^{-1}C \Rightarrow B = A^{-1}C$$

Now by equating the corresponding elements, the following simultaneous equations are obtained.

Then $B = A^{-1} \times C$

Cramer's Rule

Cramer's Rule is another method to solve the equations of the form:

$$\begin{cases} ax + cy = p \\ bx + dy = q \end{cases}$$

By using the inverse of a matrix method, we have seen that:

$$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{ad - bc} \begin{pmatrix} d & -c \\ -b & a \end{pmatrix} \begin{pmatrix} p \\ q \end{pmatrix}$$

Or:

$$x = \frac{dp - cq}{ad - bc}, \quad y = \frac{aq - bp}{ad - bc}$$

Where $ad - bc$ is the determinant of the matrix $\begin{pmatrix} a & c \\ b & d \end{pmatrix}$.

So

$$x = \frac{\begin{vmatrix} p & c \\ c & d \end{vmatrix}}{\begin{vmatrix} a & c \\ b & d \end{vmatrix}} \quad \text{and} \quad y = \frac{\begin{vmatrix} d & p \\ b & q \end{vmatrix}}{\begin{vmatrix} a & c \\ b & d \end{vmatrix}}$$

In $x = \frac{\begin{vmatrix} p & c \\ c & d \end{vmatrix}}{\begin{vmatrix} a & c \\ b & d \end{vmatrix}}$, the numerator is obtained by replacing the first column by the column $\begin{pmatrix} p \\ q \end{pmatrix}$.

In $y = \frac{\begin{vmatrix} d & p \\ b & q \end{vmatrix}}{\begin{vmatrix} a & c \\ b & d \end{vmatrix}}$, the numerator is obtained by taking the determinant of the matrix after replacing the second column by the column $\begin{pmatrix} p \\ q \end{pmatrix}$.

While the denominator in both cases is the determinant of $\begin{pmatrix} a & c \\ b & d \end{pmatrix}$.

Example

Solve

$$\begin{cases} 3x - y = 11 \\ x + 3y = -3 \end{cases}$$

Solution:

The equations can be written as

$$\begin{pmatrix} 3 & -1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 11 \\ -3 \end{pmatrix}$$

So,

$$x = \frac{\begin{vmatrix} 11 & -1 \\ -3 & 3 \end{vmatrix}}{\begin{vmatrix} 3 & -1 \\ 1 & 3 \end{vmatrix}} \quad \text{and} \quad y = \frac{\begin{vmatrix} 3 & 11 \\ 1 & -3 \end{vmatrix}}{\begin{vmatrix} 3 & -1 \\ 1 & 3 \end{vmatrix}}$$

Now, calculate:

For $x$:

$$x = \frac{11 \times 3 - (-3 \times -1)}{3 \times 3 - 1 \times (-1)} = \frac{33 - 3}{9 + 1} = \frac{30}{10} = 3$$

For $y$:

$$y = \frac{3 \times (-3) - 1 \times 11}{3 \times 3 - 1 \times (-1)} = \frac{-9 - 11}{9 + 1} = \frac{-20}{10} = -2$$

Therefore, $x = 3$ and $y = -2$