2.1.5 Inverses of Matrices

What is the Inverse of a Matrix?

infoNote

The inverse of a matrix is a matrix that, when multiplied by the original matrix, gives the identity matrix. Not all matrices have an inverse—only those with a non-zero determinant are invertible. The inverse of a matrix $A$ is usually denoted as $A^{-1}$

Definition:

For a square matrix $A$ , if there exists a matrix $A^{-1}$ such that:

A \times A^{-1} = A^{-1} \times A = I

where $I$ is the identity matrix, then $A^{-1}$ is called the inverse of matrix $A$ .

Conditions for a Matrix to Have an Inverse

infoNote

A matrix has an inverse if and only if its determinant is non-zero. If the determinant is zero, the matrix is called singular, and it does not have an inverse.

Inverse of a $2 × 2$ Matrix

For a $2 × 2$ matrix:

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

The inverse $A^{-1}$ is given by:

A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

where $\text{det}(A) = ad - bc$

lightbulbExample

Example: Find the inverse of

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

Step 1: Calculate the determinant:

\text{det}(A) = (1 \times 4) - (2 \times 3) = 4 - 6 = -2

Step 2: Apply the formula for the inverse:

A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}

Thus, the inverse of $A$ is:

A^{-1} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}

Inverse of a $3 × 3$ Matrix

For a $3 × 3$ matrix, the inverse calculation is more complex and involves finding the matrix of minors, matrix of cofactors, and the adjugate of the matrix.

Calculation Method

Steps for finding the inverse of a $3 × 3$ matrix $A$ :

Step 1: Find the Determinant

Step 2: Find the Matrix of Minors

Step 3: Form the Matrix of Cofactors

Step 4: Find the Adjugate

Step 5: Multiply by the Reciprocal of the Determinant

Step 1: Find the Determinant: Calculate the determinant $\text{det}(A)$ . If $\text{det}(A) = 0$ , the matrix is singular and does not have an inverse.

Step 2: Find the Matrix of Minors: For each element in the matrix, calculate the minor, which is the determinant of the $2 × 2$ matrix that remains after removing the row and column of that element.

Step 3: Form the Matrix of Cofactors: Apply a checkerboard pattern of signs (+ and -) to the minors to create the cofactor matrix.

Step 4: Find the Adjugate: Transpose the cofactor matrix to get the adjugate matrix.

Step 5: Multiply by the Reciprocal of the Determinant: Finally, the inverse is given by:

A^{-1} = \frac{1}{\text{det}(A)} \times \text{adjugate}(A)

Identity Matrix and Inverses

The identity matrix, denoted $I$ , is a special matrix where all diagonal elements are $1$ , and all other elements are $0$ . The identity matrix acts as the neutral element in matrix multiplication, similar to how $1$ is the neutral element in multiplication of numbers.

For a $2 × 2$ identity matrix:

I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

If $A \times A^{-1} = I$ , then $A^{-1}$ is the inverse of $A$ .

Properties of Inverses

Only Square Matrices: Only square matrices can have an inverse (i.e., matrices with the same number of rows and columns).
Uniqueness: If a matrix has an inverse, it is unique.
Product of Inverses: If $A$ and $B$ are invertible matrices, then:

(AB)^{-1} = B^{-1} A^{-1}

(Notice the order is reversed when finding the inverse of a product of matrices.)

Applications of Matrix Inverses

Solving Systems of Equations:

If $AX = B$ , where $A$ is a square matrix, $X$ is the column of unknowns, and $B$ is the result matrix, you can solve for $X$ by multiplying both sides by $A^{-1}$

X = A^{-1} B

Geometric Transformations:

In computer graphics and geometry, inverse matrices are used to reverse transformations (e.g., scaling, rotation).

Inverse of a Matrix

The inverse of a matrix $M$ is denoted $M^{-1}$ . For an inverse, the following is the case:

MM^{-1} = M^{-1}M = I

It's important to note that in this sense a power of $-1$ does not mean reciprocal. It is used in the sense of function notation to mean "inverse."

Worked Examples

lightbulbExample

Example: Find the Inverse of

A = \begin{pmatrix} 7 & 2 & 0 \\ 2 & 1 & 1 \\ 1 & 4 & 6 \end{pmatrix}

Step 1: Calculate the determinant of $A$

\det(A) = 7 \begin{vmatrix} 1 & 1 \\ 4 & 6 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ 1 & 6 \end{vmatrix} + 0 \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix}

= 7[(1)(6) - (4)(1)] - 2[(2)(6) - (1)(1)] + 0

= 7(2) - 2(11) + 0 = 14 - 22 = -8

Since $\det(A) = -8 \neq 0$ , the matrix is invertible.

Step 2: To compute the inverse, first find the matrix of minors by calculating the determinant of each $2 \times 2$ submatrix.

Minor $m_{11}$

\begin{vmatrix} 1 & 1 \\ 4 & 6 \end{vmatrix} = (1)(6) - (4)(1) = 2

Minor $m_{12}$

\begin{vmatrix} 2 & 1 \\ 1 & 6 \end{vmatrix} = (2)(6) - (1)(1) = 11

Minor $m_{13}$

\begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} = (2)(4) - (1)(1) = 7

Minor $m_{21}$

\begin{vmatrix} 2 & 1 \\ 4 & 6 \end{vmatrix} = (2)(6) - (1)(4) = 8

Minor $m_{22}$

\begin{vmatrix} 7 & 0 \\ 1 & 6 \end{vmatrix} = (7)(6) - (0)(1) = 42

Minor $m_{23}$

\begin{vmatrix} 7 & 0 \\ 1 & 4 \end{vmatrix} = (7)(4) - (0)(1) = 28

Minor $m_{31}$

\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 1

Minor $m_{32}$

\begin{vmatrix} 7 & 0 \\ 2 & 1 \end{vmatrix} = (7)(1) - (0)(2) = 7

Minor $m_{33}$

\begin{vmatrix} 7 & 2 \\ 2 & 1 \end{vmatrix} = (7)(1) - (2)(2) = 3

To create the matrix of cofactors, alternate signs in a checkerboard pattern:

\text{Matrix of Cofactors} = \begin{pmatrix} - & - & + \\ - & + & - \\ - & - & + \end{pmatrix}

Step 3: Apply these signs to the minors:

C = \begin{pmatrix} 2 & -11 & 7 \\ -8 & 42 & -28 \\ 1 & -7 & 3 \end{pmatrix}

Step 4: Transpose the matrix of cofactors, swapping rows and columns:

C^T = \begin{pmatrix} 2 & -8 & 1 \\ -11 & 42 & -7 \\ 7 & -28 & 3 \end{pmatrix}

Step 5: Multiply the transpose of the cofactor matrix by

\frac{1}{\det(A)} = -\frac{1}{8}

Thus:

A^{-1} = -\frac{1}{8} \times \begin{pmatrix} 2 & -8 & 1 \\ -11 & 42 & -7 \\ 7 & -28 & 3 \end{pmatrix}

Step 6: Simplify:

A^{-1} = \begin{pmatrix} -\frac{1}{4} & 1 & -\frac{1}{8} \\ \frac{11}{8} & -\frac{21}{4} & \frac{7}{8} \\ -\frac{7}{8} & \frac{7}{2} & -\frac{3}{8} \end{pmatrix}

The inverse of the matrix is:

A^{-1} = \begin{pmatrix} -\frac{1}{4} & 1 & -\frac{1}{8} \\ \frac{11}{8} & -\frac{21}{4} & \frac{7}{8} \\ -\frac{7}{8} & \frac{7}{2} & -\frac{3}{8} \end{pmatrix}

lightbulbExample

Example: Find the Inverse of

\begin{pmatrix} 2 & 7 \\ 3 & 6 \end{pmatrix}

Formula:

To find the inverse of a $2 \times 2$ matrix, we can use the following formula:

A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Where the matrix

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

And $\det(A)$ is the determinant of $A$ , given by:

\det(A) = ad - bc

Step 1: Calculate the Determinant

For the matrix

A = \begin{pmatrix} 2 & 7 \\ 3 & 6 \end{pmatrix}

$a = 2, b = 7, c = 3, d = 6$ The determinant is:

\det(A) = (2)(6) - (7)(3)

= 12 - 21 = -9

Step 2: Find the Adjoint Matrix

Next, apply the formula for the adjoint matrix:

\text{Adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Substituting the values of $a$ , $b$ , $c$ , and $d$

\text{Adj}(A) = \begin{pmatrix} 6 & -7 \\ -3 & 2 \end{pmatrix}

Step 3: Multiply by $\frac{1}{\det(A)}$

The inverse is:

A^{-1} = \frac{1}{\det(A)} \times \text{Adj}(A)

Substituting $\det(A) = -9$

A^{-1} = \frac{1}{-9} \begin{pmatrix} 6 & -7 \\ -3 & 2 \end{pmatrix}

Simplify:

A^{-1} = -\frac{1}{9} \begin{pmatrix} 6 & -7 \\ -3 & 2 \end{pmatrix} = \frac{1}{9} \begin{pmatrix} -6 & 7 \\ 3 & -2 \end{pmatrix}

Step 4: Verify the Result

To verify, multiply the original matrix by its inverse:

\begin{pmatrix} 2 & 7 \\ 3 & 6 \end{pmatrix} \times \frac{1}{9} \begin{pmatrix} -6 & 7 \\ 3 & -2 \end{pmatrix}

First, perform the matrix multiplication:

= \frac{1}{9} \begin{pmatrix} (2)(-6) + (7)(3) & (2)(7) + (7)(-2) \\ (3)(-6) + (6)(3) & (3)(7) + (6)(-2) \end{pmatrix}

Simplify each element:

= \frac{1}{9} \begin{pmatrix} -12 + 21 & 14 - 14 \\ -18 + 18 & 21 - 12 \end{pmatrix}

= \frac{1}{9} \begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix}

= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

The result is the identity matrix, confirming that the inverse is correct.

Final Answer

The inverse of

\begin{pmatrix} 2 & 7 \\ 3 & 6 \end{pmatrix}

Is:

A^{-1} = \frac{1}{9} \begin{pmatrix} -6 & 7 \\ 3 & -2 \end{pmatrix}

lightbulbExample

Example: Find the Inverse of the General Matrix

\begin{pmatrix} a & c \\ b & d \end{pmatrix}

To find the inverse of a $2 \times 2$ matrix using algebraic manipulation, we solve the equation:

\begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} q & s \\ r & t \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

This equation states that the product of the given matrix and its inverse must yield the identity matrix. Our goal is to find $q$ , $r$ , $s$ and $t$ in terms of $a$ , $b$ , $c$ and $d$ .

Step 1: Set Up a System of Equations

Multiplying the matrices gives the following system of equations:

$aq + cr = 1 \quad$ (Equation A)
$bq + dr = 0 \quad$ (Equation B)
$as + ct = 0 \quad$ (Equation C)
$bs + dt = 1 \quad$ (Equation D)

Step 2: Solve for $q$ and $r$ Using Equations $A$ and $B$

From Equation $A$ :

aq + cr = 1 \quad \Rightarrow \quad q = \frac{1 - cr}{a}

Substitute this expression for $q$ into Equation $B$ :

b\left(\frac{1 - cr}{a}\right) + dr = 0

Simplify:

\frac{b}{a}(1 - cr) + dr = 0

\frac{b}{a} - \frac{crb}{a} + dr = 0

Multiply through by $a$ to clear the fraction

b - crb + dra = 0

Rearrange:

r(ad - bc) = -b

Solve for $r$ :

r = \frac{-b}{ad - bc}

Substitute this value of $r$ into the expression for $q$ :

q = \frac{1}{a} - \frac{cr}{a}

q = \frac{1}{a} - \frac{c}{a} \left(\frac{-b}{ad - bc}\right)

Simplify:

q = \frac{1}{a} + \frac{bc}{a(ad - bc)}

q = \frac{ad}{a(ad - bc)} = \frac{d}{ad - bc}

Step 3: Solve for $s$ and $t$ Using Equations $C$ and $D$

From Equation $C$ :

as + ct = 0 \quad \Rightarrow \quad s = \frac{-ct}{a}

Substitute this into Equation $D$ :

b\left(\frac{-ct}{a}\right) + dt = 1

Simplify:

\frac{-bct}{a} + dt = 1

Multiply through by $a$ :

-bct + dta = a

t(ad - bc) = a

Solve for $t$ :

t = \frac{a}{ad - bc}

Substitute this value of $t$ into the expression for $s$ :

s = \frac{-ct}{a}

s = \frac{-c \left(\frac{a}{ad - bc}\right)}{a}

Simplify:

s = \frac{-c}{ad - bc}

Step 4: Write the Inverse Matrix

Now that we have all elements:

q = \frac{d}{ad - bc}, \quad r = \frac{-b}{ad - bc}

\quad s = \frac{-c}{ad - bc}, \quad t = \frac{a}{ad - bc}

Thus, the inverse matrix is:

M^{-1} = \begin{pmatrix} q & s \\ r & t \end{pmatrix} = \begin{pmatrix} \frac{d}{ad - bc} & \frac{-c}{ad - bc} \\ \frac{-b}{ad - bc} & \frac{a}{ad - bc} \end{pmatrix}

This can also be written as:

M^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -c \\ -b & a \end{pmatrix}

Final Answer

The inverse of

\begin{pmatrix} a & c \\ b & d \end{pmatrix}

is:

M^{-1} = \frac{1}{\text{det}(M)} \begin{pmatrix} d & -c \\ -b & a \end{pmatrix}

Where $\text{det}(M) = ad - bc$

lightbulbExample

Example: Find the Inverse of

\begin{pmatrix} 2 & 2 \\ 6 & 6 \end{pmatrix}

Step 1: Calculate the Determinant

The determinant of a $2 \times 2$ matrix

\begin{pmatrix} a & c \\ b & d \end{pmatrix}

is given by:

\det(M) = ad - bc

For the matrix

\begin{pmatrix} 2 & 2 \\ 6 & 6 \end{pmatrix}

$a = 2, b = 6, c = 2, d = 6$ Substitute these values into the formula:

\det(M) = (2)(6) - (6)(2) = 12 - 12 = 0

Step 2: Interpret the Result

Since the determinant is zero, the matrix is singular, meaning it does not have an inverse.

The formula for the inverse of a $2 \times 2$ matrix is:

M^{-1} = \frac{1}{\det(M)} \begin{pmatrix} d & -c \\ -b & a \end{pmatrix}

However, since $\det(M) = 0$ , this results in division by zero:

M^{-1} = \frac{1}{0} \begin{pmatrix} 6 & -2 \\ -6 & 2 \end{pmatrix}

This is undefined, so the matrix has no inverse.

Conclusion

When the determinant of a matrix is zero, the matrix is singular and does not have an inverse. Therefore, the matrix:

\begin{pmatrix} 2 & 2 \\ 6 & 6 \end{pmatrix}

is singular and its inverse is undefined.

infoNote

Key Takeaways:

The inverse of a matrix $A$ is a matrix $A^{-1}$ such that $A \times A^{-1} = I$ , where $I$ is the identity matrix.
A matrix is invertible if and only if its determinant is non-zero.
The inverse of a $2 × 2$ matrix is calculated using the formula

A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Inverses are essential for solving systems of equations, performing transformations, and more. Understanding how to calculate and apply matrix inverses is crucial in both algebra and geometry.

Inverses of Matrices Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

2.1.5 Inverses of Matrices

What is the Inverse of a Matrix?

Definition:

Conditions for a Matrix to Have an Inverse

Inverse of a $2 × 2$ Matrix

Inverse of a $3 × 3$ Matrix

Calculation Method

Identity Matrix and Inverses

Properties of Inverses

Applications of Matrix Inverses

Solving Systems of Equations:

Geometric Transformations:

Inverse of a Matrix

Worked Examples

Final Answer

Key Takeaways:

500K+ Students Use These Powerful Tools to Master Inverses of Matrices For their A-Level Exams.

Other Revision Notes related to Inverses of Matrices you should explore

Inverses of Matrices Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

2.1.5 Inverses of Matrices

What is the Inverse of a Matrix?

Definition:

Conditions for a Matrix to Have an Inverse

Inverse of a 2×22 × 22×2 Matrix

Inverse of a 3×33 × 33×3 Matrix

Calculation Method

Identity Matrix and Inverses

Properties of Inverses

Applications of Matrix Inverses

Solving Systems of Equations:

Geometric Transformations:

Inverse of a Matrix

Worked Examples

Final Answer

Key Takeaways:

500K+ Students Use These Powerful Tools to Master Inverses of Matrices For their A-Level Exams.

Other Revision Notes related to Inverses of Matrices you should explore

Inverse of a $2 × 2$ Matrix

Inverse of a $3 × 3$ Matrix