Matrix Inverse

Multiplying a Matrix by Its Inverse

In the set of real numbers, every non-zero number $a$ has a reciprocal, which is the number $a^{-1}$, satisfying the property $a \cdot a^{-1} = a^{-1} \cdot a = 1$. A similar concept applies to matrices.

If $A$ is a square matrix (e.g., of order $n \times n$) and $I$ is the identity matrix of the same order, then the inverse of matrix $A$, denoted as $A^{-1}$, is a matrix that satisfies the property:

The identity matrix $I$ is a square matrix where all main diagonal elements are $1$ and all other elements are $0$. For example, for a $2 \times 2$ order:

Nonsingular and Singular Matrices

Not all square matrices have an inverse. A matrix $A$ has an inverse if and only if the determinant of the matrix is not equal to zero ($\det(A) \neq 0$ or $|A| \neq 0$).

• Matrix $A$ is called a nonsingular matrix if $\det(A) \neq 0$. A nonsingular matrix always has an inverse.
• Matrix $A$ is called a singular matrix if $\det(A) = 0$. A singular matrix does not have an inverse.

Inverse of a Two by Two Matrix

For a $2 \times 2$ matrix $A$, let:

The inverse of matrix $A$ can be found using the following formula, provided that $\det(A) \neq 0$:

Understand each component of this formula:

1. Determinant of Matrix $A$ ($\det(A)$ or $|A|$):

   Calculated as:

   $$|A| = ad - bc$$

2. Adjugate of Matrix $A$ ($\text{Adj}(A)$):

   Obtained by swapping the main diagonal elements and changing the sign of the other diagonal elements:

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

So, the complete formula for the inverse of a $2 \times 2$ matrix is:

Example of a Two by Two Matrix Inverse

Find the inverse of the following matrix $P$:

Solution:

Step 1: Identify the elements of matrix $P$.

Step 2: Calculate the determinant of matrix $P$.

Since $\det(P) \neq 0$, matrix $P$ has an inverse.

Step 3: Determine the adjugate of matrix $P$.

Step 4: Calculate the inverse of matrix $P$.

The matrix above is $P^{-1}$.

Verify the result by multiplying the matrices:

Inverse of a Three by Three Matrix

The basic concept for finding the inverse of a $3 \times 3$ matrix is the same as for a $2 \times 2$ matrix, i.e., using the formula:

However, the calculation of the determinant ($\det(A)$) and adjugate ($\text{Adj}(A)$) for a $3 \times 3$ matrix is more complex.

• The determinant of a $3 \times 3$ matrix can be calculated using Sarrus's rule or the cofactor expansion method.
• The adjugate of a $3 \times 3$ matrix is obtained from the transpose of its cofactor matrix.

For a $3 \times 3$ matrix, the process begins by calculating the determinant and adjugate. Both results are then substituted into the inverse formula.

Solving a Linear System with an Inverse

A matrix inverse can solve a system of linear equations when its coefficient matrix is invertible. Write the system in matrix form:

In this equation, $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix. If $A$ has an inverse $A^{-1}$, multiply both sides from the left by $A^{-1}$. Keep this order because matrix multiplication is not generally commutative:

Multiplying the left side of both sides by $A^{-1}$ gives $X=A^{-1}B$. The multiplication order must stay the same because matrix multiplication is not commutative.

Exercises

Given the following matrices:

1. Determine matrices $X^{-1}$ and $Y^{-1}$.
2. Determine matrix $X^{-1} + Y^{-1}$.
3. Determine matrix $(X+Y)^{-1}$.
4. Is matrix $X^{-1} + Y^{-1}$ equal to matrix $(X+Y)^{-1}$? Explain your answer.

Worked Solutions

1. Determining $X^{-1}$:

   $$X = \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix}$$

   $$\begin{aligned} \det(X) &= (1)(4) - (-3)(-1) \\ &= 4 - 3 \\ &= 1 \end{aligned}$$

   $$\text{Adj}(X) = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}$$

   $$X^{-1} = \frac{1}{1} \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}$$

   Determining $Y^{-1}$:

   $$Y = \begin{bmatrix} -4 & 2 \\ 3 & -2 \end{bmatrix}$$

   $$\begin{aligned} \det(Y) &= (-4)(-2) - (2)(3) \\ &= 8 - 6 \\ &= 2 \end{aligned}$$

   $$\text{Adj}(Y) = \begin{bmatrix} -2 & -2 \\ -3 & -4 \end{bmatrix}$$

   $$\begin{aligned} Y^{-1} &= \frac{1}{2} \begin{bmatrix} -2 & -2 \\ -3 & -4 \end{bmatrix} \\ &= \begin{bmatrix} -1 & -1 \\ -3/2 & -2 \end{bmatrix} \end{aligned}$$

2. Determining $X^{-1} + Y^{-1}$:

   $$\begin{aligned} &X^{-1} + Y^{-1} \\ &= \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & -1 \\ -3/2 & -2 \end{bmatrix} \\ &= \begin{bmatrix} 4-1 & 3-1 \\ 1-3/2 & 1-2 \end{bmatrix} \\ &= \begin{bmatrix} 3 & 2 \\ -1/2 & -1 \end{bmatrix} \end{aligned}$$

3. Determining $(X+Y)^{-1}$:

   First, calculate $X+Y$:

   $$\begin{aligned} &X+Y \\ &= \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix} + \begin{bmatrix} -4 & 2 \\ 3 & -2 \end{bmatrix} \\ &= \begin{bmatrix} 1-4 & -3+2 \\ -1+3 & 4-2 \end{bmatrix} \\ &= \begin{bmatrix} -3 & -1 \\ 2 & 2 \end{bmatrix} \end{aligned}$$

   Let $Z = X+Y$. Calculate $Z^{-1}$:

   $$\begin{aligned} \det(Z) &= (-3)(2) - (-1)(2) \\ &= -6 - (-2) \\ &= -6 + 2 \\ &= -4 \end{aligned}$$

   $$\text{Adj}(Z) = \begin{bmatrix} 2 & 1 \\ -2 & -3 \end{bmatrix}$$

   $$\begin{aligned} (X+Y)^{-1} &= Z^{-1} \\ &= \frac{1}{-4} \begin{bmatrix} 2 & 1 \\ -2 & -3 \end{bmatrix} \\ &= \begin{bmatrix} -2/4 & -1/4 \\ 2/4 & 3/4 \end{bmatrix} \\ &= \begin{bmatrix} -1/2 & -1/4 \\ 1/2 & 3/4 \end{bmatrix} \end{aligned}$$

4. Comparison of $X^{-1} + Y^{-1}$ and $(X+Y)^{-1}$:

   From the calculations:

   $$X^{-1} + Y^{-1} = \begin{bmatrix} 3 & 2 \\ -1/2 & -1 \end{bmatrix}$$

   $$(X+Y)^{-1} = \begin{bmatrix} -1/2 & -1/4 \\ 1/2 & 3/4 \end{bmatrix}$$

   The two matrices differ, so $X^{-1} + Y^{-1} \neq (X+Y)^{-1}$. Inversion is not distributive over addition. The same rule already holds for ordinary reciprocals: in general, $1/x+1/y\neq1/(x+y)$.