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Matrix

Matrix Inverse

Understanding Matrix Inverse

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

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

A⋅A−1=A−1⋅A=IA \cdot A^{-1} = A^{-1} \cdot A = IA⋅A−1=A−1⋅A=I

The identity matrix III is a square matrix where all main diagonal elements are 1 and all other elements are 0. For example, for a 2x2 order: I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}I=[10​01​].

Nonsingular and Singular Matrices

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

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

Inverse of a 2x2 Matrix

For a 2x2 matrix AAA, let:

A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}A=[ac​bd​]

The inverse of matrix AAA can be found using the following formula, provided that det⁡(A)≠0\det(A) \neq 0det(A)=0:

A−1=1det⁡(A)Adj(A)A^{-1} = \frac{1}{\det(A)} \text{Adj}(A)A−1=det(A)1​Adj(A)

Let's understand each component of this formula:

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

    Calculated as:

    ∣A∣=ad−bc|A| = ad - bc∣A∣=ad−bc
  2. Adjoint of Matrix A (Adj(A)\text{Adj}(A)Adj(A)):

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

    Adj(A)=[d−b−ca]\text{Adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}Adj(A)=[d−c​−ba​]

So, the complete formula for the inverse of a 2x2 matrix is:

A−1=1ad−bc[d−b−ca]A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}A−1=ad−bc1​[d−c​−ba​]

Example of Inverse of a 2x2 Matrix

Find the inverse of matrix P=[3−7−12]P = \begin{bmatrix} 3 & -7 \\ -1 & 2 \end{bmatrix}P=[3−1​−72​].

Solution:

Step 1: Identify the elements of matrix PPP.

a=3,b=−7,c=−1,d=2a=3, b=-7, c=-1, d=2a=3,b=−7,c=−1,d=2

Step 2: Calculate the determinant of matrix PPP.

det⁡(P)=(3)(2)−(−7)(−1)=6−7=−1\det(P) = (3)(2) - (-7)(-1) = 6 - 7 = -1det(P)=(3)(2)−(−7)(−1)=6−7=−1

Since det⁡(P)≠0\det(P) \neq 0det(P)=0, matrix PPP has an inverse.

Step 3: Determine the adjoint of matrix PPP.

Adj(P)=[2−(−7)−(−1)3]=[2713]\text{Adj}(P) = \begin{bmatrix} 2 & -(-7) \\ -(-1) & 3 \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix}Adj(P)=[2−(−1)​−(−7)3​]=[21​73​]

Step 4: Calculate the inverse of matrix PPP.

P−1=1det⁡(P)Adj(P)P^{-1} = \frac{1}{\det(P)} \text{Adj}(P)P−1=det(P)1​Adj(P)
P−1=1−1[2713]P^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix}P−1=−11​[21​73​]
P−1=−1[2713]P^{-1} = -1 \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix}P−1=−1[21​73​]
P−1=[−2−7−1−3]P^{-1} = \begin{bmatrix} -2 & -7 \\ -1 & -3 \end{bmatrix}P−1=[−2−1​−7−3​]

Thus, the inverse of matrix PPP is P−1=[−2−7−1−3]P^{-1} = \begin{bmatrix} -2 & -7 \\ -1 & -3 \end{bmatrix}P−1=[−2−1​−7−3​].

Inverse of a 3x3 Matrix

The basic concept for finding the inverse of a 3x3 matrix is the same as for a 2x2 matrix, i.e., using the formula:

A−1=1det⁡(A)Adj(A)A^{-1} = \frac{1}{\det(A)} \text{Adj}(A)A−1=det(A)1​Adj(A)

However, the calculation of the determinant (det⁡(A)\det(A)det(A)) and adjoint (Adj(A)\text{Adj}(A)Adj(A)) for a 3x3 matrix is more complex.

  • The determinant of a 3x3 matrix can be calculated using Sarrus's rule or the cofactor expansion method.
  • The adjoint of a 3x3 matrix is obtained from the transpose of its cofactor matrix.

An in-depth discussion on how to calculate the determinant and adjoint of a 3x3 matrix will usually be studied separately as it involves more steps.

Properties of Matrix Inverse

One important use of the matrix inverse is to solve systems of linear equations. If a system of linear equations can be expressed in matrix multiplication form:

AX=BAX = BAX=B

where AAA is the coefficient matrix, XXX is the variable matrix, and BBB is the constant matrix. If matrix AAA has an inverse (A−1A^{-1}A−1), then the solution for XXX can be found by:

X=A−1BX = A^{-1}BX=A−1B

This is a very useful property in various mathematical and engineering applications.

Exercises

Given matrices X=[1−3−14]X = \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix}X=[1−1​−34​] and Y=[−423−2]Y = \begin{bmatrix} -4 & 2 \\ 3 & -2 \end{bmatrix}Y=[−43​2−2​].

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

Answer Key

  1. Determining X−1X^{-1}X−1:

    X=[1−3−14]X = \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix}X=[1−1​−34​]
    det⁡(X)=(1)(4)−(−3)(−1)=4−3=1\det(X) = (1)(4) - (-3)(-1) = 4 - 3 = 1det(X)=(1)(4)−(−3)(−1)=4−3=1
    Adj(X)=[4311]\text{Adj}(X) = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}Adj(X)=[41​31​]
    X−1=11[4311]=[4311]X^{-1} = \frac{1}{1} \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}X−1=11​[41​31​]=[41​31​]

    Determining Y−1Y^{-1}Y−1:

    Y=[−423−2]Y = \begin{bmatrix} -4 & 2 \\ 3 & -2 \end{bmatrix}Y=[−43​2−2​]
    det⁡(Y)=(−4)(−2)−(2)(3)=8−6=2\det(Y) = (-4)(-2) - (2)(3) = 8 - 6 = 2det(Y)=(−4)(−2)−(2)(3)=8−6=2
    Adj(Y)=[−2−2−3−4]\text{Adj}(Y) = \begin{bmatrix} -2 & -2 \\ -3 & -4 \end{bmatrix}Adj(Y)=[−2−3​−2−4​]
    Y−1=12[−2−2−3−4]=[−1−1−3/2−2]Y^{-1} = \frac{1}{2} \begin{bmatrix} -2 & -2 \\ -3 & -4 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -3/2 & -2 \end{bmatrix}Y−1=21​[−2−3​−2−4​]=[−1−3/2​−1−2​]
  2. Determining X−1+Y−1X^{-1} + Y^{-1}X−1+Y−1:

    X−1+Y−1=[4311]+[−1−1−3/2−2]=[4−13−11−3/21−2]=[32−1/2−1]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}X−1+Y−1=[41​31​]+[−1−3/2​−1−2​]=[4−11−3/2​3−11−2​]=[3−1/2​2−1​]
  3. Determining (X+Y)−1(X+Y)^{-1}(X+Y)−1:

    First, calculate X+YX+YX+Y:

    X+Y=[1−3−14]+[−423−2]=[1−4−3+2−1+34−2]=[−3−122]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}X+Y=[1−1​−34​]+[−43​2−2​]=[1−4−1+3​−3+24−2​]=[−32​−12​]

    Let Z=X+Y=[−3−122]Z = X+Y = \begin{bmatrix} -3 & -1 \\ 2 & 2 \end{bmatrix}Z=X+Y=[−32​−12​]. Now, calculate Z−1Z^{-1}Z−1:

    det⁡(Z)=(−3)(2)−(−1)(2)=−6−(−2)=−6+2=−4\det(Z) = (-3)(2) - (-1)(2) = -6 - (-2) = -6 + 2 = -4det(Z)=(−3)(2)−(−1)(2)=−6−(−2)=−6+2=−4
    Adj(Z)=[21−2−3]\text{Adj}(Z) = \begin{bmatrix} 2 & 1 \\ -2 & -3 \end{bmatrix}Adj(Z)=[2−2​1−3​]
    (X+Y)−1=Z−1=1−4[21−2−3]=[−2/4−1/42/43/4]=[−1/2−1/41/23/4](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}(X+Y)−1=Z−1=−41​[2−2​1−3​]=[−2/42/4​−1/43/4​]=[−1/21/2​−1/43/4​]
  4. Comparison of X−1+Y−1X^{-1} + Y^{-1}X−1+Y−1 and (X+Y)−1(X+Y)^{-1}(X+Y)−1:

    From the calculations:

    X−1+Y−1=[32−1/2−1]X^{-1} + Y^{-1} = \begin{bmatrix} 3 & 2 \\ -1/2 & -1 \end{bmatrix}X−1+Y−1=[3−1/2​2−1​]
    (X+Y)−1=[−1/2−1/41/23/4](X+Y)^{-1} = \begin{bmatrix} -1/2 & -1/4 \\ 1/2 & 3/4 \end{bmatrix}(X+Y)−1=[−1/21/2​−1/43/4​]

    Clearly, X−1+Y−1≠(X+Y)−1X^{-1} + Y^{-1} \neq (X+Y)^{-1}X−1+Y−1=(X+Y)−1. This shows that the inverse of the sum of two matrices is generally not equal to the sum of their individual inverses.

    This property differs from some algebraic operations on real numbers.

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  • Matrix InverseLearn matrix inverse formulas for 2x2 and 3x3 matrices. Master determinant calculations, adjoint methods, and solve linear equations step-by-step.
On this page
  • Understanding Matrix Inverse
  • Nonsingular and Singular Matrices
  • Inverse of a 2x2 Matrix
    • Example of Inverse of a 2x2 Matrix
  • Inverse of a 3x3 Matrix
  • Properties of Matrix Inverse
  • Exercises
    • Answer Key
  • Comments
  • Report
  • Source code