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Matrix

Matrix Inverse

Understanding Matrix Inverse

In the set of real numbers, every non-zero number aa has a reciprocal, which is the number a1a^{-1}, satisfying the property aa1=a1a=1a \cdot a^{-1} = a^{-1} \cdot a = 1. A similar concept applies to matrices.

If AA is a square matrix (e.g., of order n×nn \times n) and II is the identity matrix of the same order, then the inverse of matrix AA, denoted as A1A^{-1}, is a matrix that satisfies the property:

AA1=A1A=IA \cdot A^{-1} = A^{-1} \cdot A = I

The identity matrix II 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}.

Nonsingular and Singular Matrices

Not all square matrices have an inverse. A matrix AA has an inverse if and only if the determinant of the matrix is not equal to zero (det(A)0\det(A) \neq 0 or A0|A| \neq 0).

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

Inverse of a 2x2 Matrix

For a 2x2 matrix AA, let:

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

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

A1=1det(A)Adj(A)A^{-1} = \frac{1}{\det(A)} \text{Adj}(A)

Let's understand each component of this formula:

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

    Calculated as:

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

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

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

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

A1=1adbc[dbca]A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

Example of Inverse of a 2x2 Matrix

Find the inverse of matrix P=[3712]P = \begin{bmatrix} 3 & -7 \\ -1 & 2 \end{bmatrix}.

Solution:

Step 1: Identify the elements of matrix PP.

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

Step 2: Calculate the determinant of matrix PP.

det(P)=(3)(2)(7)(1)=67=1\det(P) = (3)(2) - (-7)(-1) = 6 - 7 = -1

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

Step 3: Determine the adjoint of matrix PP.

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}

Step 4: Calculate the inverse of matrix PP.

P1=1det(P)Adj(P)P^{-1} = \frac{1}{\det(P)} \text{Adj}(P)
P1=11[2713]P^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix}
P1=1[2713]P^{-1} = -1 \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix}
P1=[2713]P^{-1} = \begin{bmatrix} -2 & -7 \\ -1 & -3 \end{bmatrix}

Thus, the inverse of matrix PP is P1=[2713]P^{-1} = \begin{bmatrix} -2 & -7 \\ -1 & -3 \end{bmatrix}.

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:

A1=1det(A)Adj(A)A^{-1} = \frac{1}{\det(A)} \text{Adj}(A)

However, the calculation of the determinant (det(A)\det(A)) and adjoint (Adj(A)\text{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 = B

where AA is the coefficient matrix, XX is the variable matrix, and BB is the constant matrix. If matrix AA has an inverse (A1A^{-1}), then the solution for XX can be found by:

X=A1BX = A^{-1}B

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

Exercises

Given matrices X=[1314]X = \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix} and Y=[4232]Y = \begin{bmatrix} -4 & 2 \\ 3 & -2 \end{bmatrix}.

  1. Determine matrices X1X^{-1} and Y1Y^{-1}.
  2. Determine matrix X1+Y1X^{-1} + Y^{-1}.
  3. Determine matrix (X+Y)1(X+Y)^{-1}.
  4. Is matrix X1+Y1X^{-1} + Y^{-1} equal to matrix (X+Y)1(X+Y)^{-1}? Explain your answer.

Answer Key

  1. Determining X1X^{-1}:

    X=[1314]X = \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix}
    det(X)=(1)(4)(3)(1)=43=1\det(X) = (1)(4) - (-3)(-1) = 4 - 3 = 1
    Adj(X)=[4311]\text{Adj}(X) = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}
    X1=11[4311]=[4311]X^{-1} = \frac{1}{1} \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}

    Determining Y1Y^{-1}:

    Y=[4232]Y = \begin{bmatrix} -4 & 2 \\ 3 & -2 \end{bmatrix}
    det(Y)=(4)(2)(2)(3)=86=2\det(Y) = (-4)(-2) - (2)(3) = 8 - 6 = 2
    Adj(Y)=[2234]\text{Adj}(Y) = \begin{bmatrix} -2 & -2 \\ -3 & -4 \end{bmatrix}
    Y1=12[2234]=[113/22]Y^{-1} = \frac{1}{2} \begin{bmatrix} -2 & -2 \\ -3 & -4 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -3/2 & -2 \end{bmatrix}
  2. Determining X1+Y1X^{-1} + Y^{-1}:

    X1+Y1=[4311]+[113/22]=[413113/212]=[321/21]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}
  3. Determining (X+Y)1(X+Y)^{-1}:

    First, calculate X+YX+Y:

    X+Y=[1314]+[4232]=[143+21+342]=[3122]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}

    Let Z=X+Y=[3122]Z = X+Y = \begin{bmatrix} -3 & -1 \\ 2 & 2 \end{bmatrix}. Now, calculate Z1Z^{-1}:

    det(Z)=(3)(2)(1)(2)=6(2)=6+2=4\det(Z) = (-3)(2) - (-1)(2) = -6 - (-2) = -6 + 2 = -4
    Adj(Z)=[2123]\text{Adj}(Z) = \begin{bmatrix} 2 & 1 \\ -2 & -3 \end{bmatrix}
    (X+Y)1=Z1=14[2123]=[2/41/42/43/4]=[1/21/41/23/4](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}
  4. Comparison of X1+Y1X^{-1} + Y^{-1} and (X+Y)1(X+Y)^{-1}:

    From the calculations:

    X1+Y1=[321/21]X^{-1} + Y^{-1} = \begin{bmatrix} 3 & 2 \\ -1/2 & -1 \end{bmatrix}
    (X+Y)1=[1/21/41/23/4](X+Y)^{-1} = \begin{bmatrix} -1/2 & -1/4 \\ 1/2 & 3/4 \end{bmatrix}

    Clearly, X1+Y1(X+Y)1X^{-1} + Y^{-1} \neq (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.