How to Calculate the Inverse Matrix for 2×2 and 3×3 Matrices

is $$\frac{1}{8}$$, the inverse of 20 is $$\frac{1}{20}$$ and so on. Therefore, a number multiplied by its inverse will always equal 1. An inverse of a number is denoted with a $$-1$$ superscript.

Inverses of Numbers and Matrices

The inverse of a number is its reciprocal. For example, the inverse of 8 is $$\frac{1}{8}$$, the inverse of 20 is $$\frac{1}{20}$$ and so on. Therefore, a number multiplied by its inverse will always equal 1. An inverse of a number is denoted with a $$-1$$ superscript.

$$x \cdot \frac{1}{x} = x \cdot x^{-1} = x^{-1} \cdot x = 1$$

The inverse of a matrix $$A$$ is another matrix denoted by $$A^{-1}$$ and is defined as:

$$A^{-1}A = AA^{-1} = I$$

Where $$I$$ is the identity matrix. Thus, similar to a number and its inverse always equaling 1, a matrix multiplied by its inverse equals the identity.

This post will explore several concepts related to the inverse of a matrix, including linear dependence and the rank of a matrix. Afterward, the method of computing an inverse (if one exists) of a $$2 \times 2$$ or $$3 \times 3$$ matrix shall be demonstrated. Finding the inverse of a square matrix with $$\geq 4$$ columns is computationally intensive and best left to R's built-in linear algebra routines which are built on LINPACK and LAPACK. Here is an excellent resource that lists the linear algebra operations available in R. Here is a good resource on how to compute a 4x4 inverse matrix manually for those interested.

The example inverse matrix problems used in the post are from Jim Hefferon's excellent book Linear Algebra on page 249. I highly recommend the book to those learning more about linear algebra. The book is free to download and comes with many exercises and other features.

Linear Dependence of a Matrix

The following matrix A has three column vectors.

$$A = \begin{bmatrix} 2 & 2 & 3 \\ 1 & -2 & -3 \\ 4 & -2 & - 3 \end{bmatrix}$$

Notice the second column vector is a multiple of the third column. The matrix is therefore linearly dependent as the matrix contains a column vector that is a multiple of another. The matrix is linearly independent when no column vector can be expressed as a multiple of another vector in the matrix.

$$\begin{bmatrix} 2 \\ -2 \\ -2 \end{bmatrix} = \frac{3}{2} \begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix}$$

Rank of a Matrix

The rank of a matrix is the maximum number of linearly independent columns or linearly independent rows in the matrix. Therefore, the rank of a $$row \times column$$ matrix is the minimum of the two values. For example, the above matrix would have a rank of 1. Inverses only exist for a square $$r \times r$$ matrix with rank $$r$$, which is called a full rank or nonsingular matrix.

Computing an inverse matrix

Consider a 2x2 matrix:

$$\underset{2 \times 2}{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

The $$2 \times 2$$ inverse matrix is then:

$$\underset{2 \times 2}{A^{-1}} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{D} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

Where $$D = ad - bc$$. $$D$$ is called the determinant of the matrix.

The $$3 \times 3$$ matrix can be defined as:

$$\underset{3 \times 3}{B} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & k \end{bmatrix}$$

Then the inverse matrix is:

$$\underset{3 \times 3}{B^{-1}} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & k \end{bmatrix}^{-1} = \frac{1}{det(B)} \begin{bmatrix} (ek - fh) & -(bk - ch) & (bf - ce) \\ -(dk - fg) & (ak - cg) & -(af - cd) \\ (dh - eg) & -(ah - bg) & (ae - bd) \end{bmatrix}$$

Where $$det(B)$$ is equal to:

$$det(B) = a(ek - fh) - b(dk -fg) + c(dh - eg)$$

The following function implements a quick and rough routine to find the inverse of a $$2 \times 2$$ or $$3 \times 3$$ matrix should one exist.

matrix.inverse <- function(mat) {
A <- as.matrix(mat)

# If there are more than four columns in the supplied matrix, stop routine
if ((ncol(A) >= 4) | (nrow(A) >= 4)) {
stop('Matrix is not 2x2 or 3x3')
}

# Stop if matrix is a single column vector
if (ncol(A) == 1) {
stop('Matrix is a vector')
}

# 2x2 inverse matrix
if(ncol(A) == 2) {
# Determinant
a <- A[1]
b <- A[3]
c <- A[2]
d <- A[4]
det <- a * d - b * c
# Check to see if matrix is singular
if (det == 0) {
stop('Determinant of matrix equals 0, no inverse exists')
}
# Compute inverse matrix elements
a.inv <- d / det
b.inv <- -b / det
c.inv <- -c / det
d.inv <- a / det
# Collect the results into a new matrix
inv.mat <- as.matrix(cbind(c(a.inv,c.inv), c(b.inv,d.inv)))
}

# 3x3 inverse matrix
if (ncol(A) == 3) {
# Extract the entries from the matrix
a <- A[1]
b <- A[4]
c <- A[7]
d <- A[2]
e <- A[5]
f <- A[8]
g <- A[3]
h <- A[6]
k <- A[9]

# Compute the determinant and check that it is not 0
det <- a * (e * k - f * h) - b * (d * k - f * g) + c * (d * h - e * g)
if (det == 0) {
stop('Determinant of matrix equals 0, no inverse exists')
}

# Using the equations defined above, calculate the inverse matrix entries.
A.inv <- (e * k - f * h) / det
B.inv <- -(b * k - c * h) / det
C.inv <- (b * f - c * e) / det
D.inv <- -(d * k - f * g) / det
E.inv <- (a * k - c * g) / det
F.inv <- -(a * f - c * d) / det
G.inv <- (d * h - e * g) / det
H.inv <- -(a * h - b * g) / det
K.inv <- (a * e - b * d) / det

# Collect the results into a new matrix
inv.mat <- as.matrix(cbind(c(A.inv,D.inv,G.inv), c(B.inv,E.inv,H.inv), c(C.inv,F.inv,K.inv)))
}

return(inv.mat)
}


The results from the above function can be used to verify the definitions and equations of the inverse matrix above in conjunction with R's built-in methods.

A <- as.matrix(cbind(c(3,0),c(1,2)))
A

##      [,1] [,2]
## [1,]    3    1
## [2,]    0    2

A1 <- matrix.inverse(A)
A1

##           [,1]       [,2]
## [1,] 0.3333333 -0.1666667
## [2,] 0.0000000  0.5000000

solve(A)

##           [,1]       [,2]
## [1,] 0.3333333 -0.1666667
## [2,] 0.0000000  0.5000000

B <- as.matrix(cbind(c(1,0,-1), c(1,2,1), c(3,4,0)))
B

##      [,1] [,2] [,3]
## [1,]    1    1    3
## [2,]    0    2    4
## [3,]   -1    1    0

B1 <- matrix.inverse(B)
B1

##      [,1] [,2] [,3]
## [1,]    2 -1.5    1
## [2,]    2 -1.5    2
## [3,]   -1  1.0   -1

solve(B)

##      [,1] [,2] [,3]
## [1,]    2 -1.5    1
## [2,]    2 -1.5    2
## [3,]   -1  1.0   -1


Recall the product of the matrix and its inverse will always equal the identity matrix.

A %*% A1

##      [,1] [,2]
## [1,]    1    0
## [2,]    0    1

B %*% B1

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1


Matrices that are singular or not of full rank will have a determinant of 0, and thus no inverse exists.

C <- as.matrix(cbind(c(2,-1),c(-4,2)))
C

##      [,1] [,2]
## [1,]    2   -4
## [2,]   -1    2

solve(C)

## Error in solve.default(C): Lapack routine dgesv: system is exactly singular: U[2,2] = 0

D <- as.matrix(cbind(c(2,1,4),c(2,-2,-2),c(3,-3,-3)))
D

##      [,1] [,2] [,3]
## [1,]    2    2    3
## [2,]    1   -2   -3
## [3,]    4   -2   -3

solve(D)

## Error in solve.default(D): Lapack routine dgesv: system is exactly singular: U[3,3] = 0


Summary

The inverse matrix was explored by examining several concepts such as linear dependency and the rank of a matrix. The method of calculating an inverse of a $$2 \times 2$$ and $$3 \times 3$$ matrix (if one exists) was also demonstrated. As stated earlier, finding an inverse matrix is best left to a computer, especially when dealing with matrices of $$4 \times 4$$ or above.

References

Hefferon, J. (n.d.). Linear Algebra

Inverse matrix of 2x2 matrix, 3x3 matrix, 4x4 matrix. Retrieved August 10, 2016, from http://www.cg.info.hiroshima-cu.ac.jp/~miyazaki/knowledge/teche23.html

Kutner, M. H., Nachtsheim, C. J., Neter, J., Li, W., & Wasserman, W. (2004). Applied linear statistical models (5th ed.). Boston, MA: McGraw-Hill Higher Education.