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.

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

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.

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.

## 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:

The \(2 \times 2\) inverse matrix is then:

Where \(D = ad - bc\). \(D\) is called the determinant of the matrix.

The \(3 \times 3\) matrix can be defined as:

Then the inverse matrix is:

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

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