Find the determinant of a matrix

5 min read

The determinant of a matrix is a number associated with a square (nxn) matrix. The determinant can tell us if columns are linearly correlated, if a system has any nonzero solutions, and if a matrix is invertible. See the wikipedia entry for more details on this.

Computing a determinant is key to a lot of linear algebra, and by extension, to a lot of machine learning. It is easy to calculate the determinant for a 2x2 matrix:

A=[abcd]detA=abcddetA=adbc\begin{align}A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \\ det A = \begin{vmatrix} a & b \\ c & d \end{vmatrix} \\ det A = ad - bc \end{align}

Calculating the determinant for a bigger matrix is a bit more complicated, as we will see. All the code for this is available from the algorithms repository.

Laplace Expansion

Determinants for larger matrices can be recursively obtained by the Laplace Expansion. This computes the matrix determinant by making it equal to a sum of the scaled minors of the matrix. A minor is the determinant of a matrix after deleting one row and one column (so a 3x3 matrix would turn into a 2x2 matrix).

So, let’s start with this matrix:

A=[112234345]A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix}

To find the determinant of this matrix, we will first consult the formula for laplace expansion.

  • j=1n\sum_{j=1}^{n} - this means the sum from j=1 to n, in this case from the first column to the last one.
  • (1)1+j(-1)^{1+j} - this ensures that alternating entries will be added and subtracted
  • aj1a_{j1} - the element in the matrix A at position 1,j. So, A[1][j].
  • Mj1M_{j1} - the minor of matrix A after removing row 1 and column j.

So, we are taking the sum from j=1 to n of -1 to the power (1+j) times the element a at index (1,j) in the original matrix times the minor of the matrix after removing row 1 and column j.

Let’s expand this out for our matrix:

113445+112435+122334(3544)(2543)+2(2433)1+221\begin{align} 1 * 1 * \begin{vmatrix} 3 & 4 \\ 4 & 5 \end{vmatrix} + -1 * 1 * \begin{vmatrix} 2 & 4 \\ 3 & 5 \end{vmatrix} + 1 * 2 * \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} \\ (3*5 - 4*4) -(2*5 - 4*3) + 2 * (2*4 - 3*3) \\ -1 +2 -2 \\ -1 \end{align}

So, our final determinant for this matrix is -1.


Now that we know the formula, we can formalize it in pseudocode:

Suppose that we have an nxn matrix A, with number of columns j.
if the number of columns is 2, compute the determinant using ad-bc and return.
Iterate through all of the columns.
    Calculate the multiplier by taking -1 to the power (1+j) times the element at A[1][j].
    Delete row 1 and column j from A, and create a new matrix X.
    Find the determinant of X through recursion (start again at the top with A=X).
    Multiply the determinant by the multiplier.
Sum all of the values and return.

This will work by recursing through our matrix to eventually reduce it to a series of 2x2 matrices, where the minors can be calculated.

In order to implement this, we will use the Matrix class that we already developed, with one addition:

def del_column(self, key):
    Delete a specified column
    for i in xrange(0,self.rows):
        del self.X[i][key]

We can then implement a function that takes in a matrix object and computes its determinant:

def recursive_determinant(X):
    Find the determinant in a recursive fashion. Very inefficient
    X - Matrix object
    #Must be a square matrix
    assert X.rows == X.cols
    #Must be at least 2x2
    assert X.rows > 1

    term_list = []
    #If more than 2 rows, reduce and solve in a piecewise fashion
    if X.cols>2:
        for j in xrange(0,X.cols):
            #Remove i and j columns
            new_x = deepcopy(X)
            del new_x[0]
            #Find the multiplier
            multiplier = X[0][j] * math.pow(-1,(2+j))
            #Recurse to find the determinant
            det = recursive_determinant(new_x)
        return sum(term_list)
        return(X[0][0]*X[1][1] - X[0][1]*X[1][0])

We can verify if it works by testing out the matrix that we used above:

X = Matrix([[1,1,2],[2,3,4],[3,4,5]])


We can now test matrices to see if their columns are linearly dependent:

X = Matrix([[1,1,2],[2,2,4],[4,4,8]])

The zero indicates that the columns are dependent on each other.

This will also tell us if the matrix can be inverted.

X = Matrix([[1,1,2],[2,2,4],[4,4,8]])

The above should cause an error.

Performance improvements

There are certainly other, higher-performing, solutions to finding a matrix determinant, like LU Decomposition but I like this because it makes it easy to figure out what is happening under the hood.

Possible performance improvements to this algorithm could be:

  • Creating a global “minor cache” to avoid recomputing the same minors over and over for large matrices.
  • Use the 3x3 formula instead of the 2x2 formula, which will avoid some recursion steps.

Vikas Paruchuri

I’m Vikas, a developer based in Oakland. I'm a lifelong learner currently diving deep into AI. I've started an education company, won ML competitions, and served as a US diplomat.