Theorem – Cramer’s Rule

Contents
1. Theorem
2. Proof

Theorem

Let A be a matrix with dimension n\times n, and \mathbf{x}, \mathbf{x} be vectors. Then, consider the system of linear equations given by

A\mathbf{x} = \mathbf{b}

Let a_{ij} represent the elements of A, and similarly x_i, b_i represent the elements of \mathbf{x} and \mathbf{b}. Finally, let A_i be the matrix A, except where column i (which will be a column vector) of the matrix A is replaced with the vector \mathbf{b}.

Then,

x_i = \frac{det(A_i)}{det(A)}

2. Proof

Let the columns of A be denoted by c_i for i=1,..., n. Then, we can write

A\mathbf{x} = \mathbf{b} = c_1x_1 + c_2x_2 + ... + c_n x_n

The key point here is that \mathbf{b} = c_1x_1 + c_2x_2 + ... + c_n x_n. Furthermore, we can write the matrix A as A = c_1+ c_2 + ... + c_n. Then, the matrix A_i, where c_i is replaced with the vector \mathbf{b}, can be written as

A_i = \mathbf{b} + c_2 + ... + c_n

If i=1,

A_i = c_1 + ... + \mathbf{b} + c_n

if i = n, or

A_i = c_1 + ... + \mathbf{b} + ... + c_n

the determinant of the matrix A_i is therefore of the form above, ie for i=1, det(A_i) = det(\mathbf{b}~c_2~...~c_n). But, recall that

\mathbf{b} = c_1x_1 + c_2x_2 + ... + c_n x_n

So, substituting \mathbf{b} back into A_i, we end up with (in the case that i=1),

det(A_i) = det(c_1x_1 + c_2x_2 + ... + c_n ~|~ x_n ~|~ c_2~|~...~|~c_n)

But, from the laws of determinants, we have that adding multiples of columns together does not change the value of the determinant, and that multiplying one column by a constant, say x_i, multiplies the determinant by x_i. Thus,

det(A_i) = x_i \times det(A)

Which, by the assumption that A is not singular (and hence det(A)\neq 0, we have

x_i = \frac{det(A_i)}{det(A)}

And this proves the theorem. \quad \square

NB – The inverse of a matrix is a special use of cramer’s rule, where \mathbf{b} is the column of zeros and a one for each column of the identity matrix.