Cramer's Rule

Recall the following:

• The linear system $A {\mathbf x}= {\mathbf b}$ has a unique solution for every ${\mathbf b}$ if and only if $A^{-1}$ exists.
• $A$ has an inverse if and only if $\det (A) \neq 0.$

Thus, $A {\mathbf x}= {\mathbf b}$ has a unique solution FOR EVERY ${\mathbf b}$ if and only if $\det (A) \neq 0.$

The following theorem gives a direct method of finding the solution of the linear system $A {\mathbf x}= {\mathbf b}$ when $\det (A) \neq 0.$

THEOREM 2.6.18 (Cramer's Rule)   Let $A {\mathbf x}= {\mathbf b}$ be a linear system with $n$ equations in $n$ unknowns. If $\det(A) \neq 0,$ then the unique solution to this system is

$$x_j = \frac{ \det(A_j)}{\det(A)}, \;\; {\mbox{ for }} j=1, 2, \ldots, n,$$

where $A_j$ is the matrix obtained from $A$ by replacing the $j$ th column of $A$ by the column vector ${\mathbf b}.$

Proof. Since $\det(A) \neq 0,\;$ $A^{-1} = \displaystyle\frac{1}{\det(A)} Adj(A).$ Thus, the linear system $A {\mathbf x}= {\mathbf b}$ has the solution ${\mathbf x}= \displaystyle\frac{1}{\det(A)} Adj(A) {\mathbf b}.$ Hence, $x_j,$ the $j$ th coordinate of ${\mathbf x}$ is given by

$$x_j=\frac{b_1 C_{1j} + b_2 C_{2j} + \cdots + b_n C_{nj}}{\det(A)} = \frac{\det (A_j)}{\det(A)}.$$

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The theorem implies that

$$x_1 = \frac{1}{\det (A)} \; \begin{vmatrix}b_1 & a_{12} & \cdots ... ...ts & \vdots & \ddots & \vdots \\ b_n & a_{n2} & \cdots & a_{nn} \end{vmatrix},$$

and in general

$$x_j = \frac{1}{\det (A)} \; \begin{vmatrix}a_{11} & \cdots & a_{1... ... a_{1n} & \cdots & a_{n j-1} & b_n & a_{n j+1} & \cdots & a_{nn} \end{vmatrix}$$

for $j=2, 3, \ldots, n.$

EXAMPLE 2.6.19   Suppose that $A = \begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}$ and ${\mathbf b}= \begin{bmatrix}1\\ 1 \\ 1\end{bmatrix}.$ Use Cramer's rule to find a vector ${\mathbf x}$ such that $A {\mathbf x}= {\mathbf b}.$
Solution: Check that $\det(A) = 1.$ Therefore $x_1 = \begin{vmatrix}1 &2 & 3 \\ 1 & 3 & 1 \\ 1 & 2 & 2\end{vmatrix}= -1,$
$x_2 = \begin{vmatrix}1 & 1 & 3 \\ 2 & 1 & 1 \\ 1 & 1 & 2\end{vmatrix}= 1,$ and $x_3 = \begin{vmatrix}1 &2 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & 1 \end{vmatrix}=0.$ That is, ${\mathbf x}^t = (-1,1,0).$