Cramer's Rule

4.3.1 Cramer's Rule

Theorem. Consider three linear equations in three variables $x,y,z$:

where $a_i,b_i,c_i,d_i$ are constants. Provided

the (unique) solution is

where $D_x,D_y,D_z$ are obtained from $D$ by replacing the column of coefficients of $x$, $y$, $z$ respectively with the constants column $d_1,d_2,d_3$:

Remarks. (1) The full proof (obtained by solving the system algebraically and matching the result against the determinant expansions above) is referenced via a QR code in the textbook rather than spelled out; the essential idea is that eliminating two of the three unknowns from the three equations, by the usual elimination method, reproduces exactly the ratios $D_x/D, D_y/D, D_z/D$. (2) If $D=0$, Cramer's Rule does not apply — the system either has no solution or infinitely many (it is not possible to conclude a unique solution).

Worked Examples

Example 1. Solve $x+y+z=6,\ x-y+z=2,\ x+2y-z=2$ using Cramer's Rule.

Step 1: $D=\begin{vmatrix}1&1&1\\1&-1&1\\1&2&-1\end{vmatrix}=1(1-2)-1(-1-1)+1(2+1)=-1+2+3=4$.

Step 2: $D_x=\begin{vmatrix}6&1&1\\2&-1&1\\2&2&-1\end{vmatrix}=6(1-2)-1(-2-2)+1(4+2)=-6+4+6=4$.

Step 3: $D_y=\begin{vmatrix}1&6&1\\1&2&1\\1&2&-1\end{vmatrix}=1(-2-2)-6(-1-1)+1(2-2)=-4+12+0=8$.

Step 4: $D_z=\begin{vmatrix}1&1&6\\1&-1&2\\1&2&2\end{vmatrix}=1(-2-4)-1(2-2)+6(2+1)=-6+0+18=12$. …