Cramer's Rule can be used to solve systems of linear equations. For a 3x3 system, it involves calculating the determinants of the coefficient matrix (D) and matrices where one column is replaced by the constants vector (Dx, Dy, Dz). The solutions are then given by x=Dx/D, y=Dy/D, z=Dz/D. As an example, Cramer's Rule is used to solve the 3x3 system 2x-y+3z=-3, -x-y+3z=-6, -2y-z=-2, finding the solutions to be x=1, y=2, z=-1.

1. Use Cramer's Rule to solve the following matrix equation, show your work.
-1 2 0 a 1
2 1 1 b = 2
-3-1 4 c -1
Solution
o find rules (or formulas) that may be used solve any system of equations, we need to solve the
general system of the form a 1 x + b 1 y = c 1 (1) a 2 x + b 2 y = c 2 (2) We multiply equation (1)
by b 2 and equation (2) by -b 1 and add the right and left hand terms. b 2 a 1 x + b 2 b 1 y = b 2 c
1 (1) - b 1 a 2 x - b 1 b 2 y = - b 1 c 2 (2) ____________________________ b 2 a 1 x - b 1 a 2 x
= b 2 c 1 - b 1 c 2 Assuming that a 1 b 2 - a 2 b 1 is not equal to zero, solve the above equation
for x to obtain. x = ( c 1 b 2 - c 2 b 1 ) / ( a 1 b 2 - a 2 b 1 ) We can use similar steps to eliminate
x and solve for y to obtain. y = ( a 1 c 2 - a 2 c 1 ) / ( a 1 b 2 - a 2 b 1 ) Using the determinant of a
Matrix notation, the solution to the given 2 by 2 system of linear equations is given by x = D x /
D and y = D y / D where D, D x and D y are the determinants defined by For a 3 by 3 system of
linear equations of the form a 1 x + b 1 y + c 1 z = d 1 (1) a 2 x + b 2 y + c 2 z = d 2 (2) a 3 x + b
3 y + c 3 z = d 3 (3) Cramer's rule gives the solution as follows x = D x / D , y = D y / D and z =
D z / D where D, D x, D y and D z are determinants defined by Example: Use Cramer's rule for
a 3 by 3 system of linear equations to solve the following system 2 x - y + 3 z = - 3 - x - y + 3 z =
- 6 x - 2y - z = - 2 Solution: Determinants D, D x, D y and D z are given by (see how to Calculate
Determinant of a Matrix.) D = 21 D x = 21 D y = 42 D z = -21 We now use cramer's rule to find
the solution of the system of equations x = D x / D= 1 y = D y / D= 2 z = D z / D= -1 As an
exercise check that (1,2,-1) is a solution to the given system of equations.