Cramer Rule; Matrices

Question

Solve the following equations using cramer's rule. 

 

-3x+4y-x=-5

x-y-z=-8

y+z=3

 

Solve for x

Answer 1

Hello Alissa,

 

So, Cramer's Rule is being taught in Algebra II now? I feel like each new generation of students is just getting smarter and smarter!  :)

 

Given a system of equations:

 

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃,

 

where all of the as, bs, cs, and ds are constants, we can rewrite this system as a matrix equation:

 

⌈ a₁   b₁   c₁ ⌉  ⌈ x ⌉       ⌈ d₁ ⌉

| a₂   b₂   c₂ |•| y |  =  | d₂ |

⌊ a₃   b₃   c₃ ⌋  ⌊ z ⌋       ⌊ d₃ ⌋

 

Since it's really tough to type out these matrices, I'm going to call the left-hand matrix M, the middle matrix R, and the right-hand matrix D. That way we can write the above matrix equation as:

 

M • R = D.

 

Ok, cool.

 

Let's start filling in values for all of the constants in this equation. We're presented with the three equations:

1. -3x + 4y - z = -5,
2. x - y - z = -8,
3. y + z = 3.

 

This means that:

1. a₁ = -3, b₁ = 4, c₁ = -1, and d₁ = -5
2. a₂ = 1, b₂ = -1, c₂ = -1, and d₂ = -8
3. a₃ = 0, b₃ = 1, c₃ = 1, and d₃ = 3.

Now, before we can do anything else, we have to make sure that the determinant of M isn't equal to zero. So, let's calculate det(M). Since there's a zero in the bottom row of M, let's find the determinant by going across the bottom row:

 

det(M) = a₃(b₁c₂ - c₁b₂) - b₃(a₁c₂ - c₁a₂) + c₃(a₁b₂ - b₁a₂)

           = 0(doesn't matter) - 1{(-3)(-1) - (-1)(1)} + 1{(-3)(-1) - (4)(1)}

           = 0 - 1(3 + 1) + 1(3 - 4)

           = -4 - 1

det(M) = -5.

 

Okay, the determinant of M isn't zero. Great! We can move forward.

 

Cramer's Rule says that we can solve for the variables in a system of equations using only determinants. For a system of three equations and three variables,

 

x = det(M_a) / det(M)

y = det(M_b) / det(M)

z = det(M_c) / det(M)

 

The matrix M_a is the same as matrix M, except that a₁ has been replaced with d₁, a₂ with d₂, and a₃ with d₃.

 

Similarly, M_b is the same as matrix M, except that the bs have been replaced with the ds.

 

Finally, M_c is the same as matrix M, except that the cs have been replaced with the ds.

 

 

I'm going to go ahead and solve for z, as an example. I'll let you solve x on your own. First, let's write out the matrix M_c:

 

          ⌈ -3   4  -5 ⌉

M_c  =  |  1  -1  -8 |

          ⌊   0   1   3 ⌋

 

The determinant of M_c is (we'll go across the bottom row because there's that convenient zero there):

 

det(M_c) = 0(doesn't matter) - 1{(-3)(-8) - (-5)(1)} + 3{(-3)(-1) - (4)(1)}

            = 0 - 1(24 + 5) + 3(3 - 4)

            = -29 - 3

det(M_c) = -32

 

Now, find z:

 

z = det(M_c) / det(M)

   = -32 / -5

z = 6.4

 

Ok! Your turn. Solve for x. Good luck!