Question

Can someone please help me to figure out this Cramers Rule problem.

Answer 1

\[a+b+c+d=11\]
\[a+b+c+2d=10\]
\[a+b+2c+3d=7\]
\[a+2b+3c+4d=11\]

Here is the system of equations, I really don't understand Cramers Rule so if someone could help that would be great.

Answer 2

http://www.purplemath.com/modules/cramers.htm

Answer 3

using determinants to solve linear equations is a horrible (very inefficient, and difficult) idea.

But, if you have to use Cramer's Rule, the first step is make a matrix
of the coefficients.
can you do that ?

Answer 4

for example, from the 1st equation
a+b+c+d=11

the first row of the matrix is
1 1 1 1 (because the coefficients of the variables are all 1)

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @phi
using determinants to solve linear equations is a horrible (very inefficient, and difficult) idea.

But, if you have to use Cramer's Rule, the first step is make a matrix
of the coefficients.
can you do that ?
\(\color{#0cbb34}{\text{End of Quote}}\)
You're absolutely right. Moreover, these equations are not complicated so we can solve them just by making different substitutions

Answer 6

Yes @phi i know how to make it into matrix form; what do i do after that?

Answer 7

Can you show us the matrix form of the whole set of equations?

Answer 8

The next step is find the determinant of that matrix (which is a pain)

Answer 9

\[\left[\begin{matrix}1 & 1&1&1 \\ 1&1&1&2 \\ 1&1&2&3\\1&2&3&4\end{matrix}\right]\]

Answer 10

get the determinant (call it D)
to solve for "a", replace the first column (the "a" column)
with [11 10 7 11] (transposed to make a column)
to get a new matrix.
find the determinant of that matrix. (gack!)

Answer 11

\[\left[\begin{matrix}1 &1&1&1\\1&1&1&2\\1&1&2&3\\1&2&3&4\end{matrix}\right]\left(\begin{matrix}a \\ b\\c\\d\end{matrix}\right)=\left(\begin{matrix}11 \\ 10\\7\\11\end{matrix}\right)\]

Answer 12

call that detminant Da

then a = Da/ D

now go back to the original matrix, and replace the 2nd column (the "b" column)
with [11 10 7 11] (as a column)
find the determinant of that matrix and call it Db

b= Db / D

etc for c and d
it's lots of work

Answer 13

using matlab (no way I would do this by hand)
I get D= -1
Da=-7 (which means a= -7/-1 = 7)
Db= -7
Dc = 2
Dd= 1

Answer 14

contrast that with using elimination. If we reorder the rows to
1 1 1 1 11
1 2 3 4 11
1 1 2 3 7
1 1 1 2 10
multiply the first row by -1 and add to each of the rows below it. we get
1 1 1 1 11
0 1 2 3 0
0 0 1 2 -4
0 0 0 1 -1

now use "back substitution"
i.e. the last row means d= -1
the next row up means
c +2d = -4 or
c + 2(-1) = -4
c= -2

next: b-4-3 = 0 ---> b= 7
finally: a+7-2-1= 11 --> a=7

a process considerably easier than Cramer's Rule.

Answer 15

This video shows one good way to find the determinant

https://www.youtube.com/watch?v=kK-Uj2nJgus

it's not your exact matrix of course. However, hopefully the similar example will help

Answer 16

We can simplify this matrix by doing some preparations, namely transformations.
The determinant does not change if we subtract rows.
So
row 1: as is
row 2: subtract row 1 from row 2
row 3: subtract row 2 from row 3
row 4: subtract row 3 from row 4
with the resulting 4x4 determinant becomes
1 1 1 1 | 11
0 0 0 1 | -1
0 0 1 1 | -3
0 1 1 1 | -4
which can be evaluated as minor(1,1), a 3x3 matrix.
However, we skip evaluating matrix D1 (i.e. variable a). and solve for "a" in the end by back substitution of b,c and d.

To evaluate a 3x3 matrix visually, study the following diagram.

Once you grasp the procedure, you will not need even a pencil to work out determinants with simple integers.