simultaneous equation in matrix form?

Question

I have been stuck for hours, "solve the simultaneous equation give in part a in the matix form A x(bar) = b(bar). Find the inverse A^-1, use A ^-1 to solve given system

So i have done part A->

3x+2y-z=1
2x+y-2z=2
x+3y+3z=0

Got the following matrix:
(3 2 -1)...(Y).......(1)
(3 1 -2)...(X)... =..(2)
(1 3 3)...(Z)........(0)

The original determinant i got -3

I got X=3, Y= 3 and Z= 2

Now i have to work out the "simultaneous equation give in part a in the matix form A x(bar) = b(bar). Find the inverse A^-1, use A ^-1 to solve given system "

I know i should be getting the results as the inverse matrix:-
(100)
(010)
(001)

But i dont know what i have been doing wrong, so can anyone help me
Im not getting the answer above as when i calculate 1/-3 x ( the co factor i have been getting for all 9) it comes out wrong.

could someone please calculate and show how they are able to get the inverse matrix please?

Answer 1

First if you're stuck at some point use this online calculator (http://www.bluebit.gr/matrix-calculator/) to check your answers. Secondly you wrote '3' in 2nd row 1st column but it should be '2' (because equation has 2x). Thirdly, you're solving the equations wrong. Simultaneous equation can be solved by inverse method using this formula (first see its derivation):



As you know, we can write simultaneous equation into matrix form like this

AX = B



Now left multiple both sides by A^-1. it becomes

A^-1.A.X = A^-1 .B as A^-1.A = I or Identity matrix, the eq becomes

I.X = A^-1.B since IX=X because Identity matrix is like 1 of the matrix



X = A^-1.B

Now you have to first calculate A^-1 and then multiply it with matrix B to get values for x, y and z or X matrix. A^-1 is just adj(A)/det(A) OR ad-joint of a matrix divide by determinant of that matrix. Now I'm going to calculate all things using that online calculator.



[ 1.500 -1.500 -0.500]

[-1.333 1.667 0.667]= A^-1

[0.833 -1.167 -0.167]



X= A^-1. B

X= [ 1.500 -1.500 -0.500] [1]

.....[-1.333 1.667 0.667] x [2]

.....[0.833 -1.167 -0.167] [0]

X= [-1.5]

.....[2 ]

.....[-1.5]



so, x= -1.5 y= 2 and z= -1.5



Checking: putting values of x, y and z in eq 1

3x+2y-z=1

3(-1.5)+2(2)-(-1.5)=1

-4.5+4+1.5=1

1=1



putting values of x, y and z in eq 2

2x+y-2z=2

2(-1.5)+2-2(-1.5)=2

2=2



putting values of x, y and z in eq 3

x+3y+3z=0

-1.5+6-4.5=0

0=0



Remember, there are several methods to solve simultaneous equations some of them are: by subsitution, by equating the co-efficients, by inverse matrix method, cramers rule or by simply row operations (latter 3 methods are, if you're using matrix to solve those equations).



If you still didn't understand view these set of lectures of khanacadmey

https://www.khanacademy.org/math/linear-algebra/matrix_transformations/inverse_of_matrices/v/linear-algebra--deriving-a-method-for-determining-inverses

Answer 2

It seems purely approximately precisely as you have it different than you eliminate the variables and the =. a million....a million.....2| 8 2....a million.....a million| 7 2...-a million....-4|-9 i could desire to have not positioned | yet i did that to remind you that 8,7,-9 is the augmented column of the unique 3x3 matrix.

Answer 3

Haven't worked out your problem, but it appears you copied it wrong. Your matrix doesn't match the original equations in the second row.



always,

tony