inverse matrix help !?

Question

Solve the system by finding the inverse of the coefficient matrix :

3x + 4y = 3
2x + 2y = -1

The answer is X= -5
9/2

I just wanna know how to get there .. thanx

Answer 1

Basically, the above system represents this matrix:

[ 3 4 ] [x] = [3 ]
[ 2 2 ] [y] = [-1]

(Above, I want you to treat those two equal signs as one).

To solve for the inverse, what we need to do is put it side by side with the identity matrix and change it to reduced row echelon form. The resulting matrix will be the inverse. I'll show you what I mean.

[ 3 4 | 1 0 ]
[ 2 2 | 0 1 ]

First step: R1 -> 1/3 R1 [take 1/3 of row 1]

[ 1 4/3 | 1/3 0 ]
[ 2 2 | 0 1 ]

R2 -> R2 - 2R1 (Subtract twice row 1 from row 2)

[1 4/3 | 1/3 0 ]
[0 -2/3 | -2/3 1]

R2 -> (-3/2)R2 (Multiply the second row by -3/2)

[1 4/3 | 1/3 0 ]
[0 1 | 1 -3/2]

R1 -> R1 - 4/3 R2

[1 0 | -1 2 ]
[0 1 | 1 -3/2 ]

Therefore, the inverse matrix is:

[-1 2]
[1 -3/2]

To solve our equation, remember that if
Ax = b, then
x = A(inverse)b

In our case, A =
[3 4]
[2 2]

And x =
[x]
[y]

And b =
[3]
[-1]

Therefore, to solve for x, we calculate A(inverse)(b), or
[-1 2] [3]
[1 -3/2] [-1]

And we actually perform matrix multiplication to get

[-3 - 2]
[3 + 3/2]

Or

[-5]
[9/2]

[x] = [-5]
[y] = [9/2]

Therefore, x = -5 and y = 9/2

I hope it wasn't too confusing for you. Matrices are hard to do on here!

Answer 2

Write the system as
[3 4]* [x] = [3]
[2 2] [y] [-1]

This can be written as AX=B
A is the matrix containing the coefficients of x, and y in both equations.

The solution is X = A^(-1) * B

To find the inverse of A, first calculate its determinant as follows:
the product of main diagonal elements minus the product of the other elements, which is, in this case, 3*2 - 4*2 = -2

Now, make another matrix by swapping the main diagonal elements and changing the signs of the other elemtents.
Here, you get:
[2 -4]
[-2 3]

The inverse is just the final matrix divided by the determinant, which is here:
[2/-2 -4/-2] =
[-2/-2 3/-2]
[-1 2]
[1 -3/2]

Now, multiply the inverse by B to get

[-1*3 + 2*-1] = [-5=x]
[1*3 + -3/2*-1] [4.5=y = 9/2]

Answer 3

What you have is the general form Ax = b; so, to solve you need to do (A^-1)Ax = (A^-1)b and Ix = (A^-1)b

So here's what you do:

Set up your A matrix along side an identify matrix I. Thus, you'd have A | I. The A matrix is on the left of the | and the new identity matrix I is on the right.

Now, through row and column operations, begin to transform the A matrix into an identify matrix (e.g., A-->I).

As you transform the A into I, do exactly the same operations on the original identity matrix.

As you do A --> I operations, the original I becomes the A^-1 matrix; or I --> A^-1 .

Once you have A^-1 by doing the parallel operations on A and I, you can substitute it into Ix = (A^-1)b, which will give you the x's (your x and y) straight away.

Answer 4

In addition to the below, which explains the general idea by gauss-jordan, and matrix inverse calculations by hand, the idea of course is solving Az=b, where in this case:

where A = [3,4;2,2], z = [x;y], and b = [3;-1]

The solution of of that equation is z = A^(-1)*b
so once you find the inverse of the matrix A, as below, you multiply it times b, using normal matrix/vector arithmetic, and your solution vector z contains the values for x and y.

Answer 5

Solve the system by finding the inverse of the coefficient matrix :

3x + 4y = 3
2x + 2y = -1

So


(3 .. 4)(x) .. ( 3)
( ...... )( .) = (.. )
(2 .. 2)(y) .. (-1)

Let A =
(3 .. 4)
( ...... )
(2 .. 2)

Now A(-1) =

.....1 .... (2 .. -4)
------- * ( ......... ) =
detA .... (-2 .. 3)

(-1 ..... 2)
( .......... )
(1 .. -1.5)

So
..............(x) .. (-1 ..... 2)( 3) .. ( -5 )
A^(-1)A (.) = ( ............ )(. .) = (.....)
..............(y) .. (1 .. -1.5)(-1) .. (4.5)

ie x = -5 y = 4.5 (this checks out)

Answer 6

http://mathworld.wolfram.com/MatrixInverse.html

try this...

Now to fin the roots,

Covert the equations to matrix forms...

A = [ 3 4 ]
= [ 2 2 ]

X = [ x ]
= [ y ]

B = [ 3 ]
= [ -1]

We have AX =B

=> X = inverseof(A)*B


All the best.
Peace out.