Inverse?? Matrix multipication question

[markraz]

Hi, So I have this matrix/vector formula shown on top of this graphic below.
I plugged in some arbitrary numbers and got an answer. How can I go in reverse and verify the answer?

in other words how can I find the row vector [p0 p1 p2 p3] if I am given the [u^3 u^2 u^1 1] vector, the 4x4 matrix, and the 1.75 ??

I was thinking using Ax = b and solving for x thinking x would be the [p0 p1 p2 p3] row vector I am looking to find
If I can go in reverse my answer would be the row vector
[0
1
3
2]

can this be done some way using Ax = b or something similar?

Thanks in advance

 

[Steven G]

If A*b = c, then multiplying on the left on both sides by A^-1 we get A^-1(A*b) = A^-1c or b = A^-1c

 

[markraz]

If A*b = c, then multiplying on the left on both sides by A^-1 we get A^-1(A*b) = A^-1c or b = A^-1c

Thanks so much

so just to make it easy I labeled each matrix/vector for future reference


value = single double value
CV1 = column vector 1 x 4
M1 = 4x4 matrix
RV1 - row vector 4 x 1


so would I multiply CV1 and M1 together and call that A? then take the inverse? the product would then be a column Vector . Can that be inverted?


Thanks

 

[Dr.Peterson]

so just to make it easy I labeled each matrix/vector for future reference

View attachment 21277
value = single double value
CV1 = column vector 1 x 4
M1 = 4x4 matrix
RV1 - row vector 4 x 1


so would I multiply CV1 and M1 together and call that A? then take the inverse? the product would then be a column Vector . Can that be inverted?

If you know u, then you can multiply CV1 and M1 to get a row vector A in the equation B = A*CV1. (Isn't "RV1" a column vector??) Then your equation will be equivalent to

[MATH]a_0P_0 + a_1P_1 + a_2P_2 + a_3P_3 = B[/MATH]​

That's one equation in four variables, right? There will not be a unique solution.

 

[markraz]

If you know u, then you can multiply CV1 and M1 to get a row vector A in the equation B = A*CV1. (Isn't "RV1" a column vector??) Then your equation will be equivalent to
[MATH]a_0P_0 + a_1P_1 + a_2P_2 + a_3P_3 = B[/MATH]That's one equation in four variables, right? There will not be a unique solution.

Thanks yeah I made a mistake, RV1" is a column. Thanks

So when you say there will not be a unique solution is there a finite cumber of solutions I can get? is there 4??
I would just calculate for each P0 P1 P2 P3 ??
thanks

 

[Dr.Peterson]

How many solutions would you expect from an equation like w + 2x + 3y + 4z = 10? Isn't it infinite? Pick any values for w, x, and y, and you can find z.

 

[markraz]

How many solutions would you expect from an equation like w + 2x + 3y + 4z = 10? Isn't it infinite? Pick any values for w, x, and y, and you can find z.

thanks so I guess I am confused. So my question is given the information below is it possible;e for me to P0, P1, P2, P3 ?

Thanks

 

[Dr.Peterson]

Have you tried carrying out the process you described? Multiply the first two matrices, and write out the resulting equation.

I'll tell you that some of the many solutions are (writing the P vector as a row) [4,0,0,0], [0,4/3,0,0], [1,0,0,3], [0,0,1,1], ... (if I did everything right).

 

[Dr.Peterson]

Have you tried carrying out the process you described? Multiply the first two matrices, and write out the resulting equation.

I'll tell you that some of the many solutions are (writing the P vector as a row) [4,0,0,0], [0,4/3,0,0], [1,0,0,3], [0,0,1,1], ... (if I did everything right).

I accidentally used 0.5 instead of 1.75. Correct solutions include [14,0,0,0], [0,14/3,0,0], [7,0,0,7], [0,0,4,3], ..., as well as [0,1,3,2].