# Matrice Equation

• Oct 13th 2010, 03:33 AM
Goldberg Variation
Matrice Equation
Hello, I need to solve the following matrice equation. The known matrices are A & B.

AX + 3IX = B

If anyone could help me here I would appreciate it.

I figured it would look something like this, though I am not sure:

$AX + 3IX = B$

$inv(A)AX + 3IX = inv(A)B$

$X + 3IX = inv(A)B$

$X + inv(3I)(3I)X = inv(3I)inv(A)B$

$2X = inv(3I)inv(A)B$

IF this is correct how would I count the inverse of 3I ? I would be the identity matrix of course, it is all 3 in a diagonal instead of 1?

• Oct 13th 2010, 04:32 AM
HallsofIvy
Quote:

Originally Posted by Goldberg Variation
Hello, I need to solve the following matrice equation. The known matrices are A & B.

AX + 3IX = B

If anyone could help me here I would appreciate it.

I figured it would look something like this, though I am not sure:

$AX + 3IX = B$

$inv(A)AX + 3IX = inv(A)B$

You are incorrect at this point. You are multiplyin both sides by $A^{-1}$ so it should be
$A^{-1}(AX+ 3X)= A^{-1}Ax+ A^{-1}3Ix= A^{-1}B$

But then you will have trouble with $A^{-1}3Ix= 3A^{-1}x$ (since Ix= x).
Instead, write the equation as $Ax+ 3Ix= (A+ 3I)x= Bx$ and multiply by the inverse of $A+ 3I$, not A.

Of course, A+ 3I is the matrix A with "3" added to each number on the main diagonal.

By the way, the English singular of "matrices" is "matrix", not "matrice".

Quote:

$X + 3IX = inv(A)B$

$X + inv(3I)(3I)X = inv(3I)inv(A)B$

$2X = inv(3I)inv(A)B$

IF this is correct how would I count the inverse of 3I ? I would be the identity matrix of course, it is all 3 in a diagonal instead of 1?

• Oct 13th 2010, 04:53 AM
Ackbeet
Quote:

...multiply by the inverse of $A+3I$...
Of course, this only works if -3 is not an eigenvalue of A. Otherwise, said inverse does not exist! Do we know what A is?
• Oct 13th 2010, 07:20 AM
Goldberg Variation
Hello and thank you both for your replies.

HallsofIvy I am not sure I follow your steps you posted. On your last line of the equation do you break out the matrix X? If so why does it appear next to the matrix B on the right side of the equation?

Or if it was a typo I can see the answer of the equation I think. Then it would be:

$X = B inv(A + 3I)$ Correct?

Oh and Ackbeet the known matrices are the ones I mentioned in my first post, matrix A and B. I just need to figure out the equation so I can solve matrix X.

P.S Where can I find the MATH codes for specific functions and such like the inverse (power of -1). Tried search but nothing.
• Oct 13th 2010, 07:42 AM
Ackbeet
Type $(A+3I)^{-1}.$ You can double-click the code to see how I did it. Use the caret symbol ^ to get exponents, and the underscore _ to get subscripts.

I think HallsofIvy assumed that X was a vector, instead of a matrix. The general method was fine, however. That is, you have

$AX+3IX=B$

$(A+3I)X=B$

$X=(A+3I)^{-1}B,$

assuming -3 is not an eigenvalue of A, as I said before.
• Oct 13th 2010, 08:17 AM
Goldberg Variation
Thank you both HallsofIvy and Ackbeet for your answers. I get it now and can solve the X matrix now on my own.

P.S I didn't know you could double-click the MATH code to actually view the code. Really neat, thanks.
• Oct 13th 2010, 08:23 AM
Ackbeet
You're welcome. Have a good one!

Like your username, by the way. I'm partial to the Glenn Gould recording myself.