Math Help - matrix inversion

1. matrix inversion

Hi guys,

I am unable to get the answer to the following question: Find all the values of (x,y,z) which satisfy:

$\left( \begin{array}{ccc}
7 & 2 & 4 \\ 4 & 1 & 2 \\ 3 & 1 & 1 \end{array} \right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{c} p \\ q \\ r \end{array} \right)
$

if (p,q,r) satisfies

$\left( \begin{array}{ccc}
2 & -7 & 5 \\ 6 & -9 & -1 \\ -4 & 5 & 2 \end{array} \right)
\left( \begin{array}{c} p \\ q \\ r \end{array} \right) =
\left( \begin{array}{c} 5 \\ 7 \\ -4 \end{array} \right)
$

So, we have:
$\mathbf{A}x = p$

detA = 7(-1) -2(-2) + 4(1) = 1. So it has an inverse. I used two methods to calculate the inverse, the adjoint and row reduction. Both gave the same result:

$A^{-1} = \left( \begin{array}{ccc}
-1 & 2 & 0 \\ -2 & -5 & 1 \\ 0 & -1 & -1 \end{array} \right)
$

which gives:
$x = 2q - p$
$y = 2p - 5q + 2r$
$z = p - q - r$

The second matrix B is singular i.e.

$det B = 2(-13) + 7(8) + 5(-6) = 0$ But we have the following relationships:

$2p - 7q + 5r = 5$
$6p - 9q - r = 7$
$-4p + 5q + 2r = -4$

I messed around and got:

p = 7x + 2y + 4z
q = 4x + y + 2z
r = 3x + y + z

but this doesn't help. The answers are given in terms of $\lambda!$ Am I missing something here?

2. Originally Posted by s_ingram
Hi guys,

I am unable to get the answer to the following question: Find all the values of (x,y,z) which satisfy:

$\left( \begin{array}{ccc}
7 & 2 & 4 \\ 4 & 1 & 2 \\ 3 & 1 & 1 \end{array} \right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{c} p \\ q \\ r \end{array} \right)
$

if (p,q,r) satisfies

$\left( \begin{array}{ccc}
2 & -7 & 5 \\ 6 & -9 & -1 \\ -4 & 5 & 2 \end{array} \right)
\left( \begin{array}{c} p \\ q \\ r \end{array} \right) =
\left( \begin{array}{c} 5 \\ 7 \\ -4 \end{array} \right)
$

So, we have:
$\mathbf{A}x = p$

detA = 7(-1) -2(-2) + 4(1) = 1. So it has an inverse. I used two methods to calculate the inverse, the adjoint and row reduction. Both gave the same result:

$A^{-1} = \left( \begin{array}{ccc}
-1 & 2 & 0 \\ -2 & -5 & 1 \\ 0 & -1 & -1 \end{array} \right)
$

which gives:
$x = 2q - p$
$y = 2p - 5q + 2r$
$z = p - q - r$

The second matrix B is singular i.e.

$det B = 2(-13) + 7(8) + 5(-6) = 0$ But we have the following relationships:

$2p - 7q + 5r = 5$
$6p - 9q - r = 7$
$-4p + 5q + 2r = -4$

I messed around and got:

p = 7x + 2y + 4z
q = 4x + y + 2z
r = 3x + y + z

but this doesn't help. The answers are given in terms of $\lambda!$ Am I missing something here?
It follows from the first two matrix equations that $\left( \begin{array}{ccc}
2 & -7 & 5 \\ 6 & -9 & -1 \\ -4 & 5 & 2 \end{array} \right)
\left( \begin{array}{ccc}
7 & 2 & 4 \\ 4 & 1 & 2 \\ 3 & 1 & 1 \end{array} \right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{c} 5 \\ 7 \\ -4 \end{array} \right)
$
.

Do the matrix multiplication and then solve the resulting system.

3. Thanks for your reply, but.. when I do multiply the matrices I get:

$\left( \begin{array}{ccc}
1 & 2 & -1 \\ 3 & 2 & 5 \\ -2 & -1 & -4 \end{array} \right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{c} 5 \\ 7 \\ -4 \end{array} \right)$

and the matrix is singular.

4. Originally Posted by s_ingram
Thanks for your reply, but.. when I do multiply the matrices I get:

$\left( \begin{array}{ccc}
1 & 2 & -1 \\ 3 & 2 & 5 \\ -2 & -1 & -4 \end{array} \right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{c} 5 \\ 7 \\ -4 \end{array} \right)$

and the matrix is singular.
det = 0 => either no solution or infinite solutions. So, deal with this situation in the way you have been taught (refer to class notes or textbook if necessary). If you need more help, please say where exactly you're stuck.

5. Originally Posted by s_ingram
Hi guys,

I am unable to get the answer to the following question: Find all the values of (x,y,z) which satisfy:

$A= \left( \begin{array}{ccc}
7 & 2 & 4 \\ 4 & 1 & 2 \\ 3 & 1 & 1 \end{array} \right)
\vec x= \left( \begin{array}{c} x \\ y \\ z \end{array} \right)
\vec {soln}= \left( \begin{array}{c} p \\ q \\ r \end{array} \right)
$

$A\vec x= \vec {soln}$

$A^{-1}A\vec x = A^{-1}\vec {soln}$

$\vec x = A^{-1}\vec {soln}$

Now via Texas-instrument

$[2nd]$ $[matrix]$ then scroll over to edit.

Create the two arrays $A$ and $\vec {soln}.$

Then, $(A^{-1})*\vec {soln}.$ gives the desired result

6. Thanks guys.

I took the matrix:

$
\left( \begin{array}{ccc}
1 & 2 & -1 \\ 3 & 2 & 5 \\ -2 & -1 & -4 \end{array} \right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{c} 5 \\ 7 \\ -4 \end{array} \right)
$

Then using row reduction, I converted it into an equivalent triangular matrix:
$
\left( \begin{array}{ccc}
1 & 2 & -1 \\ 0 & -4 & 8 \\ 0 & 0 & 0 \end{array} \right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{c} 5 \\ -8 \\ 0 \end{array} \right)
$

which gives a consistent result:

x + 2y - z = 5 (i)
-4y + 8z = -8 (ii)
0z = 0 (iii)

The last equation is true for any z, so I can set $z = \lambda$ and everything is fine.

Previously I made an arithmetic error so that the equations appeared inconsistent with no solutions. I checked the result once and then twice and then posted. I think the best advice when doing matrix questions is not to post after a late night. Try again in the morning and things are usually a lot clearer! Anyway, thanks again guys.