Matrix question

**worc3247** · January 23rd 2011, 10:26 AM

Say you have a 4x3 matrix P, whose entries are all integers.
What is a necessary and sufficient condition on $\mathbf b$ such that $P \mathbf x = \mathbf b$ (where $x = \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}$ and $b = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{pmatrix}$ ) has a solution?

**TheEmptySet** · January 23rd 2011, 10:54 AM

Originally Posted by worc3247

Say you have a 4x3 matrix P, whose entries are all integers.
What is a necessary and sufficient condition on $\mathbf b$ such that $P \mathbf x = \mathbf b$ (where $x = \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}$ and $b = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$ ) has a solution?

First as a side note the column vector $\vec{b}$ should be a $4 \times 1$ .

If you write this out as a linear system it is overdetermined. It is saying that you have 4 equations in 3 unknowns. So a necessary condition would be that two of the rows (in the augmented) matrix must be identical. This alone is not enough sufficient. Now if you eliminate this 4th row from both $\vec{p},\vec{b}$ you will now have a system of equations in 3 variables with 3 unknowns. Now what has to be true for this system to have a solution for the new column vector $\hat{b}$ .

**worc3247** · January 23rd 2011, 11:16 AM

They all have to be different equations so that you don't actually have two equations for 3 unknowns?

**TheEmptySet** · January 23rd 2011, 11:36 AM

Originally Posted by worc3247

They all have to be different equations so that you don't actually have two equations for 3 unknowns?

If you have a $3 \times 3$ matrix $A$ and two $x \times 1$ columns $\vec{x},\vec{b}$

This would give the linear system

$A\vec{x}=\vec{b}$ . What property must $A$ have for this system to be consistent for every vector $\vec{b}$ ?

**Prove It** · January 23rd 2011, 03:20 PM

You want $\displaystyle \mathbf{Px} = \mathbf{b}$ to have solution for $\displaystyle \mathbf{x}$ .

Using some matrix algebra...

$\displaystyle \mathbf{P}^T\mathbf{Px} = \mathbf{Pb}$ now gives you a square matrix on the LHS ( $\displaystyle \mathbf{P}^T\mathbf{P}$ )

$\displaystyle (\mathbf{P}^T\mathbf{P})^{-1}\mathbf{P}^T\mathbf{Px} = (\mathbf{P}^T\mathbf{P})^{-1}\mathbf{Pb}$

$\displaystyle \mathbf{Ix} = (\mathbf{P}^T\mathbf{P})^{-1}\mathbf{Pb}$

$\displaystyle \mathbf{x} = (\mathbf{P}^T\mathbf{P})^{-1}\mathbf{Pb}$ .

Of course, this solution will only exist if $\displaystyle |\mathbf{P}^T\mathbf{P}| \neq 0$ , so this is a necessary and sufficient condition.

**worc3247** · January 24th 2011, 07:20 AM

Sorry i'm not entirely sure. A is invertible?

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