MHB A and solution are known find B matrix

[TomSavage]

I have the matrix of A

1 2 -1
2 -1 1

and i am asked if there is any B matrix that can make AB = 1-1
1 1

I assume that this is not possible because if we follow the law of Ax=B then {A}^{-1} * B =x and since matrix a is singular then it cannot be inverted and thus this operation is impossible. Am I wrong in thinking this?

 

[I like Serena]

I have the matrix of A

1 2 -1
2 -1 1

and i am asked if there is any B matrix that can make AB = 1-1
1 1

I assume that this is not possible because if we follow the law of Ax=B then {A}^{-1} * B =x and since matrix a is singular then it cannot be inverted and thus this operation is impossible. Am I wrong in thinking this?

Hi TomSavage! Welcome to MHB! ;)

I'm afraid that's not quite true.

We can solve
$$\begin{pmatrix}
1 &2 &-1\\
2 &-1 &1
\end{pmatrix}\begin{pmatrix}
b_{11}\\b_{21}\\b_{31}
\end{pmatrix} = \begin{pmatrix}
1\\1
\end{pmatrix}$$
can't we?

It is under determined so that it has infinitely many solutions.
We can solve it with Gaussian elimination and pick for instance:
$$\begin{pmatrix}
b_{11}\\b_{21}\\b_{31}
\end{pmatrix} = \begin{pmatrix}
3/5\\1/5\\0
\end{pmatrix}$$

Similarly we can solve:
$$\begin{pmatrix}
1 &2 &-1\\
2 &-1 &1
\end{pmatrix}\begin{pmatrix}
b_{12}\\b_{22}\\b_{32}
\end{pmatrix} = \begin{pmatrix}
-1\\1
\end{pmatrix}$$

Combining the result with the first solution we get:
$$\begin{pmatrix}
1 &2 &-1\\
2 &-1 &1
\end{pmatrix}\begin{pmatrix}
3/5 &0 \\1/5 &0 \\0 & 1
\end{pmatrix} = \begin{pmatrix}
1 & -1\\1 & 1
\end{pmatrix}$$

There you go, a solution for $B$.

More generally, we can find a solution in the least-square sense with:
$$B=A^+ \begin{pmatrix}
1 & -1\\1 & 1
\end{pmatrix}$$
where $A^+$ is the Moore-Penrose pseudoinverse.
The wiki article also explains how we can use it to find all solutions.

 

[TomSavage]

Hi TomSavage! Welcome to MHB! ;)

I'm afraid that's not quite true.

We can solve
$$\begin{pmatrix}
1 &2 &-1\\
2 &-1 &1
\end{pmatrix}\begin{pmatrix}
b_{11}\\b_{21}\\b_{31}
\end{pmatrix} = \begin{pmatrix}
1\\1
\end{pmatrix}$$
can't we?

It is under determined so that it has infinitely many solutions.
We can solve it with Gaussian elimination and pick for instance:
$$\begin{pmatrix}
b_{11}\\b_{21}\\b_{31}
\end{pmatrix} = \begin{pmatrix}
3/5\\1/5\\0
\end{pmatrix}$$

Similarly we can solve:
$$\begin{pmatrix}
1 &2 &-1\\
2 &-1 &1
\end{pmatrix}\begin{pmatrix}
b_{12}\\b_{22}\\b_{32}
\end{pmatrix} = \begin{pmatrix}
-1\\1
\end{pmatrix}$$

Combining the result with the first solution we get:
$$\begin{pmatrix}
1 &2 &-1\\
2 &-1 &1
\end{pmatrix}\begin{pmatrix}
3/5 &0 \\1/5 &0 \\0 & 1
\end{pmatrix} = \begin{pmatrix}
1 & -1\\1 & 1
\end{pmatrix}$$

There you go, a solution for $B$.

More generally, we can find a solution in the least-square sense with:
$$B=A^+ \begin{pmatrix}
1 & -1\\1 & 1
\end{pmatrix}$$
where $A^+$ is the Moore-Penrose pseudoinverse.
The wiki article also explains how we can use it to find all solutions.

Hey, I now understand mostly everything but what I don't know is how you got the solutions for B31 and B32. Did you get the other 4 solutions from gauss jordan and then just put the other two in that satisfy the solution or is there a direct mathematical method to get them because wouldn't adding the (3X1) solution set into the (2x3) matrix mess things up?

 

[I like Serena]

Hey, I now understand mostly everything but what I don't know is how you got the solutions for B31 and B32. Did you get the other 4 solutions from gauss jordan and then just put the other two in that satisfy the solution or is there a direct mathematical method to get them because wouldn't adding the (3X1) solution set into the (2x3) matrix mess things up?

Let me show you how to solve:
$$\begin{pmatrix}
1 &2 &-1\\
2 &-1 &1
\end{pmatrix}\begin{pmatrix}
b_{11}\\b_{21}\\b_{31}
\end{pmatrix} = \begin{pmatrix}
1\\1
\end{pmatrix}$$
with Gaussian elimination.

Take the first row, multiply it with $-2$, and add it to the second row.
Consequently we get a new system that has the same solutions as the previous system.
The result is:
$$\begin{pmatrix}
1 &2 &-1\\
0 &-5 &3
\end{pmatrix}\begin{pmatrix}
b_{11}\\b_{21}\\b_{31}
\end{pmatrix} = \begin{pmatrix}
1\\-1
\end{pmatrix}$$

Starting from the bottom, we pick $b_{31}=0$, and see what happens afterwards.
To solve the second equation, we need $b_{21}=\frac 15$ now.
Then we solve the first equation, using the values we found so far, and we can with $b_{11}=\frac 35$.

There you go. We found one of the infinite number of solutions: $(b_{11},b_{21},b_{31})=(\frac 35, \frac 15, 0)$.Alternatively, we might have started with $b_{21}=0$.
Then we need $b_{31}=-\frac 13$ to solve the second equation.
And finally $b_{11}=\frac 23$ to solve the first equation for the solution $(b_{11},b_{21},b_{31})=(\frac 23, 0, -\frac 13)$.
Note that these 2 solutions together 'span' the solution space.