If $A =\begin{pmatrix} -1 & 0 & 1\\ 0 & 1 & 1\end{pmatrix}$ and $AB = I$ find the $3\times 2$ matrix $B$.

Question

Alright so you multiply $A$ and $B$ and you get four equations. Then you do $\det[AB] = \det[I] = 1$ and you get a fifth. I'm stuck here now. What else can I do to find $B$?

I'm trying to get this result:

$$\mathbf B=\begin{bmatrix} a & b\\ -a-1 & 1-b\\ a+1 & b \end{bmatrix}$$

Answer 1

Take a general $3\times 2$ matrix $B=\begin{pmatrix} a & b\\ c & d\\ e & f \end{pmatrix}$. Now you require $$ \begin{pmatrix} -1 & 0 & 1\\ 0 & 1 & 1\\ \end{pmatrix} \cdot \begin{pmatrix} a & b\\ c & d\\ e & f \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$

which gives you the set of four equations: $$ \begin{cases} -a+e=1\\ c+e=0\\ -b+f=0\\ d+f=1 \end{cases}. $$

So the first equation says $e=1+a$, the second says $c=-e$, and plugging in the first equality you get $c=-e=-1-a$. Similarly you get $f=b$ from the third equation, and $d=1-f=1-b$ from the third and fourth equations. Thus you obtain that $$ B=\begin{pmatrix} a & b\\ c & d\\ e & f \end{pmatrix} = \begin{pmatrix} a & b\\ -1-a & 1-b\\ 1+a & b \end{pmatrix}. $$

Answer 2

$$ B = \left( \begin{array}{cc} 0 & 0\\ -1 & 1\\ 1 & 0 \end{array} \right) $$

When you set

$$ B = \left( \begin{array}{cc} a & b\\ c & d\\ e & f \end{array} \right) $$

You get

$$ A B = \left( \begin{array}{cc} e-a & f-b\\ c+e & d+f\\ \end{array} \right) $$

So

$$ e-a=1, f-b=0,c+e=0, d+f=1 $$

Thus we can write

$$ B = \left( \begin{array}{cc} a & b\\ -1-a & 1-b\\ 1+a & b \end{array} \right) $$

Simplest case is $a=0$, $b=0$.

---

General case

$$ B = \left( \begin{array}{cc} 0 & 0\\ -1 & 1\\ 1 & 0 \end{array} \right) + a \left( \begin{array}{cc} 1 & 0\\ -1 & 0\\ 1 & 0 \end{array} \right) + b \left( \begin{array}{cc} 0 & 1\\ 0 & -1\\ 0 & 1 \end{array} \right) $$

Answer 3

Just consider for instance the matrix $$B=\left(\begin{array}{ccc}-1&0\\0&1\\0&0\end{array}\right)\,.$$ Actually, you have 4 equations and 6 coefficients. There are plenty of solutions depending on 2 parameters.

Answer 4

You don't really have to resort to using tons of unknowns in a big linear system. It is very easy to do by inspection.

The first column of B creates a linear combination of the columns of A which is $\begin{bmatrix}1\\0\end{bmatrix}$. This can be done in multiple ways. Some examples are : $\begin{bmatrix}-1\\0\\0\end{bmatrix}$ and $\begin{bmatrix}0\\-1\\1\end{bmatrix}$.

The same can be done for the second column trying to get a linear combination that makes $\begin{bmatrix}0\\1\end{bmatrix}$.

---

You can actually consider this as solving the system of $6$ unknowns. In effect what you're doing is considering two smaller systems with three unknowns. The fact that there are "not enough equations" is only a problem if you seek a unique solution: if you don't care about that, then it's a boon because it says there are many solutions :)

Answer 5

If you denote the rows of the matrix $B$ as $\vec\alpha$, $\vec\beta$ and $\vec\gamma$, i.e., $$B=\begin{pmatrix} \vec\alpha\\\vec\beta\\\vec\gamma \end{pmatrix},$$ then you can rewrite your equation as $$ \begin{pmatrix} -1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} \vec\alpha\\\vec\beta\\\vec\gamma \end{pmatrix}= \begin{pmatrix} -\vec\alpha+\vec\gamma\\ \vec\beta+\vec\gamma \end{pmatrix}= \begin{pmatrix}\vec e_1\\\vec e_2\end{pmatrix} $$

So you get two equations $$ \begin{align*} -\vec\alpha+\vec\gamma&=\vec e_1\\ \vec\beta+\vec\gamma&=\vec e_2 \end{align*} $$ which leads to $$ \begin{align*} \vec\alpha&=\vec\gamma-\vec e_1\\ \vec\beta&=\vec e_2-\vec\gamma& \end{align*} $$ If I denote $\vec\gamma=(x,y)$, then the result is $$B=\begin{pmatrix} \vec\alpha\\\vec\beta\\\vec\gamma \end{pmatrix}= \begin{pmatrix} x-1 & y\\ -x & 1-y\\ x & y \end{pmatrix}.$$ After the substitution $x=a+1$, this is the same solution as you have given in your post.