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Can someone help?

$W\in R^{2\times 3}$ with first row $\pmatrix {1&1&1}$

$D\in R^{3\times 2}$ with first row $\pmatrix {2&2}$

Can the above matrices exist so that $WD=I_2$ ?

my ideas...

$rankW$ is either $1$ or $2$

$rankD$ is either $1$ or $2$

$rankI_2=2$

$rankWD\le min(rankW,rankD) \implies rankWD\le2$

Help...

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    Basically, if we find two matrices, we will have solved it. – 2017-02-06

2 Answers 2

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Your idea to check ranks first is good. Since by that check the existence of the matrices seems plausible, try to find the matrices. There are a lot of degrees of freedom, so let's erase some of them and fashion arbitrarily a row-rank 2 matrix for $W$ and then use that to fill out $D$. This isn't sketchy because we leave ourselves four variables in $D$, which will result in four equations in $4$ variables so long as $W$ has independent rows. So, let $$ W = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} $$

Now to fill out $D$ you can just write $$D = \begin{bmatrix} 2 & 2 \\ a & b \\ c & d \end{bmatrix} $$ and solve the resulting system of linear equations. The system will be solvable!

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    Well, I was trying to solve this without putting 'convenient' numbers in matrix $A$. This really helped. Thank you very much @BadamBaplan – 2017-02-06
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For example, we can take $$ W = \pmatrix{1&1&1\\1&2&0}, \quad D = \pmatrix{2&2\\-1 &-1/2\\0&-3/2} $$

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    $WD=I_2$ it's OK, thank you !!! – 2017-02-06