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I am getting nowhere with this question. I found a similar question on this site but I need help in a more step by step way. I would be very grateful.

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5 Answers 5

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Consider matrices as models of linear maps. Then the product of two matrices stands for the composition of the maps. The first matrix (right hand side of the multiplication) maps $\mathbb R^3$ to $\mathbb R^2.$ The second matrix (left hand side of the multiplication) maps $\mathbb R^2$ to $\mathbb R^3.$ The latter map can never be onto, so neither can the composed map be invertible.

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Let $A=\begin{bmatrix} a_1 & a_2 \\ b_1 & b_2 \\ c_1 & c_2 \end{bmatrix}$ and $B=\begin{bmatrix} a'_1 & b'_1 & c'_1\\ a'_2 & b'_2 & c'_2 \end{bmatrix}$.

Also let $A'=\begin{bmatrix} a_1 & a_2 & 0 \\ b_1 & b_2 & 0\\ c_1 & c_2 & 0 \end{bmatrix}$ and $B'=\begin{bmatrix} a'_1 & b'_1 & c'_1\\ a'_2 & b'_2 & c'_2\\ 0 & 0 & 0 \end{bmatrix}$. So we have $A\times B = A' \times B'$, but we have: $$det(A\times B)=det(A'\times B')=det(A')\cdot det(B')=0$$, therefore $A\times B$ is not invertible.

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  • $\begingroup$ Well, not strong objection, but $B = 2x3$ $\endgroup$ Commented Dec 11, 2015 at 12:01
  • $\begingroup$ @IlanAizelmanWS: Thank you for comment. I edited my post. $\endgroup$ Commented Dec 11, 2015 at 12:05
  • $\begingroup$ I like your answer. But it would be much easier to understand if you kept the OP's notation. The original given matrices were called $A$ and $B$, not $A$ and $A'$. $\endgroup$
    – bubba
    Commented Dec 11, 2015 at 13:36
  • $\begingroup$ @bubba: Yes, you are right. thank you for your comment. I change their names. $\endgroup$ Commented Dec 11, 2015 at 13:44
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There are at most 2 columns which are not linearly independent in matrix $A$. Therefore, as multiplication with $B$ is just a linear combination of the columns of $A$, then it cannot be that now we have 3 linearly independent columns. Hence the matrix $AB$ cannot be invertible.

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HINT:

Each of the $3$ columns of $AB$ is a linear combinations of the $2$ columns of $A$. Use the multilinearity properties of the determinant we conclude that $\det(AB) = 0$, and so $AB$ cannot be invertible.

Obs: This proof works for matrices over a commutative ring with $1$.

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AB is linearly dependent by A's columns. and A has only 2 columns, which means AB will have 2 columns which are linearly dependent. Thus, AB can't be linearly independent. Which means, $|AB| = 0$. and from here it is pretty clear it can't be invertible.

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