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If A and B are any matrices of order 2x1, how can we show that the product AB^t has no inverse?

Any guidance is much appreciated!

jaykirby
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    It is easy to show that the columns of $AB^T$ are linearly dependent since both consist of scaled copies of $A$. In other words $AB^T= [b_1 A, b_2 A]$. – Tpofofn Aug 11 '13 at 01:10

2 Answers2

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The rank of $A$ and the rank of $B$ are both equal to 1. since $AB^{T}$ is a 2x2 matrix its rank must be 2 for it to be invertible. but $rank(AB)\leq min(rank(A),rank(B))$ see http://en.wikipedia.org/wiki/Rank_(linear_algebra) . Thus $AB^{T}$ cannot have an inverse

hchengz
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If $$A=\begin{pmatrix} a \\ b\\ \end{pmatrix},B=\begin{pmatrix} c \\ d\\ \end{pmatrix}$$ then $$AB^t=\begin{pmatrix} ac & ad\\ bc & bd\\ \end{pmatrix}$$ Since $|AB^t|=0$ so it has no inverse.

Mikasa
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