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Q39E

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Found in: Page 144

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# We are told that a certain${\mathbf{5}}{\mathbf{×}}{\mathbf{5}}$ matrix${\mathbit{A}}$ can be written as${\mathbit{A}}{\mathbf{=}}{\mathbit{B}}{\mathbit{C}}$,where $\mathbit{B}$is ${\mathbf{5}}{\mathbf{×}}{\mathbf{4}}$and${\mathbit{C}}$ is ${\mathbf{4}}{\mathbf{×}}{\mathbf{5}}$. Explain how you know that${\mathbit{A}}$ is not invertible.

Thus, it is proved matrix$A$ is not invertible.

See the step by step solution

## Step 1: Given in the question.

Let matrix $A$is $5×5$which is $A=BC$where$B$ is $5×4$and$C$ is$4×5$ .

Let assume that$colspace\left(A\right)\subseteq colspace\left(B\right)$ . There are different ways and one way is as the linear transformation${T}_{A}$ is the composition of ${T}_{B}$and${T}_{C}$ gives $im\left({T}_{A}\right)\subseteq im\left({T}_{B}\right)$.

## Step 2: Explanation of A is not invertible.

Take dimensions where$rank\left(A\right)\le rank\left(B\right)$ . Now by the rank-nullity theorem applied to $B$, gives$rank\left(B\right)\le 4$.

Which implies that$rank\left(A\right)$ is never 5, that is needed for $A$to be invertible.

Hence, matrix$A$is not invertible.