Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 13

MATH152 April 2017

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Question A 13
Circle each of the following statements that is true for all invertible $n\times n$ matrices $A$ : (a) The rank of $A$ is $n$ . (b) $\det A=0$ . (c) The reduced row echelon form of $A$ is the $n\times n$ identity matrix. (d) $A^{-1}$ exists. (e) Columns of $A$ are linearly independent. (f) $A=A^{T}$ . .

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall the definition of an invertible matrix and its properties.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that we say a ${\textstyle n\times n}$ matrix invertible if ${\textstyle A^{-1}}$ exists. This implies that ${\textstyle \det(A)\neq 0}$ as ${\textstyle \det(A)\cdot \det(A^{-1})=\det(AA^{-1})=\det(I)=1.}$ Thus, ${\textstyle A}$ has full rank ${\textstyle n}$ , and the columns of ${\textstyle A}$ are linear independent. Note that any invertible matrix has the identity as its reduced row echelon form, thus $(c)$ is also correct. However, this does not necessarily imply that any invertible matrix is symmetric. Consider $A={\begin{bmatrix}2&2\\1&0\end{bmatrix}}.$ ${\textstyle A}$ is invertible but ${\textstyle A\neq A^{T}}$ .The answer is $\color {blue}{(a),(c),(d),(e)}$ .