Science:Math Exam Resources/Courses/MATH152/April 2010/Question A 28

MATH152 April 2010

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Question A 28
Circle all of the following statements that are true for all invertible matrices A, and all matrices B and C of the same size as A and all scalars r ≠ 0: (AB)C = A(BC) (A+B)(A-B) = A2-B2 ATBT = (AB)T (A-1)T = (AT)-1 (rA)-1 = ${\frac {1}{r}}$A-1

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
Carefully review your linear algebra rules. One very important difference from numbers to matrices is that in general matrices do not commute, i.e. $AB\neq {}BA$

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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. a. $\displaystyle (AB)C=A(BC)$ This is true and is known as the distributive property. Notice we are not changing the order in which the multiplication occurs, just which multiplications we are performing first. b. $\displaystyle (A+B)(A-B)=A^{2}-B^{2}$ This is false. Let's carefully expand out the left-hand side, $\displaystyle {}A^{2}-AB+BA-B^{2}$ however, since matrices do not in general commute then AB is not the same as BA and those terms do not cancel like they do with real numbers. c. $\displaystyle A^{T}B^{T}=(AB)^{T}$ This is false. The property of transposes is that $\displaystyle {}(AB)^{T}=B^{T}A^{T}$ however even without remembering the correct form of the property we can conclude that the presented form is wrong. Let A be size n x m and B be size m x n such that $m\neq {}n$. Notice that AB is a sensible operation here and produces a matrix n x n. If we then transpose the resulting matrix it is also n x n. However, notice that $A^{T}$ is m x n and $B^{T}$ is n x m. Therefore, $A^{T}B^{T}$ is m x m which since $m\neq {n}$ is not the same size as n x n and therefore the property can not hold for all matrices. d. $\displaystyle (A^{T})^{-1}=(A^{-1})^{T}$ This is true and is a consequence of the transpose property, $\displaystyle {}(AB)^{T}=B^{T}A^{T}.$ e. $(rA)^{-1}={\frac {1}{r}}A^{-1}$ This is true (as long as the scalar is non-zero which is given) and is a property of inverses. Therefore, a,d, and e are true.