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Matrices · Question 78 of 106

Question 78

MediumOriginal question

If A′=[3−124]A' = \begin{bmatrix}3 & -1 \\ 2 & 4\end{bmatrix} and B=[10−25]B = \begin{bmatrix}1 & 0 \\ -2 & 5\end{bmatrix}, what is (A+2B)′(A + 2B)'?

Options for question 78

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. (A+2B)′=A′+2B′(A + 2B)' = A' + 2B', and A′A' is given.

  2. B′=[1−205]B' = \begin{bmatrix}1 & -2 \\ 0 & 5\end{bmatrix}, so A′+2B′=[3−124]+[2−4010]=[5−5214]A' + 2B' = \begin{bmatrix}3 & -1 \\ 2 & 4\end{bmatrix} + \begin{bmatrix}2 & -4 \\ 0 & 10\end{bmatrix} = \begin{bmatrix}5 & -5 \\ 2 & 14\end{bmatrix}.

Worth noting

[5−1−214]\begin{bmatrix}5 & -1 \\ -2 & 14\end{bmatrix} adds 2B2B instead of 2B′2B': only AA was transposed.

Concept
Properties of the transpose
Typical time
90 s
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The rule this testsThe transpose
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Ten questions from Matrices, at your own pace.

Practise this chapter →

Ten questions from Matrices, at your own pace.