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Chapter 3 · Matrices

Exercise 3.3 · Q4: Find (A+2B)′(A+2B)' given A′A' and BB

Exercise 3.3Transpose of a matrix · Section 3.5
The question

If

A′=[−2312]A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}

and

B=[−1012],B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix},

then find (A+2B)′(A + 2B)'⁠.

Exam answer A=(A′)′=[−2132]A = (A')' = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} 2B=[−2024]2B = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix} A+2B=[−4156]A + 2B = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}

∴ (A+2B)′=[−4516]\therefore\ (A + 2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The question gives A′A'⁠, so first undo the transpose: A=(A′)′A = (A')'⁠. Then build A+2BA + 2B and transpose the result. There is also a shortcut that needs no AA at all: by the properties of the transpose, (A+2B)′=A′+2B′(A+2B)' = A' + 2B'⁠. Doing it both ways checks the answer.

Given, to find, and the plan
Given
A′A' and BB⁠, both of order 2×22\times2⁠.
To find
(A+2B)′(A+2B)'⁠.
Plan
(1) Recover AA⁠. (2) Compute 2B2B⁠. (3) Add to get A+2BA+2B⁠. (4) Transpose. (5) Confirm with A′+2B′A' + 2B'⁠.

Basics you need

1Transposing twice

(A′)′=A(A')' = A⁠, so AA is the transpose of the given A′A'⁠. Basics: reading a table sideways

2Scalar multiples and sums

kB=[k bij]kB = [k\,b_{ij}]⁠: multiply every entry by kk⁠. Matrices of the same order add entry by entry. Basics: adding only like with like

3Transpose rules

For matrices of suitable orders, (A+B)′=A′+B′(A+B)' = A'+B' and (kA)′=kA′(kA)' = kA'⁠. Together they give (A+2B)′=A′+2B′(A+2B)' = A' + 2B'⁠. Lesson: transpose

Solution

  1. Recover AA from A′A'

    The rows of A′A' are −2,3-2, 3 and 1,21, 2⁠. Writing them as columns gives A=(A′)′A = (A')'⁠:

    A=[−2132].A = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}.

    AA is 2×22\times2⁠, the same order as BB⁠, so A+2BA + 2B will be defined.

    The question is about A+2BA + 2B⁠, so we need AA itself. For a 2×22\times2 matrix, transposing simply swaps the two off-diagonal entries: the 3 and the 1 trade places.
    (A′)′=A(A')' = A⁠, Section 3.5.1, property (i). Lesson: transpose

  2. Compute 2B2B

    2B=[2(−1)2(0)2(1)2(2)]=[−2024].2B = \begin{bmatrix} 2(-1) & 2(0) \\ 2(1) & 2(2) \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}.
    A scalar multiplies every entry, not just the first row or the diagonal.
    Multiplication of a matrix by a scalar, Section 3.4.2. Basics: scalar multiples

  3. Add: A+2BA + 2B

    Both are 2×22\times2⁠:

    A+2B=[−2+(−2)1+03+22+4]=[−4156].\begin{aligned} A + 2B &= \begin{bmatrix} -2+(-2) & 1+0 \\ 3+2 & 2+4 \end{bmatrix} \\[1mm] &= \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}. \end{aligned}
    Sum entry by entry, Section 3.4.1.

  4. Transpose the result

    Rows −4,1-4, 1 and 5,65, 6 become columns:

    (A+2B)′=[−4516].(A + 2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}.
    The dash is outside the bracket, so it acts on the finished matrix A+2BA + 2B⁠. Stopping at Step 3 would give the wrong answer.

  5. Confirm by the second route, A′+2B′A' + 2B'

    Transpose BB⁠: its rows −1,0-1, 0 and 1,21, 2 become columns, and then double:

    B′=[−1102],2B′=[−2204].B' = \begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix}, \qquad 2B' = \begin{bmatrix} -2 & 2 \\ 0 & 4 \end{bmatrix}.

    Add the given A′A'⁠:

    A′+2B′=[−2+(−2)3+21+02+4]=[−4516].\begin{aligned} A' + 2B' &= \begin{bmatrix} -2+(-2) & 3+2 \\ 1+0 & 2+4 \end{bmatrix} \\[1mm] &= \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}. \end{aligned}

    This matches Step 4. ✓

    Using (A+2B)′=A′+(2B)′=A′+2B′(A+2B)' = A' + (2B)' = A' + 2B'⁠, the question can be answered without ever finding AA⁠. Getting the same matrix by two independent routes is strong evidence that no entry was miscopied.
    (A+B)′=A′+B′(A+B)' = A'+B' and (kA)′=kA′(kA)' = kA'⁠, Section 3.5.1, properties (ii), (iii) (Basic 3).
Result (A+2B)′=[−4516](A + 2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}

(Textbook answer: the same matrix.)

Check it with numbers

The (1,2) entry of (A+2B)′(A+2B)' is the (2,1) entry of A+2BA+2B⁠, that is a21+2b21=3+2(1)=5a_{21} + 2b_{21} = 3 + 2(1) = 5⁠. ✓ The (2,1) entry is a12+2b12=1+2(0)=1a_{12} + 2b_{12} = 1 + 2(0) = 1⁠. ✓ Diagonal: −2+2(−1)=−4-2 + 2(-1) = -4 and 2+2(2)=62 + 2(2) = 6⁠. ✓

Common mistakes
  1. Using A′A' as if it were AA⁠. Then A′+2BA' + 2B gives entries −4,3,3,6-4, 3, 3, 6⁠, and its transpose is wrong.
  2. Forgetting the final transpose. A+2BA + 2B itself has 1 in position (1,2), not 5.
  3. Doubling only some entries of BB⁠. 2B2B doubles all four.
What to take away
  1. (A′)′=A(A')' = A lets you recover a matrix from its transpose.
  2. (A+kB)′=A′+kB′(A + kB)' = A' + kB'⁠: transposing passes through sums and scalar multiples.
  3. When two routes are available, use the second as a check.

Every fact used, and where it comes from

FactUsed inSource
Transpose (A′)ij=aji(A')_{ij} = a_{ji}Steps 1, 4, 5Definition 3, Section 3.5 (Lesson: transpose)
(A′)′=A(A')' = AStep 1Section 3.5.1, property (i)
Scalar multiple kB=[k bij]kB = [k\,b_{ij}]Steps 2, 5Section 3.4.2 (Lesson: addition and scalar multiplication)
Sum entry by entrySteps 3, 5Section 3.4.1
(A+B)′=A′+B′(A+B)' = A'+B'⁠, (kA)′=kA′(kA)' = kA'Step 5Section 3.5.1, properties (ii), (iii)
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