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Chapter 3 · Matrices

Exercise 3.3 · Q2: Verify (A+B)′=A′+B′(A+B)'=A'+B' and (A−B)′=A′−B′(A-B)'=A'-B'

Exercise 3.3Transpose of a matrix · Section 3.5
The question

If

A=[−123579−211]A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}

and

B=[−41−5120131],B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix},

then verify that

(i) (A+B)′=A′+B′(A + B)' = A' + B'⁠,   (ii) (A−B)′=A′−B′(A - B)' = A' - B'

Exam answer A′=[−15−2271391]A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix} B′=[−411123−501]B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}

(i)

A+B=[−53−2699−142]A + B = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix} LHS=(A+B)′=[−56−1394−292]\begin{aligned} \text{LHS} &= (A + B)' \\ &= \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix} \end{aligned} RHS=A′+B′=[−56−1394−292]\begin{aligned} \text{RHS} &= A' + B' \\ &= \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix} \end{aligned}

∴\therefore LHS = RHS, so (A+B)′=A′+B′(A + B)' = A' + B'⁠.

(ii)

A−B=[318459−3−20]A - B = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix} LHS=(A−B)′=[34−315−2890]\begin{aligned} \text{LHS} &= (A - B)' \\ &= \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix} \end{aligned} RHS=A′−B′=[34−315−2890]\begin{aligned} \text{RHS} &= A' - B' \\ &= \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix} \end{aligned}

∴\therefore LHS = RHS, so (A−B)′=A′−B′(A - B)' = A' - B'⁠. Hence verified.

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

The one idea this problem is built on

"Verify" means: compute the left side, compute the right side, and see that they are the same matrix. The two routes are "add (or subtract) first, then transpose" and "transpose first, then add (or subtract)". They must agree, because transposing only moves entries, and adding happens entry by entry, so it does not matter whether the entries are moved before or after they are added.

Given, to find, and the plan
Given
Two 3×33\times3 matrices AA and BB⁠.
To show
(i) (A+B)′=A′+B′(A+B)' = A'+B'⁠; (ii) (A−B)′=A′−B′(A-B)' = A'-B'⁠.
Plan
(1) Check the orders, so every sum and difference is defined. (2) Find A+BA+B and transpose it. (3) Find A′A' and B′B' and add them. (4) Compare. (5) Repeat with subtraction. (6) Explain why the result had to come out this way.

Basics you need

1Adding and subtracting matrices

Two matrices of the same order are added by adding entries in the same position: [aij]+[bij]=[aij+bij][a_{ij}] + [b_{ij}] = [a_{ij} + b_{ij}]⁠. Subtraction works the same way: A−B=[aij−bij]A - B = [a_{ij} - b_{ij}]⁠. Matrices of different orders cannot be added. Basics: adding only like with like

2The transpose

A′A' is AA with rows written as columns: (A′)ij=aji(A')_{ij} = a_{ji}⁠. An m×nm\times n matrix has an n×mn\times m transpose. Basics: reading a table sideways

3Equal matrices

Two matrices are equal when they have the same order and every pair of corresponding entries is equal. So to verify an equation of matrices we compare all entries. Basics: rows, columns and addresses

Part (i): (A+B)′=A′+B′(A+B)' = A'+B'

  1. (i) Check the orders

    AA and BB are both 3×33\times3⁠, so A+BA+B is defined and is 3×33\times3⁠; its transpose is 3×33\times3⁠. Also A′A' and B′B' are both 3×33\times3⁠, so A′+B′A'+B' is defined and is 3×33\times3⁠. Both sides have the same order, so comparing them makes sense.

    An addition is only defined for matrices of the same order (Basic 1). Checking first means we never write down a meaningless sum.

  2. (i) Left side: add, then transpose

    Add entry by entry:

    A+B=[−1+(−4)2+13+(−5)5+17+29+0−2+11+31+1]=[−53−2699−142].\begin{aligned} &A + B \\ &= \begin{bmatrix} -1+(-4) & 2+1 & 3+(-5) \\ 5+1 & 7+2 & 9+0 \\ -2+1 & 1+3 & 1+1 \end{bmatrix} \\[1mm] &= \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}. \end{aligned}

    Now write the rows −5,3,−2-5, 3, -2⁠; 6,9,96, 9, 9⁠; −1,4,2-1, 4, 2 as columns:

    (A+B)′=[−56−1394−292].(A + B)' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}.
    The left side tells us the order of operations: the bracket (A+B)(A+B) is formed first, and the dash acts on the whole result.
    Sum of matrices, Section 3.4.1; transpose, Definition 3. Basics: adding like with like

  3. (i) Right side: transpose, then add

    Rows of AA become columns of A′A'⁠, and rows of BB become columns of B′B'⁠:

    A′=[−15−2271391],A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}, B′=[−411123−501].B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}.

    Add entry by entry:

    A′+B′=[−1+(−4)5+1−2+12+17+21+33+(−5)9+01+1]=[−56−1394−292].\begin{aligned} &A' + B' \\ &= \begin{bmatrix} -1+(-4) & 5+1 & -2+1 \\ 2+1 & 7+2 & 1+3 \\ 3+(-5) & 9+0 & 1+1 \end{bmatrix} \\[1mm] &= \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}. \end{aligned}
    Here the dashes act on AA and BB separately, before the addition. We get the same nine numbers as in the previous step, but by a different route.
    (A′)ij=aji(A')_{ij} = a_{ji} (Basic 2). Lesson: transpose

  4. (i) Compare

    Part (i), Steps 2 and 3 give matrices of the same order 3×33\times3 whose nine entries agree position by position: −5,6,−1-5, 6, -1 in row 1, 3,9,43, 9, 4 in row 2, −2,9,2-2, 9, 2 in row 3. Hence (A+B)′=A′+B′(A+B)' = A'+B'.

    Equality of matrices: same order and equal corresponding entries, Section 3.3.1 (Basic 3).

Part (ii): (A−B)′=A′−B′(A-B)' = A'-B'

  1. (ii) Left side: subtract, then transpose

    All four matrices are 3×33\times3⁠, so both differences are defined. Subtract entry by entry, taking care with the minus signs:

    A−B=[−1−(−4)2−13−(−5)5−17−29−0−2−11−31−1]=[318459−3−20].\begin{aligned} &A - B \\ &= \begin{bmatrix} -1-(-4) & 2-1 & 3-(-5) \\ 5-1 & 7-2 & 9-0 \\ -2-1 & 1-3 & 1-1 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}. \end{aligned}

    Rows become columns:

    (A−B)′=[34−315−2890].(A - B)' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}.
    Subtracting a negative number adds it: −1−(−4)=−1+4=3-1-(-4) = -1+4 = 3 and 3−(−5)=83-(-5) = 8⁠. These two entries are where sign slips usually happen.
    Difference A−B=[aij−bij]A - B = [a_{ij} - b_{ij}]⁠, Section 3.4.2. Basics: adding like with like

  2. (ii) Right side: subtract the transposes

    Using A′A' and B′B' from Part (i), Step 3:

    A′−B′=[−1−(−4)5−1−2−12−17−21−33−(−5)9−01−1]=[34−315−2890].\begin{aligned} &A' - B' \\ &= \begin{bmatrix} -1-(-4) & 5-1 & -2-1 \\ 2-1 & 7-2 & 1-3 \\ 3-(-5) & 9-0 & 1-1 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}. \end{aligned}

  3. (ii) Compare

    Both sides are 3×33\times3 with the same nine entries, so (A−B)′=A′−B′(A-B)' = A'-B'.

  4. Why it had to work

    Take any position (i,j)(i,j)⁠. On the left, the (i,j)(i,j) entry of (A+B)′(A+B)' is the (j,i)(j,i) entry of A+BA+B⁠, which is aji+bjia_{ji} + b_{ji}⁠. On the right, the (i,j)(i,j) entry of A′+B′A'+B' is (A′)ij+(B′)ij=aji+bji(A')_{ij} + (B')_{ij} = a_{ji} + b_{ji}⁠. These are the same number for every ii and jj⁠. The argument with −- in place of ++ gives (A−B)′=A′−B′(A-B)' = A'-B'⁠.

    The verification above checks one pair of matrices. This short argument shows the rule holds for every pair of matrices of the same order, which is the property the textbook states in Section 3.5.1.
    Properties of transpose, Section 3.5.1 (iii): (A+B)′=A′+B′(A+B)' = A'+B'⁠. Lesson: transpose
Result

Both sides of (i) equal

[−56−1394−292],\begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix},

and both sides of (ii) equal

[34−315−2890].\begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}.

So (A+B)′=A′+B′(A+B)' = A'+B' and (A−B)′=A′−B′(A-B)' = A'-B' are verified.

Check it with numbers

Pick one position, say row 3, column 1 of (A+B)′(A+B)'⁠. It should be the row 1, column 3 entry of A+BA+B⁠, that is a13+b13=3+(−5)=−2a_{13} + b_{13} = 3 + (-5) = -2⁠. The matrix in Part (i), Step 2 has −2-2 there. ✓ For (A−B)′(A-B)'⁠, position (1,3) should be a31−b31=−2−1=−3a_{31} - b_{31} = -2 - 1 = -3⁠, and it is. ✓

Common mistakes
  1. Transposing only one matrix. A′+BA'+B is not the right side; both AA and BB must be transposed.
  2. Sign slips in subtraction. −1−(−4)-1-(-4) is 33⁠, not −5-5⁠.
  3. Calling it verified after one entry. Two matrices are equal only when all corresponding entries agree.
What to take away
  1. Transposing and adding can be done in either order: (A±B)′=A′±B′(A\pm B)' = A' \pm B'⁠.
  2. To verify a matrix identity, compute both sides completely and compare every entry.
  3. An entry-level argument, (A+B)ij′=aji+bji(A+B)'_{ij} = a_{ji} + b_{ji}⁠, proves the rule for all matrices of the same order.

Every fact used, and where it comes from

FactUsed inSource
Sum [aij+bij][a_{ij}+b_{ij}] of matrices of the same orderPart (i), Steps 2–3Section 3.4.1 (Lesson: addition)
Difference A−B=[aij−bij]A-B = [a_{ij}-b_{ij}]Part (ii), Steps 1–2Section 3.4.2, "Difference of matrices"
Transpose (A′)ij=aji(A')_{ij} = a_{ji}ThroughoutDefinition 3, Section 3.5 (Lesson: transpose)
Equality of matricesPart (i), Step 4; Part (ii), Step 3Definition 2, Section 3.3.1
(A+B)′=A′+B′(A+B)' = A'+B' in generalPart (ii), Step 4Section 3.5.1, property (iii); Example 20
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