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Chapter 3 · Matrices

Exercise 3.2 · Q4: Find A+BA+B⁠, B−CB-C and verify A+(B−C)=(A+B)−CA+(B-C)=(A+B)-C

Exercise 3.2Addition of matrices and its properties · Sections 3.4.1 to 3.4.3
The question

If

A=[12−35021−11]A=\begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix} B=[3−12425203]B=\begin{bmatrix}3&-1&2\\4&2&5\\2&0&3\end{bmatrix} andC=[4120321−23],\text{and}\quad C=\begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix},

then compute (A+B)(A+B) and (B−C)(B-C)⁠. Also, verify that A+(B−C)=(A+B)−CA+(B-C)=(A+B)-C⁠.

Exam answer

Adding or subtracting corresponding entries,

A+B=[1+32−1−3+25+40+22+51+2−1+01+3]=[41−19273−14]\begin{aligned} &A + B \\ &= \begin{bmatrix} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix} \\ &= \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} \end{aligned} B−C=[3−4−1−12−24−02−35−22−10+23−3]=[−1−204−13120]\begin{aligned} &B - C \\ &= \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0+2 & 3-3 \end{bmatrix} \\ &= \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} \end{aligned} LHS=A+(B−C)=[1−12−2−3+05+40−12+31+1−1+21+0]=[00−39−15211]\begin{aligned} &\text{LHS} = A + (B - C) \\ &= \begin{bmatrix} 1-1 & 2-2 & -3+0 \\ 5+4 & 0-1 & 2+3 \\ 1+1 & -1+2 & 1+0 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix} \end{aligned} RHS=(A+B)−C=[4−41−1−1−29−02−37−23−1−1+24−3]=[00−39−15211]\begin{aligned} &\text{RHS} = (A + B) - C \\ &= \begin{bmatrix} 4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1+2 & 4-3 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix} \end{aligned}

∴\therefore LHS = RHS, that is, A+(B−C)=(A+B)−CA + (B - C) = (A + B) - C⁠. Hence verified.

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

The one idea this problem is built on

Every entry of these sums and differences is computed in its own position, so in each position the claim is a statement about three ordinary numbers: a+(b−c)=(a+b)−ca+(b-c)=(a+b)-c⁠. That is true for all numbers, so the matrix equation is true for all matrices of the same order. In the language of Section 3.4.3 it is the associative law of matrix addition applied to AA⁠, BB and −C-C⁠. Here we confirm it by computing both sides in full.

Given, to find, and the plan
Given
Three 3×33\times3 matrices AA⁠, BB⁠, CC⁠.
To find
A+BA+B and B−CB-C⁠; and to show A+(B−C)=(A+B)−CA+(B-C)=(A+B)-C⁠.
Plan
(1) Check the orders. (2) Compute A+BA+B⁠. (3) Compute B−CB-C⁠. (4) Compute the left side A+(B−C)A+(B-C)⁠. (5) Compute the right side (A+B)−C(A+B)-C⁠. (6) Compare entry by entry, and explain why the equality always holds.

Basics you need

1Sum and difference of matrices

For matrices of the same order, A+B=[aij+bij]A+B=[a_{ij}+b_{ij}] and B−C=[bij−cij]B-C=[b_{ij}-c_{ij}]⁠: add or subtract in each position. Basics: adding like with like

2Equal matrices

Two matrices are equal when they have the same order and every entry of one equals the entry in the same position of the other. So "verify" means: compute both sides and compare all nine positions. Basics: positions aija_{ij}

3Brackets with numbers

For ordinary numbers, a+(b−c)=a+b−c=(a+b)−ca+(b-c)=a+b-c=(a+b)-c⁠. Subtracting a negative adds: 2−(−2)=42-(-2)=4⁠.

Solution

  1. Orders

    AA⁠, BB⁠, CC each have 3 rows and 3 columns, so each is 3×33\times3⁠. Every sum and difference in the question is therefore defined, and every result is 3×33\times3⁠.

    Addition and subtraction need matching positions. With all three matrices 3×33\times3⁠, every intermediate matrix (A+BA+B⁠, B−CB-C⁠) is also 3×33\times3⁠, so the later sums are defined too.
    Lesson: addition of matrices

  2. Compute A+BA+B

    Add the entries in each of the nine positions:

    A+B=[1+32+(−1)−3+25+40+22+51+2−1+01+3]=[41−19273−14]\begin{aligned}A+B&=\begin{bmatrix}1+3&2+(-1)&-3+2\\5+4&0+2&2+5\\1+2&-1+0&1+3\end{bmatrix}\\&=\begin{bmatrix}4&1&-1\\9&2&7\\3&-1&4\end{bmatrix}\end{aligned}
    Each position is independent. The entries to watch are those with negatives: 2+(−1)=12+(-1)=1⁠, −3+2=−1-3+2=-1⁠, −1+0=−1-1+0=-1⁠.

  3. Compute B−CB-C

    B−C=[3−4−1−12−24−02−35−22−10−(−2)3−3]=[−1−204−13120]\begin{aligned}B-C&=\begin{bmatrix}3-4&-1-1&2-2\\4-0&2-3&5-2\\2-1&0-(-2)&3-3\end{bmatrix}\\&=\begin{bmatrix}-1&-2&0\\4&-1&3\\1&2&0\end{bmatrix}\end{aligned}
    The difference is defined position by position, bij−cijb_{ij}-c_{ij} (Section 3.4.2). The (3,2)(3,2) entry is the one sign trap: c32=−2c_{32}=-2⁠, so 0−(−2)=20-(-2)=2⁠.

  4. Left side: A+(B−C)A+(B-C)

    Add AA to the matrix found in Step 3:

    [12−35021−11]+[−1−204−13120]=[1−12−2−3+05+40−12+31+1−1+21+0]=[00−39−15211]\begin{aligned} &\begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix}+\begin{bmatrix}-1&-2&0\\4&-1&3\\1&2&0\end{bmatrix}\\ &=\begin{bmatrix}1-1&2-2&-3+0\\5+4&0-1&2+3\\1+1&-1+2&1+0\end{bmatrix}\\ &=\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix} \end{aligned}
    The bracket in A+(B−C)A+(B-C) says: do B−CB-C first (done in Step 3), then add AA⁠. Using the already computed B−CB-C is exactly what the bracket asks for.

  5. Right side: (A+B)−C(A+B)-C

    Subtract CC from the matrix found in Step 2:

    [41−19273−14]−[4120321−23]=[4−41−1−1−29−02−37−23−1−1−(−2)4−3]=[00−39−15211]\begin{aligned} &\begin{bmatrix}4&1&-1\\9&2&7\\3&-1&4\end{bmatrix}-\begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix}\\ &=\begin{bmatrix}4-4&1-1&-1-2\\9-0&2-3&7-2\\3-1&-1-(-2)&4-3\end{bmatrix}\\ &=\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix} \end{aligned}
    Now the bracket says: do A+BA+B first (Step 2), then subtract CC⁠. Again the (3,2)(3,2) position has −(−2)=+2-(-2)=+2⁠, giving −1+2=1-1+2=1⁠.

  6. Compare, and see why it had to work

    Both sides are 3×33\times3⁠, and the nine entries agree position by position:

    A+(B−C)=[00−39−15211]=(A+B)−C.\begin{aligned}A+(B-C)&=\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix}\\&=(A+B)-C.\end{aligned}

    So A+(B−C)=(A+B)−CA+(B-C)=(A+B)-C is verified for these matrices.

    Why it always holds. Take any position (i,j)(i,j)⁠. The left side has entry aij+(bij−cij)a_{ij}+(b_{ij}-c_{ij})⁠, the right side has (aij+bij)−cij(a_{ij}+b_{ij})-c_{ij}⁠. These are equal for all numbers, so the two matrices agree in every position. Equivalently, writing B−C=B+(−C)B-C=B+(-C)⁠, the claim is A+(B+(−C))=(A+B)+(−C)A+(B+(-C))=(A+B)+(-C)⁠, which is the associative law of matrix addition.

    "Verify" asks only for the check with these particular matrices, which Steps 4 and 5 provide. The general reason explains why the check could not have failed (unless an arithmetic slip was made), and links the exercise to the property it illustrates.
    Associative law: (A+B)+C=A+(B+C)(A+B)+C=A+(B+C) for matrices of the same order. Lesson: properties of the operations
Result A+B=[41−19273−14]A+B=\begin{bmatrix}4&1&-1\\9&2&7\\3&-1&4\end{bmatrix} B−C=[−1−204−13120]B-C=\begin{bmatrix}-1&-2&0\\4&-1&3\\1&2&0\end{bmatrix}

and both A+(B−C)A+(B-C) and (A+B)−C(A+B)-C equal

[00−39−15211],\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix},

so A+(B−C)=(A+B)−CA+(B-C)=(A+B)-C⁠. (Textbook answer: the same A+BA+B and B−CB-C⁠.)

Check it with numbers

A third route to the same matrix: (A−C)+B(A-C)+B⁠. First

A−C=[−31−55−3001−2],A-C=\begin{bmatrix}-3&1&-5\\5&-3&0\\0&1&-2\end{bmatrix},

and adding BB gives rows (−3+3, 1−1, −5+2)=(0,0,−3)(-3+3,\ 1-1,\ -5+2)=(0,0,-3)⁠, (5+4, −3+2, 0+5)=(9,−1,5)(5+4,\ -3+2,\ 0+5)=(9,-1,5)⁠, (0+2, 1+0, −2+3)=(2,1,1)(0+2,\ 1+0,\ -2+3)=(2,1,1)⁠: the same matrix. ✓

Entry totals: AA totals 8, BB totals 20, CC totals 14, and 8+20−14=148+20-14=14⁠, while the answer totals 0+0−3+9−1+5+2+1+1=140+0-3+9-1+5+2+1+1=14⁠. ✓

Common mistakes
  1. 0−(−2)0-(-2) written as −2-2⁠. Subtracting a negative adds; the (3,2)(3,2) entry of B−CB-C is 22⁠.
  2. Mixing the bracket order. A+(B−C)A+(B-C) uses B−CB-C first; (A+B)−C(A+B)-C uses A+BA+B first. They agree in the end, but the working must follow each bracket.
  3. Checking only a few entries. Equality of matrices needs all nine positions to agree.
What to take away
  1. A matrix identity built from ++ and −- holds because it holds for numbers in each position.
  2. B−CB-C is B+(−C)B+(-C)⁠, so rules for addition (here the associative law) cover subtraction too.
  3. Bracket every negative entry when you subtract it.

Every fact used, and where it comes from

FactUsed inSource
Sum of matrices of the same orderSteps 1, 2, 4Section 3.4.1 (Lesson: addition)
Negative −C-C and difference B−C=[bij−cij]B-C=[b_{ij}-c_{ij}]Steps 3, 5Section 3.4.2
Equality of matrices: same order, equal corresponding entriesStep 6Section 3.3.1, Definition 2 (Lesson: equality)
Associative law (A+B)+C=A+(B+C)(A+B)+C=A+(B+C)Step 6Section 3.4.3 (ii) (Lesson: properties)
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