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Chapter 3 · Matrices

Example 6: Find A+BA+B

ExamplesAddition of matrices · Section 3.4.1
The question

Given

A=[31−1230]A = \begin{bmatrix} \sqrt3 & 1 & -1 \\ 2 & 3 & 0 \end{bmatrix}

and

B=[251−2312],B =\begin{bmatrix} 2 & \sqrt5 & 1 \\ -2 & 3 & \tfrac12 \end{bmatrix},

find A+BA + B⁠.

Exam answer

AA and BB are both 2×32\times3⁠, so A+BA + B is defined.

A+B=[3+21+5−1+12−23+30+12]\begin{aligned} &A + B \\ &= \begin{bmatrix} \sqrt3 + 2 & 1 + \sqrt5 & -1 + 1 \\ 2 - 2 & 3 + 3 & 0 + \frac12 \end{bmatrix} \end{aligned}

∴ A+B=[2+31+500612]\begin{aligned} &\therefore\ A + B \\ &= \begin{bmatrix} 2 + \sqrt3 & 1 + \sqrt5 & 0 \\ 0 & 6 & \dfrac12 \end{bmatrix} \end{aligned}

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

The one idea this problem is built on

Matrices are added position by position: the entry of A+BA + B in row ii⁠, column jj is aij+bija_{ij} + b_{ij}⁠. This only makes sense when AA and BB have the same shape, so that every position in AA has a partner in BB⁠. Surds such as 3\sqrt3 and 5\sqrt5 are just numbers; they are added like any other numbers, and unlike surds stay as a sum.

Given, to find, and the plan
Given
Two matrices AA and BB⁠, each with 2 rows and 3 columns.
To find
A+BA + B⁠.
Plan
Confirm both are 2×32\times3 (so the sum is defined and is 2×32\times3⁠). Add the six pairs of corresponding entries, then assemble.

Basics you need

1Sum of two matrices

If A=[aij]A = [a_{ij}] and B=[bij]B = [b_{ij}] are both m×nm\times n⁠, then A+B=[aij+bij]A + B = [a_{ij} + b_{ij}]⁠, again m×nm\times n⁠. If the orders differ, A+BA + B is not defined. Basics: adding only like with like

2Adding surds

2+32 + \sqrt3 cannot be simplified further: 3≈1.732\sqrt3 \approx 1.732 is irrational and is not "like" the whole number 2. So 2+32 + \sqrt3 is left as it is, just as 2+x2 + x is.

3The address aija_{ij}

Row first, column second. Basics: rows, columns and the address a_ij

Solution

  1. Check that the sum is defined

    AA has 2 rows and 3 columns, so it is 2×32\times3⁠. BB also has 2 rows and 3 columns, so it is 2×32\times3⁠. The orders are equal, so A+BA + B is defined, and it is a 2×32\times3 matrix.

    Addition pairs each entry of AA with the entry of BB in the same place. If the shapes differed, some entries would have no partner, and the sum would have no meaning. That is why the order check comes first.
    Addition is defined only for matrices of the same order (Section 3.4.1, and Note 1 after Example 6). Basics: adding only like with like

  2. Add the entries of row 1

    Writing cijc_{ij} for the entries of A+BA + B⁠, so that cij=aij+bijc_{ij} = a_{ij} + b_{ij}⁠:

    c11=3+2=2+3,c12=1+5,c13=−1+1=0.\begin{aligned} c_{11} &= \sqrt3 + 2 = 2 + \sqrt3,\\ c_{12} &= 1 + \sqrt5,\\ c_{13} &= -1 + 1 = 0. \end{aligned}
    Each entry of the sum uses the two entries in exactly that position. 3+2\sqrt3 + 2 and 1+51 + \sqrt5 cannot be combined into a single number, so they stay as sums. −1-1 and 11 are opposites, so they add to 0.
    Only like terms combine: whole numbers with whole numbers, multiples of 3\sqrt3 with multiples of 3\sqrt3⁠.

  3. Add the entries of row 2

    c21=2+(−2)=0,c22=3+3=6,c23=0+12=12.\begin{aligned} c_{21} &= 2 + (-2) = 0,\\ c_{22} &= 3 + 3 = 6,\\ c_{23} &= 0 + \tfrac12 = \tfrac12. \end{aligned}
    Same rule, second row. Adding −2-2 is the same as subtracting 2. Adding 0 changes nothing, so c23=12c_{23} = \tfrac12⁠.

  4. Assemble the sum

    Entry by entry first, then simplified:

    A+B=[2+31+51−12−23+30+12]A + B = \begin{bmatrix} 2+\sqrt3 & 1+\sqrt5 & 1-1 \\ 2-2 & 3+3 & 0+\tfrac12 \end{bmatrix} =[2+31+500612].= \begin{bmatrix} 2+\sqrt3 & 1+\sqrt5 & 0 \\ 0 & 6 & \tfrac12 \end{bmatrix}.
    The result is 2×32\times3⁠, as Step 1 predicted. The two zeros are entries in their own right and must be written in place; leaving them out would change the shape.
    A+B=[aij+bij]A + B = [a_{ij} + b_{ij}]⁠. Lesson: addition and scalar multiplication
Result A+B=[2+31+500612]A + B = \begin{bmatrix} 2+\sqrt3 & 1+\sqrt5 & 0 \\ 0 & 6 & \tfrac12 \end{bmatrix}

This agrees with the textbook's solution.

Check it with numbers

With decimals, 3≈1.7320508\sqrt3 \approx 1.7320508 and 5≈2.2360680\sqrt5 \approx 2.2360680⁠. So c11≈1.7320508+2=3.7320508c_{11} \approx 1.7320508 + 2 = 3.7320508 and c12≈1+2.2360680=3.2360680c_{12} \approx 1 + 2.2360680 = 3.2360680⁠. Adding the other way, B+AB + A⁠, gives the same six entries, as it must, because addition of numbers is commutative.

Common mistakes
  1. Adding surds to whole numbers. 2+32 + \sqrt3 is not 5\sqrt5 and is not 232\sqrt3⁠. It stays as 2+32 + \sqrt3⁠.
  2. Skipping the order check. Before adding, confirm both matrices are the same order; otherwise the sum is not defined.
  3. Dropping zero entries. The sum must still be 2×32\times3⁠, with the zeros shown.
What to take away
  1. Check orders first: A+BA + B exists only when AA and BB have the same order, and then it has that order too.
  2. Add entry by entry: (A+B)ij=aij+bij(A + B)_{ij} = a_{ij} + b_{ij}⁠.
  3. Entries can be any real numbers (surds, fractions); the arithmetic inside each entry follows the usual rules.

Every fact used, and where it comes from

FactUsed inSource
A+B=[aij+bij]A + B = [a_{ij} + b_{ij}]⁠, defined only for the same orderSteps 1–4Section 3.4.1 and the Note after Example 6 (Lesson: addition and scalar multiplication)
3\sqrt3⁠, 5\sqrt5 are irrational, so sums with integers do not simplifyStep 2Number systems
A+B=B+AA + B = B + ACheckSection 3.4.3 (i), commutative law
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