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Chapter 3 · Matrices

Example 7: Find 2A−B2A-B

ExamplesScalar multiple and difference · Section 3.4.2
The question

If

A=[123231]A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}

and

B=[3−13−102],B = \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix},

then find 2A−B2A - B⁠.

Exam answer 2A=[246462]2A = \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \end{bmatrix} 2A−B=[2−34+16−34+16−02−2]\begin{aligned} &2A - B \\ &= \begin{bmatrix} 2 - 3 & 4 + 1 & 6 - 3 \\ 4 + 1 & 6 - 0 & 2 - 2 \end{bmatrix} \end{aligned}

∴ 2A−B=[−153560]\therefore\ 2A - B = \begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}

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

The one idea this problem is built on

Two operations, both done entry by entry. Multiplying by a scalar kk multiplies every entry by kk⁠. Subtracting BB means adding −B=(−1)B-B = (-1)B⁠, the matrix with every sign of BB flipped. So the (i,j)(i,j) entry of 2A−B2A - B is simply 2aij−bij2a_{ij} - b_{ij}⁠.

Given, to find, and the plan
Given
Two 2×32\times3 matrices AA and BB⁠.
To find
2A−B2A - B⁠.
Plan
Check orders. Compute 2A2A⁠, then −B-B⁠, then add them entry by entry.

Basics you need

1Scalar multiple

For a number kk and a matrix A=[aij]m×nA = [a_{ij}]_{m\times n}⁠, kA=[k aij]m×nkA = [k\,a_{ij}]_{m\times n}⁠: every entry is multiplied by kk⁠, and the order does not change. Lesson: addition and scalar multiplication

2Negative and difference

−B=(−1)B-B = (-1)B flips the sign of every entry. For matrices of the same order, A−B=A+(−B)A - B = A + (-B)⁠, so (A−B)ij=aij−bij(A - B)_{ij} = a_{ij} - b_{ij}⁠. Basics: adding only like with like

3Sum

Matrices of the same order are added entry by entry.

Solution

  1. Check the orders

    AA and BB each have 2 rows and 3 columns, so both are 2×32\times3⁠. A scalar multiple keeps the order, so 2A2A is 2×32\times3 and −B-B is 2×32\times3⁠. Their sum is defined and is 2×32\times3⁠.

    The difference 2A−B2A - B is a sum of 2A2A and −B-B⁠, and a sum needs equal orders. Checking now saves us from a meaningless calculation.
    kAkA has the same order as AA (Section 3.4.2).

  2. Compute 2A2A

    Double every entry of AA⁠:

    2A=[2⋅12⋅22⋅32⋅22⋅32⋅1]=[246462].2A = \begin{bmatrix} 2\cdot1 & 2\cdot2 & 2\cdot3 \\ 2\cdot2 & 2\cdot3 & 2\cdot1 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \end{bmatrix}.
    "2A2A" means each of the six entries is multiplied by 2. Doubling only one row, or only the first entry, is a common slip; the scalar reaches every entry.
    kA=[k aij]kA = [k\,a_{ij}]⁠. Lesson: addition and scalar multiplication

  3. Compute −B-B

    Multiply every entry of BB by −1-1⁠:

    −B=[−31−310−2].-B = \begin{bmatrix} -3 & 1 & -3 \\ 1 & 0 & -2 \end{bmatrix}.
    Subtracting BB is the same as adding its negative. Writing −B-B out separately makes the signs visible before we add, which is where most errors happen. Note −(−1)=1-(-1) = 1 and −0=0-0 = 0⁠.
    −B=(−1)B-B = (-1)B and A−B=A+(−B)A - B = A + (-B) (Section 3.4.2). Basics: adding only like with like

  4. Add 2A2A and −B-B entry by entry

    Writing dijd_{ij} for the entries of 2A−B2A - B⁠, so dij=2aij−bijd_{ij} = 2a_{ij} - b_{ij}⁠:

    d11=2−3=−1,d12=4+1=5,d13=6−3=3,d21=4+1=5,d22=6+0=6,d23=2−2=0.\begin{aligned} d_{11} &= 2 - 3 = -1, & d_{12} &= 4 + 1 = 5,\\ d_{13} &= 6 - 3 = 3, & d_{21} &= 4 + 1 = 5,\\ d_{22} &= 6 + 0 = 6, & d_{23} &= 2 - 2 = 0. \end{aligned}

    Assembled:

    2A−B=[−153560].2A - B = \begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}.
    Each entry of 2A2A is added to the entry of −B-B in the same place. The result is 2×32\times3⁠, as Step 1 predicted.
Result 2A−B=[−153560]2A - B = \begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}

This agrees with the textbook's solution.

Check it with numbers

If 2A−B2A - B is right, adding BB back must give 2A2A⁠. Entry by entry: −1+3=2-1 + 3 = 2⁠, 5+(−1)=45 + (-1) = 4⁠, 3+3=63 + 3 = 6⁠, 5+(−1)=45 + (-1) = 4⁠, 6+0=66 + 0 = 6⁠, 0+2=20 + 2 = 2⁠. That is 2A2A exactly.

Common mistakes
  1. Doubling BB too. The 2 belongs to AA only: 2A−B2A - B⁠, not 2(A−B)2(A - B)⁠.
  2. Sign errors with negative entries. b12=−1b_{12} = -1⁠, so 2a12−b12=4−(−1)=52a_{12} - b_{12} = 4 - (-1) = 5⁠, not 3.
  3. Scaling only some entries. 2A2A doubles all six entries.
What to take away
  1. Scalar multiplication: multiply every entry; the order stays the same.
  2. Subtraction is addition of the negative: A−B=A+(−1)BA - B = A + (-1)B⁠, so (A−B)ij=aij−bij(A - B)_{ij} = a_{ij} - b_{ij}⁠.
  3. Check a difference by adding back what was subtracted.

Every fact used, and where it comes from

FactUsed inSource
kA=[k aij]kA = [k\,a_{ij}]⁠, same order as AASteps 1, 2Section 3.4.2 (Lesson: addition and scalar multiplication)
−B=(−1)B-B = (-1)B⁠; A−B=A+(−1)BA - B = A + (-1)BSteps 3, 4Section 3.4.2, "Negative of a matrix" and "Difference of matrices"
Sum defined entry by entry for equal ordersSteps 1, 4Section 3.4.1
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