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Chapter 3 · Matrices

Example 4: Equal 3×33\times3 matrices: find a,b,c,x,y,za, b, c, x, y, z

ExamplesEquality of matrices · Section 3.3.1
The question

If

[x+3z+42y−7−6a−10b−3−210]\begin{bmatrix} x+3 & z+4 & 2y-7 \\ -6 & a-1 & 0 \\ b-3 & -21 & 0 \end{bmatrix} =[063y−2−6−32c+22b+4−210]= \begin{bmatrix} 0 & 6 & 3y-2 \\ -6 & -3 & 2c+2 \\ 2b+4 & -21 & 0 \end{bmatrix}

Find the values of aa⁠, bb⁠, cc⁠, xx⁠, yy and zz⁠.

Exam answer

The matrices are equal, so their corresponding entries are equal:

x+3=0⇒x=−3z+4=6⇒z=22y−7=3y−2⇒y=−5a−1=−3⇒a=−20=2c+2⇒c=−1b−3=2b+4⇒b=−7\begin{aligned} x + 3 = 0 &\Rightarrow x = -3 \\ z + 4 = 6 &\Rightarrow z = 2 \\ 2y - 7 = 3y - 2 &\Rightarrow y = -5 \\ a - 1 = -3 &\Rightarrow a = -2 \\ 0 = 2c + 2 &\Rightarrow c = -1 \\ b - 3 = 2b + 4 &\Rightarrow b = -7 \end{aligned}

∴ a=−2, b=−7, c=−1\therefore\ a = -2,\ b = -7,\ c = -1⁠, x=−3, y=−5, z=2x = -3,\ y = -5,\ z = 2⁠.

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

The one idea this problem is built on

Two matrices are equal only when they have the same order and every entry of one equals the entry in the same place in the other. So one matrix equation between two 3×33\times3 matrices is really nine ordinary equations, one per position. Each unknown here appears in just one position, so each equation can be solved on its own.

Given, to find, and the plan
Given
The equality of two 3×33\times3 matrices whose entries contain the unknowns.
To find
a,b,c,x,y,za, b, c, x, y, z⁠.
Plan
Check the orders agree. Write the nine equations "entry (i,j)(i,j) on the left = entry (i,j)(i,j) on the right". Solve each one that contains an unknown; check that the others are true. Finally substitute back.

Basics you need

1Equality of matrices

A=[aij]A = [a_{ij}] and B=[bij]B = [b_{ij}] are equal when (i) they have the same order and (ii) aij=bija_{ij} = b_{ij} for every ii and jj⁠. Lesson: equality of matrices

2Solving a linear equation in one unknown

Do the same thing to both sides: add or subtract to collect the unknown on one side and numbers on the other, then divide by the coefficient. Basics: solving equations, one unknown at a time

3The address aija_{ij}

Entry (i,j)(i, j) is in row ii⁠, column jj⁠. We compare entries with the same address. Basics: rows, columns and the address a_ij

Solution

  1. Check the orders and list the nine entry equations

    Both matrices have 3 rows and 3 columns, so both are 3×33\times3⁠: condition (i) of equality holds. Condition (ii) says the entries in each of the 9 positions agree:

    PositionEquationContains
    (1, 1)x+3=0x + 3 = 0xx
    (1, 2)z+4=6z + 4 = 6zz
    (1, 3)2y−7=3y−22y - 7 = 3y - 2yy
    (2, 1)−6=−6-6 = -6no unknown
    (2, 2)a−1=−3a - 1 = -3aa
    (2, 3)0=2c+20 = 2c + 2cc
    (3, 1)b−3=2b+4b - 3 = 2b + 4bb
    (3, 2)−21=−21-21 = -21no unknown
    (3, 3)0=00 = 0no unknown

    The three equations with no unknown are true, so they put no condition on the unknowns (and they do not make the equality impossible).

    A matrix equation cannot be "solved" as a whole; it has to be broken into its entries. We must compare entries in the same position, because that is what the definition of equality says. The entries without unknowns still need a glance: if, say, position (2, 1) had read −6=6-6 = 6⁠, no values of the unknowns could make the matrices equal.
    Definition 2, Section 3.3.1: same order and aij=bija_{ij} = b_{ij} for all i,ji, j⁠. Lesson: equality of matrices

  2. Solve for xx and zz (row 1)

    x+3=0 ⇒ x=0−3=−3,z+4=6 ⇒ z=6−4=2.\begin{aligned} x + 3 = 0 &\ \Rightarrow\ x = 0 - 3 = -3,\\ z + 4 = 6 &\ \Rightarrow\ z = 6 - 4 = 2. \end{aligned}
    Each equation has the unknown plus a number. Subtracting that number from both sides leaves the unknown alone, and doing the same thing to both sides keeps the equation true.
    Subtract the same number from both sides. Basics: solving equations

  3. Solve for yy (position (1, 3))

    2y−7=3y−2.2y - 7 = 3y - 2.

    Subtract 2y2y from both sides, then add 2 to both sides:

    −7=y−2,−7+2=y,y=−5.\begin{aligned} -7 &= y - 2,\\ -7 + 2 &= y,\\ y &= -5. \end{aligned}
    Here yy is on both sides, so we first gather the yy terms on one side. Subtracting 2y2y (the smaller multiple) leaves a plain yy on the right, with no fraction to handle later.
    Collect the unknown on one side, numbers on the other. Basics: solving equations

  4. Solve for aa and cc (row 2)

    a−1=−3 ⇒ a=−3+1=−2,0=2c+2 ⇒ 2c=−2 ⇒ c=−1.\begin{aligned} a - 1 = -3 &\ \Rightarrow\ a = -3 + 1 = -2,\\ 0 = 2c + 2 &\ \Rightarrow\ 2c = -2\ \Rightarrow\ c = -1. \end{aligned}
    For aa we add 1 to both sides. For cc we subtract 2 from both sides, giving 2c=−22c = -2⁠, and then divide both sides by 2 (allowed because 2≠02\ne0⁠).

  5. Solve for bb (position (3, 1))

    b−3=2b+4.b - 3 = 2b + 4.

    Subtract bb from both sides, then subtract 4:

    −3=b+4,−3−4=b,b=−7.\begin{aligned} -3 &= b + 4,\\ -3 - 4 &= b,\\ b &= -7. \end{aligned}
    As with yy⁠, the unknown appears on both sides, so we move all the bb terms to one side first. Subtracting bb leaves a single bb on the right.

  6. Substitute back and check

    Put a=−2a = -2⁠, b=−7b = -7⁠, c=−1c = -1⁠, x=−3x = -3⁠, y=−5y = -5⁠, z=2z = 2 into each matrix. Left side, entry by entry:

    x+3=0,z+4=6,2y−7=−17,a−1=−3,b−3=−10.\begin{aligned} x + 3 &= 0, & z + 4 &= 6, & 2y - 7 &= -17,\\ a - 1 &= -3, & b - 3 &= -10. \end{aligned}

    Right side: 3y−2=−173y - 2 = -17⁠, 2c+2=02c + 2 = 0⁠, 2b+4=−102b + 4 = -10⁠. Both matrices become

    [06−17−6−30−10−210].\begin{bmatrix} 0 & 6 & -17 \\ -6 & -3 & 0 \\ -10 & -21 & 0 \end{bmatrix}.
    Substituting back is the honest test: if any of the nine positions disagreed, we would know a sign slipped. Each unknown appeared in only one equation, and each of those equations is linear with a nonzero coefficient of its unknown, so it has exactly one solution. These values are therefore the only ones that work.
Result

a=−2, b=−7, c=−1a = -2,\ b = -7,\ c = -1 and x=−3, y=−5, z=2x = -3,\ y = -5,\ z = 2. This agrees with the textbook's solution.

Check it with numbers

The hardest two equations, rechecked: 2y−7=2(−5)−7=−172y - 7 = 2(-5) - 7 = -17 and 3y−2=3(−5)−2=−173y - 2 = 3(-5) - 2 = -17⁠, equal. b−3=−7−3=−10b - 3 = -7 - 3 = -10 and 2b+4=2(−7)+4=−102b + 4 = 2(-7) + 4 = -10⁠, equal.

Common mistakes
  1. Comparing entries in different positions. Only entries with the same address are equal; x+3x + 3 is matched with 0, not with anything else.
  2. Sign slips when the unknown is on both sides. In b−3=2b+4b - 3 = 2b + 4⁠, moving terms carelessly can give b=7b = 7⁠. Substitute back to catch it.
  3. Ignoring the constant entries. Positions like (2,1)(2, 1) must also agree; here they do, but you should say so.
What to take away
  1. An equation between two m×nm\times n matrices is mnmn ordinary equations, one per position.
  2. Check the orders first, then match entry (i,j)(i, j) with entry (i,j)(i, j)⁠.
  3. Always substitute the answers back into both matrices.

Every fact used, and where it comes from

FactUsed inSource
Equal matrices: same order and aij=bija_{ij} = b_{ij} for all i,ji, jStep 1Definition 2, Section 3.3.1 (Lesson: equality of matrices)
Adding or subtracting the same number on both sides; dividing by a nonzero numberSteps 2–5Rules for equations (Basics: solving equations)
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