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Chapter 3 · Matrices

Miscellaneous · Q6: Find xx from a product equal to OO

Miscellaneous ExerciseMatrix multiplication · Section 3.4.5
The question

Find xx⁠, if

[x−5−1][102021203][x41]=O\begin{aligned} &\begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \\ &\qquad \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = O \end{aligned}
Exam answer [x−5−1][102021203]=[x−2−102x−5−3]=[x−2−102x−8]\begin{aligned} &\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \\ &= \begin{bmatrix} x - 2 & -10 & 2x - 5 - 3 \end{bmatrix} \\ &= \begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix} \end{aligned} [x−2−102x−8][x41]=[x(x−2)−40+2x−8]=[x2−48]\begin{aligned} &\begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} \\ &= [x(x - 2) - 40 + 2x - 8] \\ &= [x^2 - 48] \end{aligned}

So the given equation is [x2−48]=[0][x^2 - 48] = [0]⁠.

x2−48=0⇒x2=48x^2 - 48 = 0 \Rightarrow x^2 = 48

∴ x=±43\therefore\ x = \pm4\sqrt3⁠.

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

The one idea this problem is built on

Row ×\times matrix ×\times column is a 1×11\times1 matrix. Here xx appears in both the row and the column, so the single entry is a quadratic in xx⁠: it works out to x2−48x^2 - 48⁠. Setting it equal to 0 gives two solutions, x=43x = 4\sqrt3 and x=−43x = -4\sqrt3⁠, and both must be kept.

Given, to find, and the plan
Given
The row R=[x  −5  −1]R = [x\ \ {-5}\ \ {-1}] (1×31\times3⁠), the 3×33\times3 matrix M above, and the column C with entries x,4,1x, 4, 1 (3×13\times1⁠).
To find
All real xx with RMC=ORMC = O⁠.
Plan
(1) Check the orders; O is 1×11\times1⁠. (2) Compute RMRM entry by entry. (3) Multiply by C and simplify. (4) Solve the quadratic. (5) Check both roots, and recompute by the other bracketing.

Basics you need

1Orders in a product

(1×3)(3×3)(1\times3)(3\times3) gives 1×31\times3⁠; then (1×3)(3×1)(1\times3)(3\times1) gives 1×11\times1⁠. Basics: rows, columns and the address aija_{ij}

2Row times column

Pair entries, multiply each pair, add. With letters in the entries, expand and collect like terms. Basics: pair up, multiply, add

3Associativity

(RM)C=R(MC)(RM)C = R(MC)⁠; the order R, M, C must be kept. Lesson: properties of the operations

4Solving x2=cx^2 = c

For c>0c > 0⁠, x=±cx = \pm\sqrt c⁠. Simplify the surd: 48=16⋅3=43\sqrt{48} = \sqrt{16\cdot3} = 4\sqrt3⁠. Basics: solving equations

Solution

  1. Check the orders

    • RMRM⁠: (1×3)(3×3)(1\times3)(3\times3)⁠, defined, of order 1×31\times3⁠.
    • (RM)C(RM)C⁠: (1×3)(3×1)(1\times3)(3\times1)⁠, defined, of order 1×11\times1⁠.

    So O here is the 1×11\times1 zero matrix [0][0]⁠, and the problem will reduce to one equation in xx⁠.

    Equal matrices must have the same order, so the O on the right takes the order of the left side.
    Section 3.3 (vii): the order of O is clear from the context. Lesson: types of matrices

  2. Compute RMRM⁠, entry by entry

    The columns of M are (1,0,2)(1, 0, 2)⁠, (0,2,0)(0, 2, 0) and (2,1,3)(2, 1, 3)⁠. Multiply the row (x,−5,−1)(x, -5, -1) by each:

    (RM)11=x⋅1+(−5)⋅0+(−1)⋅2=x−2(RM)12=x⋅0+(−5)⋅2+(−1)⋅0=−10(RM)13=x⋅2+(−5)⋅1+(−1)⋅3=2x−8\begin{aligned} (RM)_{11} &= x\cdot1 + (-5)\cdot0 + (-1)\cdot2 \\ &= x - 2 \\ (RM)_{12} &= x\cdot0 + (-5)\cdot2 + (-1)\cdot0 \\ &= -10 \\ (RM)_{13} &= x\cdot2 + (-5)\cdot1 + (-1)\cdot3 \\ &= 2x - 8 \end{aligned} RM=[x−2−102x−8].RM = \begin{bmatrix} x-2 & -10 & 2x-8 \end{bmatrix}.
    Writing the negative entries in brackets, (−5)(-5) and (−1)(-1)⁠, keeps the signs visible. In (RM)13(RM)_{13}⁠, the constant terms are −5-5 and −3-3⁠, which add to −8-8⁠.
    Entry formula of Section 3.4.5. Basics: pair up, multiply, add

  3. Multiply by the column and simplify

    Now the row (x−2,−10,2x−8)(x-2, -10, 2x-8) times the column (x,4,1)(x, 4, 1)⁠:

    (RM)C=(x−2)⋅x+(−10)⋅4+(2x−8)⋅1=x2−2x−40+2x−8=x2−48.\begin{aligned} (RM)C &= (x-2)\cdot x + (-10)\cdot4 \\ &\quad + (2x-8)\cdot1 \\ &= x^2 - 2x - 40 + 2x - 8 \\ &= x^2 - 48. \end{aligned}

    So the left side of the equation is the 1×11\times1 matrix [x2−48][x^2 - 48]⁠.

    The xx-terms −2x-2x and +2x+2x cancel, which is why the quadratic has no middle term and is easy to solve. The constants −40-40 and −8-8 combine to −48-48⁠.
    Expanding brackets: (x−2)x=x2−2x(x-2)x = x^2 - 2x⁠. Basics: solving equations

  4. Solve x2−48=0x^2 - 48 = 0

    Equating the single entries of [x2−48][x^2 - 48] and [0][0]⁠:

    x2−48=0x2=48x=±48=±43.\begin{aligned} x^2 - 48 &= 0 \\ x^2 &= 48 \\ x &= \pm\sqrt{48} = \pm4\sqrt3. \end{aligned}

    A quadratic has at most two roots, and we have found two, so these are all the solutions: x=43x = 4\sqrt3 or x=−43x = -4\sqrt3⁠.

    48=16×348 = 16\times3 and 16=4\sqrt{16} = 4⁠, so 48=43\sqrt{48} = 4\sqrt3⁠. Both signs work because only x2x^2 remains in the equation. The question says "find xx", but there are two values, and both must be given.
    If x2=cx^2 = c with c>0c > 0⁠, then x=cx = \sqrt c or x=−cx = -\sqrt c⁠. Basics: solving equations

  5. Check by the other bracketing

    Each entry of MCMC is a row of M times the column (x,4,1)(x, 4, 1)⁠:

    row 1:1⋅x+0⋅4+2⋅1=x+2row 2:0⋅x+2⋅4+1⋅1=9row 3:2⋅x+0⋅4+3⋅1=2x+3\begin{aligned} \text{row 1} &: 1\cdot x + 0\cdot4 + 2\cdot1 = x + 2 \\ \text{row 2} &: 0\cdot x + 2\cdot4 + 1\cdot1 = 9 \\ \text{row 3} &: 2\cdot x + 0\cdot4 + 3\cdot1 = 2x + 3 \end{aligned}

    Then the row (x,−5,−1)(x, -5, -1) times this column:

    R(MC)=x(x+2)+(−5)⋅9+(−1)(2x+3)=x2+2x−45−2x−3=x2−48. ✓\begin{aligned} R(MC) &= x(x+2) + (-5)\cdot9 \\ &\quad + (-1)(2x+3) \\ &= x^2 + 2x - 45 - 2x - 3 \\ &= x^2 - 48. \ ✓ \end{aligned}
    The intermediate matrices are different (RMRM is a row, MCMC is a column), but associativity guarantees the same final entry. Agreement by both routes makes an arithmetic slip very unlikely.
    Associative law, Section 3.4.6. Lesson: properties of the operations
Result

The product equals [x2−48][x^2 - 48]⁠, which is O exactly when x=43x = 4\sqrt3 or x=−43x = -4\sqrt3, that is x=±43x = \pm4\sqrt3⁠. (Textbook answer: x=±43x = \pm4\sqrt3⁠.)

Check it with numbers

43≈6.928203234\sqrt3 \approx 6.92820323⁠, and 6.928203232≈48.00000006.92820323^2 \approx 48.0000000⁠, so x2−48≈0x^2 - 48 \approx 0 ✓. Exactly: (±43)2=16⋅3=48(\pm4\sqrt3)^2 = 16\cdot3 = 48⁠. For contrast, x=0x = 0 gives −48-48 and x=7x = 7 gives 11⁠, neither 0.

Common mistakes
  1. Dropping the negative root. x2=48x^2 = 48 has two solutions.
  2. Sign slips with −5-5 and −1-1⁠, for example getting 2x−22x - 2 instead of 2x−82x - 8 in RMRM⁠.
  3. Leaving 48\sqrt{48} unsimplified. It is correct but should be written 434\sqrt3⁠.
  4. Using rows of M instead of columns when computing RMRM⁠.
What to take away
  1. When the unknown appears in both outer factors, the 1×11\times1 result is a quadratic; expect up to two solutions.
  2. Computing (RM)C(RM)C and R(MC)R(MC) is a reliable self-check.
  3. This is the same pattern as Miscellaneous Q4, where xx appeared only once and the equation was linear.

Every fact used, and where it comes from

FactUsed inSource
Order and entries of a productSteps 1–3, 5Section 3.4.5 (Lesson: matrix multiplication)
Zero matrix, order from context; equality of matricesSteps 1, 4Section 3.3 (vii); Definition 2, Section 3.3.1
Associative lawStep 5Section 3.4.6 (Lesson: properties of the operations)
48=43\sqrt{48} = 4\sqrt3⁠; a quadratic has at most two rootsStep 4Algebra (surds, quadratic equations)
All products, the roots and decimalsSteps 2–5, CheckRecomputed with sympy in the site's verification script
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