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Chapter 3 · Matrices

Miscellaneous · Q4: Find xx from a row-matrix-column product

Miscellaneous ExerciseMatrix multiplication · Section 3.4.5
The question

For what values of xx⁠:

[121][120201102][02x]=O ?\begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O\,?
Exam answer [121][120201102]=[1+4+122+2]=[624]\begin{aligned} &\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \\ &= \begin{bmatrix} 1 + 4 + 1 & 2 & 2 + 2 \end{bmatrix} \\ &= \begin{bmatrix} 6 & 2 & 4 \end{bmatrix} \end{aligned} [624][02x]=[0+4+4x]=[4+4x]\begin{aligned} &\begin{bmatrix} 6 & 2 & 4 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} \\ &= [0 + 4 + 4x] = [4 + 4x] \end{aligned}

So the given equation is [4+4x]=[0][4 + 4x] = [0]⁠.

4+4x=0⇒x=−14 + 4x = 0 \Rightarrow x = -1

∴ x=−1\therefore\ x = -1⁠.

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

The one idea this problem is built on

A row (1×31\times3⁠) times a square matrix (3×33\times3⁠) times a column (3×13\times1⁠) collapses to a single number, a 1×11\times1 matrix. Multiply two factors at a time, keeping the order, and the left side becomes [4+4x][4 + 4x]⁠. The zero matrix of that order is [0][0]⁠, so the problem reduces to the ordinary equation 4+4x=04 + 4x = 0⁠.

Given, to find, and the plan
Given
The row R=[1  2  1]R = [1\ \ 2\ \ 1] (1×31\times3⁠), the 3×33\times3 matrix M above, and the column C with entries 0,2,x0, 2, x (3×13\times1⁠).
To find
All real xx with RMC=ORMC = O⁠.
Plan
(1) Check the orders and find the order of the result (so we know which O is meant). (2) Compute RMRM⁠, entry by entry. (3) Multiply by C. (4) Solve. (5) Check, including by the other bracketing R(MC)R(MC)⁠.

Basics you need

1Orders in a product

(m×n)(n×p)(m\times n)(n\times p) is defined and has order m×pm\times p⁠. Basics: rows, columns and the address aija_{ij}

2Row times column

Pair the entries of a row with the entries of a column, multiply each pair, and add. Basics: pair up, multiply, add

3Associativity

(RM)C=R(MC)(RM)C = R(MC)⁠, so either pair may be multiplied first; the order of the factors, however, must stay R, M, C. Lesson: properties of the operations

4Solving a linear equation

4+4x=04 + 4x = 0⁠: subtract 4, then divide by 4. Basics: solving equations

Solution

  1. Check the orders, and decide which O is meant

    • RMRM⁠: (1×3)(3×3)(1\times3)(3\times3)⁠; the inner numbers agree (3 and 3), so it is 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 the left side is a 1×11\times1 matrix, and the O on the right must be the 1×11\times1 zero matrix [0][0]⁠.

    The book writes O for "the zero matrix of whatever order fits". Equality of matrices needs equal orders, so here O can only be [0][0]⁠. Knowing this tells us the final answer will be one equation in xx⁠.
    Section 3.3 (vii): "Its order will be clear from the context." Lesson: types of matrices

  2. Compute RMRM⁠, entry by entry

    RMRM has one row and three columns. Entry jj is the row (1,2,1)(1, 2, 1) times column jj of M:

    (RM)11=1⋅1+2⋅2+1⋅1=6(RM)12=1⋅2+2⋅0+1⋅0=2(RM)13=1⋅0+2⋅1+1⋅2=4\begin{aligned} (RM)_{11} &= 1\cdot1 + 2\cdot2 + 1\cdot1 = 6 \\ (RM)_{12} &= 1\cdot2 + 2\cdot0 + 1\cdot0 = 2 \\ (RM)_{13} &= 1\cdot0 + 2\cdot1 + 1\cdot2 = 4 \end{aligned} RM=[624].RM = \begin{bmatrix} 6 & 2 & 4 \end{bmatrix}.
    The columns of M are (1,2,1)(1, 2, 1)⁠, (2,0,0)(2, 0, 0) and (0,1,2)(0, 1, 2)⁠. Reading them as columns (down, not across) is the step where slips happen most. Doing RMRM first is a natural choice because it involves only numbers; the unknown xx enters only in the last step.
    Entry formula of Section 3.4.5: cik=∑jaijbjkc_{ik} = \sum_j a_{ij} b_{jk}⁠. Basics: pair up, multiply, add

  3. Multiply by the column

    (RM)C=6⋅0+2⋅2+4⋅x=0+4+4x\begin{aligned} (RM)C &= 6\cdot0 + 2\cdot2 + 4\cdot x \\ &= 0 + 4 + 4x \end{aligned}

    so the left side of the equation is the 1×11\times1 matrix [4+4x][4 + 4x]⁠.

    A row of length 3 times a column of length 3 is a single row-times-column sum, so the result has exactly one entry.
    Row times column. Basics: pair up, multiply, add

  4. Solve [4+4x]=[0][4 + 4x] = [0]

    Two 1×11\times1 matrices are equal exactly when their single entries are equal:

    4+4x=04x=−4x=−1.\begin{aligned} 4 + 4x &= 0 \\ 4x &= -4 \\ x &= -1. \end{aligned}

    The equation is linear with a non-zero coefficient of xx⁠, so this is the only solution: x=−1x = -1⁠.

    The question says "values", but a linear equation ax+b=0ax + b = 0 with a≠0a \ne 0 has exactly one solution. Subtracting 4 and dividing by 4 are both reversible, so no solution is lost or gained.
    Definition 2 (Section 3.3.1), equality of matrices. Basics: solving equations

  5. Check by the other bracketing

    Associativity says R(MC)R(MC) must give the same answer. Each entry of MCMC is a row of M times the column (0,2,x)(0, 2, x)⁠:

    row 1:1⋅0+2⋅2+0⋅x=4row 2:2⋅0+0⋅2+1⋅x=xrow 3:1⋅0+0⋅2+2⋅x=2x\begin{aligned} \text{row 1} &: 1\cdot0 + 2\cdot2 + 0\cdot x = 4 \\ \text{row 2} &: 2\cdot0 + 0\cdot2 + 1\cdot x = x \\ \text{row 3} &: 1\cdot0 + 0\cdot2 + 2\cdot x = 2x \end{aligned}

    so MCMC is the column with entries 4,x,2x4, x, 2x⁠, and

    R(MC)=1⋅4+2⋅x+1⋅2x=4+4x. ✓\begin{aligned} R(MC) &= 1\cdot4 + 2\cdot x + 1\cdot2x \\ &= 4 + 4x. \ ✓ \end{aligned}
    Getting the same expression by a different route is strong evidence that no entry was miscomputed. It also illustrates the associative law: we may choose which pair to multiply first, but we may never swap the order of R, M and C.
    Associative law, Section 3.4.6. Lesson: properties of the operations
Result

The product equals [4+4x][4 + 4x]⁠, which is the 1×11\times1 zero matrix exactly when x=−1x = -1. (Textbook answer: x=−1x = -1⁠.)

Check it with numbers

Put x=−1x = -1⁠: the column becomes (0,2,−1)(0, 2, -1)⁠, and

(RM)C=6⋅0+2⋅2+4⋅(−1)=0+4−4=0. ✓\begin{aligned} (RM)C &= 6\cdot0 + 2\cdot2 + 4\cdot(-1) \\ &= 0 + 4 - 4 = 0. \ ✓ \end{aligned}

For contrast, x=0x = 0 gives 4≠04 \ne 0 and x=1x = 1 gives 8≠08 \ne 0⁠.

Common mistakes
  1. Multiplying the row by the rows of M instead of its columns, which gives (1+4+0,… )(1+4+0, \dots) and a wrong RMRM⁠.
  2. Changing the order, for example computing MRMR or CRCR⁠. MRMR is (3×3)(1×3)(3\times3)(1\times3)⁠, which is not even defined.
  3. Treating O as the number 0 without comment. It is the 1×11\times1 zero matrix; equating entries is what turns it into the equation 4+4x=04 + 4x = 0⁠.
What to take away
  1. Row ×\times matrix ×\times column is always a 1×11\times1 matrix: a single number, built from the entries of all three factors.
  2. Multiply two factors at a time, keeping their order. Doing it both ways (RM)C(RM)C and R(MC)R(MC) is a free check.
  3. Always find the order of the result first; it tells you what O means and how many equations you will get.

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 O, order from contextStep 1Section 3.3 (vii) (Lesson: types of matrices)
Equality of matricesStep 4Definition 2, Section 3.3.1 (Lesson: equality of matrices)
Associative law (AB)C=A(BC)(AB)C = A(BC)Step 5Section 3.4.6 (Lesson: properties of the operations)
All products and the solutionSteps 2–5, CheckRecomputed with sympy in the site's verification script
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