Chapter 3 · Matrices

Exercise 3.2 · Q11: Find $x$ and $y$ from a column-matrix equation

Exercise 3.2Equality of matrices, addition and scalar multiples · Sections 3.3.1, 3.4

The question

x\begin{bmatrix} 2 \\ 3 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix},

find the values of $x$ and $y$ ⁠.

Exam answer

\begin{aligned} &\begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\ \Rightarrow &\begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \end{aligned}

Equating corresponding entries,

\begin{aligned} 2x - y &= 10 \quad\ldots(1) \\ 3x + y &= 5 \quad\ldots(2) \end{aligned}

Adding (1) and (2): $5x = 15$ ⁠, so $x = 3$ ⁠.

From (2): $y = 5 - 3x = 5 - 9 = -4$ ⁠.

$\therefore\ x = 3$ ⁠, $y = -4$ ⁠.

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

The one idea this problem is built on

Equating the two entries of the column matrices turns the single matrix equation into a pair of simultaneous linear equations, one from each row. Unlike Questions 9 and 10, each equation now contains both unknowns, so we solve the pair together, by elimination.

Given, to find, and the plan

Given: Three column matrices, each of order $2\times1$ ⁠, and the equation above with unknown scalars $x$ and $y$ ⁠.
To find: The numbers $x$ and $y$ ⁠.
Plan: (1) Check the orders. (2) Form the scalar multiples and add them. (3) Equate the two entries to get two equations. (4) Eliminate $y$ by adding the equations, then find $y$ ⁠. (5) Check in both equations.

Basics you need

1Column matrices

A matrix with one column; here each has 2 rows, so the order is $2\times1$ ⁠. Basics: rows, columns and the address a_ij

2Scalar multiples and sums

$x$ times a column multiplies both its entries by $x$ ⁠; columns of the same order are added entry by entry. Basics: adding only like with like

3Equal matrices

Same order and equal corresponding entries. Lesson: equality of matrices

4Simultaneous equations by elimination

If one unknown has opposite coefficients in two equations (like $-y$ and $+y$ ⁠), adding the equations removes it, leaving one equation in one unknown. Basics: solving equations, one unknown at a time

Solution

Check the orders

All three matrices are column matrices of order $2\times1$ ⁠. Multiplying by the numbers $x$ and $y$ keeps the order $2\times1$ ⁠, so the sum on the left is defined, is $2\times1$ ⁠, and can be compared with the $2\times1$ right side.

Here $x$ and $y$ are scalars (numbers) multiplying whole matrices, not entries. Recognising that tells us which operation is happening: scalar multiplication.
Form the scalar multiples and add
$x\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 2x \\ 3x \end{bmatrix}, \qquad y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} -y \\ y \end{bmatrix}$
Adding entry by entry,
$\begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 2x + (-y) \\ 3x + y \end{bmatrix}$ $= \begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix}.$
The scalar multiplies every entry of its column. Then entries in the same row are added: row 1 collects $2x$ and $-y$ ⁠; row 2 collects $3x$ and $y$ ⁠.

$kA = [k\,a_{ij}]$ and $[a_{ij}] + [b_{ij}] = [a_{ij}+b_{ij}]$ ⁠. Lesson: addition and scalar multiplication
Equate corresponding entries

The equation is now
$\begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix},$
so, row by row,
$\begin{aligned} 2x - y &= 10 \qquad (1)\\ 3x + y &= 5 \qquad\ \ (2) \end{aligned}$
Two $2\times1$ matrices are equal exactly when their first entries agree and their second entries agree. Each equation involves both unknowns, so neither can be solved alone.

Lesson: equality of matrices
Eliminate $y$ and find $x$

The coefficients of $y$ in (1) and (2) are $-1$ and $+1$ ⁠. Adding the left sides and adding the right sides:
$(2x - y) + (3x + y) = 10 + 5$ $5x = 15 \quad\Rightarrow\quad x = \frac{15}{5} = 3.$
If two equations are true, the equation obtained by adding them is true too. Because $-y + y = 0$ ⁠, the sum contains only $x$ ⁠, which we can solve directly.

Elimination: add (or subtract) equations so that one unknown cancels. Basics: solving equations
Find $y$

Put $x = 3$ into (2):
$\begin{aligned} 3\cdot 3 + y &= 5\\ 9 + y &= 5\\ y &= 5 - 9 = -4. \end{aligned}$
Once one unknown is known, either original equation gives the other. We used (2) and keep (1) for the check, so the check is independent.
Check, and why the answer is the only one

In (1): $2\cdot3 - (-4) = 6 + 4 = 10$ ✓. In (2): $3\cdot3 + (-4) = 9 - 4 = 5$ ✓.

The solution is unique: any pair satisfying (1) and (2) also satisfies their sum $5x = 15$ ⁠, which forces $x = 3$ ⁠, and then (2) forces $y = -4$ ⁠.

Elimination gives a necessary condition (any solution must have $x = 3$ ⁠, $y = -4$ ⁠); the check shows it is also sufficient (that pair really works). Together: exactly one solution.

Read each column, top entry

a

and bottom entry

b

⁠, as the point

(a, b)

⁠: 3 copies of

(2,3)

(blue) followed by

-4

copies of

(-1,1)

(gold) land exactly on

(10, 5)

(green). The dashed path adds them in the other order and lands in the same place.

Result

$x = 3$ and $y = -4$ . (Textbook answer: $x = 3,\ y = -4$ ⁠.)

Check it with numbers

Substitute into the original matrix equation: $3\cdot2 + (-4)\cdot(-1) = 6 + 4 = 10$ for the top entry ✓, and $3\cdot3 + (-4)\cdot1 = 9 - 4 = 5$ for the bottom entry ✓.

Common mistakes

Sign of $y$ in row 1. $y$ times $-1$ is $-y$ ⁠, so row 1 is $2x - y = 10$ ⁠, not $2x + y = 10$ ⁠.
Dropping the minus in the answer. $9 + y = 5$ gives $y = -4$ ⁠; it is not 4.
Checking only in the equation used to find $y$ ⁠. That equation is satisfied automatically; check in the other one too.

What to take away

A column-matrix equation is a compact way of writing simultaneous equations: one equation per row.
When each entry equation has several unknowns, solve them together (elimination or substitution).
Geometrically, the question asks how many copies of each column are needed to reach the right side.

Every fact used, and where it comes from

Fact	Used in	Source
Column matrix of order $m\times1$	Step 1	Section 3.3 (i)
$kA = [k\,a_{ij}]$	Step 2	Section 3.4.2
Sum of matrices of the same order, entry by entry	Step 2	Section 3.4.1
Equal matrices: equal corresponding entries	Step 3	Definition 2, Section 3.3.1 (Lesson: equality)
Adding two true equations gives a true equation	Step 4	Algebra of equations
Matrix addition is commutative (figure: both paths meet)	Figure	Section 3.4.3 (i)

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Exercise 3.2 · Q11: Find xxx and yyy from a column-matrix equation

Basics you need

Solution

Check the orders

Form the scalar multiples and add

Equate corresponding entries

Eliminate yyy and find xxx

Find yyy

Check, and why the answer is the only one

Every fact used, and where it comes from

Exercise 3.2 · Q11: Find $x$ and $y$ from a column-matrix equation

Eliminate $y$ and find $x$

Find $y$