Chapter 3 · Matrices

Exercise 3.2 · Q22 (MCQ): Order of $7X-5Z$

Exercise 3.2Orders of sums and scalar multiples · Section 3.4

The question

Assume X, Y, Z, W and P are matrices of order $2\times n$ ⁠, $3\times k$ ⁠, $2\times p$ ⁠, $n\times3$ and $p\times k$ ⁠, respectively.

If $n = p$ ⁠, then the order of the matrix $7X - 5Z$ is:

(A) $p\times2$ (B) $2\times n$ (C) $n\times3$ (D) $p\times n$

Exam answer

$7X$ is of order $2\times n$ ⁠, and $5Z$ is of order $2\times p$ ⁠, that is $2\times n$ (as $p = n$ ⁠).

So $7X - 5Z$ is defined and is of order $2\times n$ ⁠.

Answer: (B) $2\times n$

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

The one idea this problem is built on

Multiplying a matrix by a number does not change its order, and a difference of two matrices is defined only when they have the same order, which is then the order of the answer. So $7X - 5Z$ has the common order of $X$ and $Z$ ⁠, which is $2\times n$ once $n = p$ ⁠.

Given, to find, and the plan

Given: $X$ is $2\times n$ ⁠, $Z$ is $2\times p$ ⁠, and $n = p$ ⁠. ( $Y$ ⁠, $W$ ⁠, $P$ are not involved.)
To find: The order of $7X - 5Z$ ⁠.
Plan: (1) Orders of $7X$ and $5Z$ ⁠. (2) Check that the difference is defined and read off its order. (3) Examine every option.

Basics you need

1Scalar multiples keep the order

$kA$ multiplies each entry of $A$ by $k$ ⁠; it has exactly as many rows and columns as $A$ ⁠. Lesson: scalar multiplication

2Differences need equal orders

$A - B = A + (-1)B$ is defined only when $A$ and $B$ have the same order, and then has that order. Basics: adding only like with like

3Order "rows × columns"

$2\times n$ means 2 rows and $n$ columns; $n\times2$ is a different shape. Basics: rows, columns and addresses

Solution

Orders of $7X$ and $5Z$

$X$ is $2\times n$ ⁠, so $7X$ is also $2\times n$ ⁠. $Z$ is $2\times p$ ⁠, so $5Z$ is also $2\times p$ ⁠. With $n = p$ ⁠, $5Z$ is $2\times n$ ⁠.

$7X$ is formed entry by entry: $7X = [7x_{ij}]$ ⁠. No rows or columns are added or removed, so the shape is unchanged.

$kA = [k\,a_{ij}]$ has the same order as $A$ ⁠. Lesson: addition and scalar multiplication
The difference is defined, of order $2\times n$

$7X$ and $5Z$ are both $2\times n$ ⁠, so the difference $7X - 5Z$ is defined, entry by entry:
$7X - 5Z = [\,7x_{ij} - 5z_{ij}\,]$
with $i = 1, 2$ and $j = 1, \dots, n$ ⁠. That is 2 rows and $n$ columns: the order is $2\times n$ .

The condition $n = p$ is exactly what the question supplies to make the difference possible. Without it, $7X$ would be $2\times n$ and $5Z$ would be $2\times p$ with $p \ne n$ ⁠, and the difference would not be defined. Since $n = p$ ⁠, the answer can be written $2\times n$ or $2\times p$ ⁠; only $2\times n$ appears among the options.

Difference of matrices of the same order, Section 3.4.2. Basics: adding only like with like

Examine every option

Option	Claim	Verdict
(A)	$p\times2$	Wrong in general. This swaps rows and columns: the result has 2 rows, not 2 columns. For example with $n = p = 3$ ⁠, the result is $2\times3$ ⁠, not $3\times2$ ⁠. (Only when $p = 2$ does $p\times2$ happen to equal $2\times n$ ⁠.)
(B)	$2\times n$	Correct. As found in Step 2.
(C)	$n\times3$	Wrong. That is the order of $W$ ⁠, which does not appear in $7X - 5Z$ ⁠. With $n = p = 4$ ⁠, the result is $2\times4$ ⁠, not $4\times3$ ⁠.
(D)	$p\times n$	Wrong in general. The result has 2 rows, so $p\times n$ is right only in the special case $p = 2$ ⁠, which is not given. With $n = p = 3$ ⁠, the result is $2\times3$ ⁠, not $3\times3$ ⁠.

Each wrong option is shown wrong by one allowed choice of

n = p

⁠. Options (A) and (D) agree with (B) only in the special case

p = n = 2

(both then read

2\times2

⁠), so neither is the order in general. Option (C) never agrees:

n\times3

equals

2\times n

only if

n = 2

and

n = 3

at once.

Result

The correct answer is (B): when $n = p$ ⁠, $7X - 5Z$ is of order $2\times n$ ⁠. (Textbook answer: B.)

Check it with numbers

Take $n = p = 3$ ⁠. Let $X$ be the $2\times3$ matrix with rows $(1, 2, 3)$ ⁠, $(4, 5, 6)$ and $Z$ the $2\times3$ matrix with rows $(1, 0, 1)$ ⁠, $(0, 1, 0)$ ⁠. Then $7X$ has rows $(7, 14, 21)$ ⁠, $(28, 35, 42)$ ⁠; $5Z$ has rows $(5, 0, 5)$ ⁠, $(0, 5, 0)$ ⁠; and $7X - 5Z$ has rows $(2, 14, 16)$ and $(28, 30, 42)$ ⁠: 2 rows and 3 columns, that is $2\times n$ ✓.

Common mistakes

Thinking the scalars 7 and 5 change the order. They only scale entries.
Swapping rows and columns and choosing (A).
Choosing (D) because both $n$ and $p$ appear in the question; the number of rows is fixed at 2 by $X$ and $Z$ ⁠.

What to take away

Scalar multiples keep the order; sums and differences need equal orders and keep that order.
Read an order as rows first, columns second.
Compare Exercise 3.2 Q21, where products of the same family of matrices are tested.

Every fact used, and where it comes from

Fact	Used in	Source
Order $m\times n$ ⁠: $m$ rows, $n$ columns	Steps 1–3	Section 3.2.1 (Lesson: what a matrix is)
$kA = [k\,a_{ij}]$ has the order of $A$	Step 1	Section 3.4.2
Difference defined only for matrices of the same order: $A - B = A + (-1)B$	Step 2	Section 3.4.2 (difference of matrices) and Section 3.4.1
The numerical example and the orders	Step 3, Check	Recomputed in the site's verification script

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Exercise 3.2 · Q22 (MCQ): Order of 7X−5Z7X-5Z7X−5Z

Basics you need

Solution

Orders of 7X7X7X and 5Z5Z5Z

The difference is defined, of order 2×n2\times n2×n

Examine every option

Every fact used, and where it comes from

Exercise 3.2 · Q22 (MCQ): Order of $7X-5Z$

Orders of $7X$ and $5Z$

The difference is defined, of order $2\times n$