Chapter 3 · Matrices

Example 25: Find $D$ with $CD-AB=O$

ExamplesMiscellaneous examples · Chapter 3

The question

Let

A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix},\quad B = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix},

C = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}.

Find a matrix $D$ such that $CD - AB = O$ ⁠.

Exam answer

$A$ ⁠, $B$ ⁠, $C$ are $2\times2$ ⁠, so $AB$ is $2\times2$ and $D$ must be $2\times2$ ⁠. Let

D = \begin{bmatrix} a & b \\ c & d \end{bmatrix}

\begin{aligned} &AB \\ &= \begin{bmatrix} 2{\cdot}5 + (-1){\cdot}7 & 2{\cdot}2 + (-1){\cdot}4 \\ 3{\cdot}5 + 4{\cdot}7 & 3{\cdot}2 + 4{\cdot}4 \end{bmatrix} \\ &= \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} \end{aligned}

CD = \begin{bmatrix} 2a + 5c & 2b + 5d \\ 3a + 8c & 3b + 8d \end{bmatrix}

$CD - AB = O \Rightarrow CD = AB$ ⁠. Equating the corresponding entries,

\begin{aligned} 2a + 5c &= 3 \quad\ldots(\text{i}) \\ 3a + 8c &= 43 \quad\ldots(\text{ii}) \\ 2b + 5d &= 0 \quad\ldots(\text{iii}) \\ 3b + 8d &= 22 \quad\ldots(\text{iv}) \end{aligned}

$2\times$ (ii) $- 3\times$ (i) (the $a$ terms cancel): $c = 2(43) - 3(3) = 77$ ⁠. Then from (i), $2a = 3 - 385$ ⁠, so $a = -191$ ⁠.

$2\times$ (iv) $- 3\times$ (iii) (the $b$ terms cancel): $d = 2(22) - 3(0) = 44$ ⁠. Then from (iii), $2b = -220$ ⁠, so $b = -110$ ⁠.

$\therefore\ D = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}$

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

The one idea this problem is built on

An unknown matrix is just a grid of unknown numbers. Once we know its order, we write $D$ with letters for its entries, compute $CD$ entry by entry, and use equality of matrices: $CD - AB = O$ means every entry of $CD - AB$ is 0. That gives four ordinary linear equations, which split into two independent pairs, one pair for each column of $D$ ⁠.

Given, to find, and the plan

Given: $A$ ⁠, $B$ ⁠, $C$ ⁠, each of order $2\times2$ ⁠, and the condition $CD - AB = O$ ⁠, where $O$ is a zero matrix.
To find: The matrix $D$ ⁠.
Plan: (1) Find the order of $D$ ⁠. (2) Compute $AB$ ⁠. (3) Write $D$ with unknown entries and compute $CD$ ⁠. (4) Set each entry of $CD - AB$ equal to 0. (5) Solve the two pairs of equations. (6) Check.

Basics you need

1Orders in a product

If $C$ is $m\times n$ and $D$ is $n\times p$ ⁠, then $CD$ is $m\times p$ ⁠. A product is defined only when the inner numbers match. Basics: pair up, multiply, add

2Equality of matrices

Two matrices are equal when they have the same order and every pair of corresponding entries is equal. So a matrix equation of order $2\times2$ is really four number equations. Basics: rows, columns and the address

3Solving two linear equations in two unknowns

Eliminate one unknown by making its coefficients equal and subtracting, then substitute back. Always check the solution in both original equations. Basics: solving equations, one unknown at a time

4Subtracting matrices

Matrices of the same order are subtracted entry by entry. Basics: adding only like with like

Solution

Work out the order of $D$

$A$ and $B$ are $2\times2$ ⁠, so $AB$ is defined and is $2\times2$ ⁠. For $CD - AB$ to be defined, $CD$ must have the same order as $AB$ ⁠, namely $2\times2$ ⁠.
- $C$ is $2\times2$ ⁠, so for $CD$ to be defined, $D$ must have 2 rows (the columns of $C$ ⁠).
- $CD$ has as many columns as $D$ ⁠; since $CD$ must be $2\times2$ ⁠, $D$ must have 2 columns.
So $D$ is a $2\times2$ matrix. Write
$D = \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$
The question does not say what size $D$ is, so we must deduce it before we can write it down. Each condition (the product is defined; the difference is defined) fixes one of the two numbers in the order. The letters $a, b, c, d$ are just names for the four unknown entries.

Section 3.4.5 (when a product is defined and its order); Section 3.4.1 (a sum or difference needs equal orders). Lesson: matrix multiplication
Compute $AB$ ⁠, entry by entry

Row $i$ of $A$ times column $j$ of $B$ ⁠:
$\begin{aligned} (AB)_{11} &= 2\cdot5 + (-1)\cdot7 = 10 - 7 = 3,\\ (AB)_{12} &= 2\cdot2 + (-1)\cdot4 = 4 - 4 = 0,\\ (AB)_{21} &= 3\cdot5 + 4\cdot7 = 15 + 28 = 43,\\ (AB)_{22} &= 3\cdot2 + 4\cdot4 = 6 + 16 = 22. \end{aligned}$ $AB = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix}.$
$AB$ contains no unknowns, so we compute it fully first. That keeps the equations in Step 4 simple.

$(AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j}$ for $2\times2$ matrices. Basics: pair up, multiply, add
Compute $CD$ ⁠, entry by entry

Row $i$ of $C$ times column $j$ of $D$ (column 1 of $D$ is $a, c$ ⁠; column 2 is $b, d$ ⁠):
$\begin{aligned} (CD)_{11} &= 2a + 5c,\\ (CD)_{12} &= 2b + 5d,\\ (CD)_{21} &= 3a + 8c,\\ (CD)_{22} &= 3b + 8d. \end{aligned}$ $CD = \begin{bmatrix} 2a + 5c & 2b + 5d \\ 3a + 8c & 3b + 8d \end{bmatrix}.$
Notice that column 1 of $CD$ involves only $a$ and $c$ (column 1 of $D$ ⁠), and column 2 only $b$ and $d$ ⁠. This is always so: column $j$ of a product $CD$ depends only on column $j$ of $D$ ⁠. It is why the problem splits into two separate pairs of equations.
Turn $CD - AB = O$ into four equations

Subtract entry by entry:
$CD - AB = \begin{bmatrix} 2a + 5c - 3 & 2b + 5d - 0 \\ 3a + 8c - 43 & 3b + 8d - 22 \end{bmatrix}.$
This equals the $2\times2$ zero matrix, so by equality of matrices each entry is 0:
$\begin{aligned} 2a + 5c - 3 &= 0 &&\text{(1)}\\ 3a + 8c - 43 &= 0 &&\text{(2)}\\ 2b + 5d &= 0 &&\text{(3)}\\ 3b + 8d - 22 &= 0 &&\text{(4)} \end{aligned}$
Two matrices are equal exactly when all corresponding entries are equal. The zero matrix has every entry 0, so we get one equation per position. Equations (1) and (2) involve only $a$ ⁠, $c$ ⁠; equations (3) and (4) only $b$ ⁠, $d$ ⁠.

Definition 2, Section 3.3.1 (equality of matrices). Lesson: equality of matrices
Solve (1) and (2) for $a$ and $c$

Rewrite them as $2a + 5c = 3$ and $3a + 8c = 43$ ⁠. To eliminate $a$ ⁠, multiply (1) by 3 and (2) by 2, so both have $6a$ ⁠:
$\begin{aligned} 6a + 15c &= 9,\\ 6a + 16c &= 86. \end{aligned}$
Subtract the first from the second: $16c - 15c = 86 - 9$ ⁠, so $c = 77$ ⁠. Substitute into (1): $2a + 5(77) = 3$ ⁠, so $2a = 3 - 385 = -382$ and $a = -191$ ⁠.

Check in (2): $3(-191) + 8(77) = -573 + 616 = 43$ ⁠. Correct.

Multiplying so that one unknown has the same coefficient, and then subtracting, removes that unknown and leaves one equation in one unknown. Checking in the equation we did not substitute into catches arithmetic slips.

Elimination method for two linear equations (Class X). Basics: solving equations
Solve (3) and (4) for $b$ and $d$

Rewrite them as $2b + 5d = 0$ and $3b + 8d = 22$ ⁠. Multiply (3) by 3 and (4) by 2:
$\begin{aligned} 6b + 15d &= 0,\\ 6b + 16d &= 44. \end{aligned}$
Subtract: $d = 44$ ⁠. Substitute into (3): $2b + 5(44) = 0$ ⁠, so $2b = -220$ and $b = -110$ ⁠.

Check in (4): $3(-110) + 8(44) = -330 + 352 = 22$ ⁠. Correct.

The coefficients $2, 5, 3, 8$ are the same as in Step 5, because both pairs come from multiplying by the same matrix $C$ ⁠. Only the right-hand sides differ.
Assemble $D$ ⁠, check, and note that it is the only answer
$D = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}.$
Check by computing $CD$ directly:
$\begin{aligned} (CD)_{11} &= 2(-191) + 5(77)\\ &= -382 + 385 = 3,\\ (CD)_{12} &= 2(-110) + 5(44)\\ &= -220 + 220 = 0,\\ (CD)_{21} &= 3(-191) + 8(77)\\ &= -573 + 616 = 43,\\ (CD)_{22} &= 3(-110) + 8(44)\\ &= -330 + 352 = 22. \end{aligned}$
So every entry of $CD$ equals the corresponding entry of $AB$ ⁠, and $CD - AB = O$ ⁠:
$CD = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} = AB.$
Each pair of equations has exactly one solution, because the elimination in Steps 5 and 6 left $c$ (or $d$ ⁠) with coefficient $16 - 15 = 1 \ne 0$ ⁠. So this $D$ is the only matrix with $CD - AB = O$ ⁠.

Substituting the answer back into the original condition is the surest check: it does not depend on how the equations were solved.

Result

D = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}.

This is the unique $2\times2$ matrix with $CD - AB = O$ ⁠. (The textbook solution gives the same $D$ ⁠.)

Check it with numbers

A different route to the same answer. The matrix $E$ with rows $8, -5$ and $-3, 2$ satisfies $EC = I$ ⁠: the entries of $EC$ are $8\cdot2 + (-5)\cdot3 = 1$ ⁠, $8\cdot5 + (-5)\cdot8 = 0$ ⁠, $(-3)\cdot2 + 2\cdot3 = 0$ ⁠, $(-3)\cdot5 + 2\cdot8 = 1$ ⁠. Multiplying $CD = AB$ on the left by $E$ gives $D = E(AB)$ ⁠, whose entries are $8\cdot3 - 5\cdot43 = -191$ ⁠, $8\cdot0 - 5\cdot22 = -110$ ⁠, $-3\cdot3 + 2\cdot43 = 77$ ⁠, $-3\cdot0 + 2\cdot22 = 44$ ⁠. The same $D$ ⁠.

Common mistakes

Writing $D$ without finding its order. The order is not given; it has to be deduced from the requirement that $CD$ and $CD - AB$ are defined.
Multiplying in the wrong order. $CD$ is not $DC$ ⁠. In $CD$ the rows of $C$ meet the columns of $D$ ⁠, giving $2a + 5c$ ⁠, not $2a + 3b$ ⁠.
Sign slips in $AB$ ⁠. $(-1)\cdot 7 = -7$ ⁠, so $(AB)_{11} = 10 - 7 = 3$ ⁠, and $(-1)\cdot4 = -4$ ⁠, so $(AB)_{12} = 0$ ⁠.
Not checking. Large values like $-191$ look suspicious; substituting back shows they are right.

What to take away

To find an unknown matrix: fix its order, name its entries, multiply out, and compare entries.
Column $j$ of $CD$ depends only on column $j$ of $D$ ⁠, so the equations split by columns of $D$ ⁠.
Here $D$ is " $AB$ divided by $C$ on the left", that is, $D = C^{-1}(AB)$ ⁠; you will meet the systematic way to find $C^{-1}$ in Chapter 4.

Every fact used, and where it comes from

Fact	Used in	Source
Product defined when inner orders match; order $m\times p$	Step 1	Section 3.4.5 (Lesson: matrix multiplication)
Difference needs equal orders; entrywise subtraction	Steps 1, 4	Section 3.4.1 (Lesson: addition)
Entry rule for products	Steps 2, 3, 7; Check	Section 3.4.5
Equality of matrices, entry by entry	Step 4	Definition 2, Section 3.3.1 (Lesson: equality)
Elimination for two linear equations	Steps 5, 6	Class X, pair of linear equations
$EC = I$ and $D = E(AB)$	Check	Section 3.7 (inverse), associative law Section 3.4.6 (Lesson: invertible matrices)

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Example 25: Find DDD with CD−AB=OCD-AB=OCD−AB=O

Basics you need

Solution

Work out the order of DDD

Compute ABABAB⁠, entry by entry

Compute CDCDCD⁠, entry by entry

Turn CD−AB=OCD - AB = OCD−AB=O into four equations

Solve (1) and (2) for aaa and ccc

Solve (3) and (4) for bbb and ddd

Assemble DDD⁠, check, and note that it is the only answer

Every fact used, and where it comes from

Example 25: Find $D$ with $CD-AB=O$

Work out the order of $D$

Compute $AB$ ⁠, entry by entry

Compute $CD$ ⁠, entry by entry

Turn $CD - AB = O$ into four equations

Solve (1) and (2) for $a$ and $c$

Solve (3) and (4) for $b$ and $d$

Assemble $D$ ⁠, check, and note that it is the only answer