Chapter 4 · Determinants

Example 18: Three numbers from three conditions

ExamplesSolving systems AX = B · Section 4.6

The question

The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.

Exam answer

Let the numbers be $x$ ⁠, $y$ ⁠, $z$ ⁠. Then $x + y + z = 6$ ⁠, $y + 3z = 11$ and $x + z = 2y$ ⁠, that is,

\begin{aligned} x + y + z &= 6 \\ y + 3z &= 11 \\ x - 2y + z &= 0 \end{aligned}

This is $AX = B$ ⁠, where

\begin{aligned} A &= \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}, \\ X &= \begin{bmatrix} x \\ y \\ z \end{bmatrix},\ B = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix} \end{aligned}

Expanding along $C_1$ ⁠:

\begin{aligned} |A| &= 1\begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} - 0\begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} \\ &\quad + 1\begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} \\ &= 1(1 + 6) - 0 + 1(3 - 1) \\ &= 9 \end{aligned}

$|A| = 9 \ne 0$ ⁠. So $A$ is non-singular and the system has a unique solution, $X = A^{-1}B$ ⁠.

Cofactors:

\begin{aligned} A_{11} &= \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} = 7 \\ A_{12} &= -\begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = 3 \\ A_{13} &= \begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix} = -1 \end{aligned}

\begin{aligned} A_{21} &= -\begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} = -3 \\ A_{22} &= \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 0 \\ A_{23} &= -\begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = 3 \end{aligned}

\begin{aligned} A_{31} &= \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} = 2 \\ A_{32} &= -\begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = -3 \\ A_{33} &= \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1 \end{aligned}

\begin{aligned} \operatorname{adj}A &= \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix} \\ A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= \frac19\begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix} \end{aligned}

\begin{aligned} X &= A^{-1}B \\ &= \frac19\begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}\begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix} \\ &= \frac19\begin{bmatrix} 42 - 33 + 0 \\ 18 + 0 - 0 \\ -6 + 33 + 0 \end{bmatrix} \\ &= \frac19\begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \end{aligned}

$\therefore\ x = 1,\ y = 2,\ z = 3$ ⁠. The numbers are 1, 2 and 3.

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

The one idea this problem is built on

Each sentence is one linear equation in the three unknown numbers. Once the words are turned into equations, the problem is a $3\times3$ system $AX = B$ ⁠, solved by $X = A^{-1}B$ when $|A| \ne 0$ ⁠. The only new skill is the translation, especially the third sentence, which has to be rearranged so that all unknowns are on the left.

Given, to find, and the plan

Given: Three facts about three numbers: their sum; "3 × third + second"; and "first + third = 2 × second".
To find: The three numbers, by the matrix method.
Plan: Name the numbers; write one equation per sentence; form $AX = B$ ⁠; compute $|A|$ ⁠, the cofactors, $\operatorname{adj}A$ and $A^{-1}$ ⁠; check $A(\operatorname{adj}A) = |A|\,I$ ⁠; compute $X = A^{-1}B$ ⁠; check the numbers against the original sentences.

Basics you need

1Turning words into equations

"Sum" means add; "multiply by 3" means $3\times$ ⁠; "double of" means $2\times$ ⁠; "we get" means "equals". Move every unknown to the left side before writing the matrix. Basics: algebra the chapter leans on

2A missing unknown has coefficient 0

In $y + 3z = 11$ there is no $x$ ⁠, so its coefficient is 0. Every row of $A$ must have one entry for each unknown. Basics: rows, columns and positions

3Cofactors, adjoint, inverse

$A_{ij} = (-1)^{i+j}M_{ij}$ ⁠; $\operatorname{adj}A$ is the transpose of $[A_{ij}]$ ⁠; $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ if $|A| \ne 0$ ⁠. Basics: why the inverse divides by |A|

Solution

Translate the words into equations

Let the first, second and third numbers be $x$ ⁠, $y$ and $z$ ⁠. Sentence by sentence:

Words Equation

The sum of the three numbers is 6 $x + y + z = 6$

Third times 3, plus the second, is 11 $y + 3z = 11$

First plus third is double the second $x + z = 2y$

Move $2y$ to the left in the third equation: $x - 2y + z = 0$ ⁠. The system is
$\begin{aligned} x + y + z &= 6\\ y + 3z &= 11\\ x - 2y + z &= 0 \end{aligned}$
The matrix method needs every equation in the form "(coefficients times unknowns) = number". " $x + z = 2y$ " has an unknown on the right, so subtracting $2y$ from both sides is necessary before reading off coefficients. This is the "represent it algebraically" part of the question.

Subtracting the same quantity from both sides keeps an equation true. Basics: algebra the chapter leans on
Write the system as $AX = B$
$A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}.$
$A$ is $3\times3$ ⁠, $X$ and $B$ are $3\times1$ ⁠. Row by row, $AX$ has entries $x + y + z$ ⁠, $0x + y + 3z$ and $x - 2y + z$ ⁠, which matches the system.

Row 2 starts with 0 because the second equation has no $x$ ⁠. Row 3 has $-2$ in the middle because of the rearrangement in Step 1. These two entries are where set-up errors usually happen.

Section 4.6.1. Lesson: solving systems AX = B
Compute $|A|$ ⁠, expanding along column 1

Column 1 contains a zero ( $a_{21} = 0$ ⁠), so expanding along it needs only two minors. The signs down column 1 are $+,-,+$ ⁠:
$\begin{aligned} |A| &= 1\begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} - 0\begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix}\\ &\quad + 1\begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix}\\ &= 1(1 + 6) - 0 + 1(3 - 1) = 7 + 2 = 9. \end{aligned}$
The textbook expands along row 1 instead and gets the same value:
$|A| = 1(1 + 6) - 1(0 - 3) + 1(0 - 1) = 7 + 3 - 1 = 9.$
Since $|A| = 9 \ne 0$ ⁠, $A^{-1}$ exists and the numbers are uniquely determined.

A determinant can be expanded along any row or column with the same result (Section 4.2.3); choosing the line with a zero saves one $2\times2$ determinant. Getting 9 both ways is a check on this key number.

Expansion along a column, Section 4.2.3. Lesson: expanding a 3 × 3 determinant
Cofactors of row 1

Signs $+,-,+$ ⁠:
$\begin{aligned} A_{11} &= +\begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} = 1 + 6 = 7,\\ A_{12} &= -\begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = -(0 - 3) = 3,\\ A_{13} &= +\begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix} = 0 - 1 = -1. \end{aligned}$
Each minor deletes row 1 and one column, leaving rows 2 and 3. In $M_{11}$ ⁠, $1\cdot1 - 3\cdot(-2) = 1 + 6 = 7$ ⁠.

Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors
Cofactors of row 2

Signs $-,+,-$ ⁠:
$\begin{aligned} A_{21} &= -\begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} = -(1 + 2) = -3,\\ A_{22} &= +\begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 1 - 1 = 0,\\ A_{23} &= -\begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = -(-2 - 1) = 3. \end{aligned}$
$M_{22}$ is 0 because rows 1 and 3 of $A$ agree in columns 1 and 3 (both have $1, 1$ there), so the minor has two equal rows.
Cofactors of row 3

Signs $+,-,+$ ⁠:
$\begin{aligned} A_{31} &= +\begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} = 3 - 1 = 2,\\ A_{32} &= -\begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = -(3 - 0) = -3,\\ A_{33} &= +\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1. \end{aligned}$
These agree with the textbook's nine values: $7, 3, -1$ ⁠; $-3, 0, 3$ ⁠; $2, -3, 1$ ⁠. As a check, row 1 with its own cofactors gives $1(7) + 1(3) + 1(-1) = 9 = |A|$ ⁠.
$\operatorname{adj}A$ and $A^{-1}$
$[A_{ij}] = \begin{bmatrix} 7 & 3 & -1 \\ -3 & 0 & 3 \\ 2 & -3 & 1 \end{bmatrix}.$
Transpose (rows become columns):
$\operatorname{adj}A = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}.$
Since $|A| = 9 \ne 0$ ⁠,
$A^{-1} = \frac{1}{9}\begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}.$
The adjoint is the transpose: the cofactors of row 1 ( $7, 3, -1$ ⁠) form its column 1. Then $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ with $|A| = 9$ ⁠.

Definition 3, Section 4.5.1; Theorem 4, Section 4.5. Lesson: the adjoint
Check: $A(\operatorname{adj}A) = 9I$ ⁠, so $AA^{-1} = I$

Row $i$ of $A$ times column $j$ of $\operatorname{adj}A$ ⁠:
$\begin{aligned} c_{11} &= 1\cdot7 + 1\cdot3 + 1(-1) = 9,\\ c_{12} &= 1(-3) + 1\cdot0 + 1\cdot3 = 0,\\ c_{13} &= 1\cdot2 + 1(-3) + 1\cdot1 = 0,\\ c_{21} &= 0\cdot7 + 1\cdot3 + 3(-1) = 0,\\ c_{22} &= 0(-3) + 1\cdot0 + 3\cdot3 = 9,\\ c_{23} &= 0\cdot2 + 1(-3) + 3\cdot1 = 0,\\ c_{31} &= 1\cdot7 + (-2)\cdot3 + 1(-1) = 0,\\ c_{32} &= 1(-3) + (-2)\cdot0 + 1\cdot3 = 0,\\ c_{33} &= 1\cdot2 + (-2)(-3) + 1\cdot1 = 9. \end{aligned}$
So $A(\operatorname{adj}A) = 9I$ and $AA^{-1} = \frac{1}{9}(9I) = I$ ⁠.

Diagonal entries are rows with their own cofactors ( $|A| = 9$ ⁠); off-diagonal entries are rows with other rows' cofactors (0). The pattern confirms every cofactor before we use $A^{-1}$ ⁠.

Theorem 1, Section 4.5. Basics: matrix facts the chapter uses
Compute $X = A^{-1}B$
$X = \frac{1}{9}\begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}\begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}.$
Row times column:
$\begin{aligned} \text{row 1}&: \ 7\cdot6 + (-3)\cdot11 + 2\cdot0\\ &= 42 - 33 + 0 = 9,\\ \text{row 2}&: \ 3\cdot6 + 0\cdot11 + (-3)\cdot0\\ &= 18 + 0 + 0 = 18,\\ \text{row 3}&: \ (-1)\cdot6 + 3\cdot11 + 1\cdot0\\ &= -6 + 33 + 0 = 27. \end{aligned}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{9}\begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.$
So $x = 1$ ⁠, $y = 2$ ⁠, $z = 3$ : the first number is 1, the second 2 and the third 3.

$X = A^{-1}B$ is the matrix method (Case I, Section 4.6.1). The 0 at the bottom of $B$ makes every third product vanish, which is why each row sum ends in " $+\,0$ ".

Case I, Section 4.6.1. Lesson: solving systems AX = B

Words	Equation
The sum of the three numbers is 6	$x + y + z = 6$
Third times 3, plus the second, is 11	$y + 3z = 11$
First plus third is double the second	$x + z = 2y$

Check against the original sentences

Condition	With 1, 2, 3	Holds?
Sum is 6	$1 + 2 + 3 = 6$	yes
3 × third + second = 11	$3\cdot3 + 2 = 11$	yes
First + third = double the second	$1 + 3 = 4 = 2\cdot2$	yes

Checking against the words, not just the equations, also tests the translation in Step 1, which no amount of matrix arithmetic can check.

Result

Algebraically: $x + y + z = 6$ ⁠, $y + 3z = 11$ ⁠, $x - 2y + z = 0$ ⁠. Here $|A| = 9$ and

X = A^{-1}B = \frac{1}{9}\begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.

The numbers are 1, 2 and 3 (first, second, third). (This agrees with the textbook's own solution.)

Check it with numbers

Solve by substitution, independently. From the third equation, $x + z = 2y$ ⁠; put this into the first: $2y + y = 6$ ⁠, so $y = 2$ ⁠. Then the second gives $3z = 11 - 2 = 9$ ⁠, so $z = 3$ ⁠, and the first gives $x = 6 - 2 - 3 = 1$ ⁠. Same numbers.

Common mistakes

Reading "add second number to it" as $3(z + y)$ ⁠. The sentence multiplies the third number first, then adds the second: $3z + y$ ⁠.
Leaving $2y$ on the right. Writing row 3 of $A$ as $1, 0, 1$ with right-hand side $2y$ is not a valid matrix equation.
Forgetting the 0 coefficient. Row 2 of $A$ is $0, 1, 3$ ⁠, not $1, 3$ ⁠.
Using $[A_{ij}]$ for the adjoint. The transpose is needed; Step 8 would expose the error.

What to take away

Word problem: name the unknowns, write one equation per condition, move all unknowns to the left, then use $X = A^{-1}B$ ⁠.
Expand $|A|$ along a row or column containing zeros; any line gives the same value.
Check the answer against the original words.

Every fact used, and where it comes from

Fact	Used in	Source
Writing a system as $AX = B$	Step 2	Section 4.6.1 (Lesson: solving systems AX = B)
Expansion along any row or column	Step 3	Section 4.2.3 (Lesson: expanding a 3 × 3 determinant)
Minors and cofactors	Steps 4 to 6	Definitions 1 and 2, Section 4.4 (Lesson: minors and cofactors)
$\operatorname{adj}A$ ⁠; $A^{-1} = \frac{1}{\|A\|}\operatorname{adj}A$	Step 7	Definition 3, Section 4.5.1; Theorem 4, Section 4.5
$A(\operatorname{adj}A) = \|A\|\,I$	Step 8	Theorem 1, Section 4.5
$\|A\| \ne 0$ ⁠: unique solution $X = A^{-1}B$	Steps 3, 9	Case I, Section 4.6.1

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Example 18: Three numbers from three conditions

Basics you need

Solution

Translate the words into equations

Write the system as AX=BAX = BAX=B

Compute ∣A∣|A|∣A∣⁠, expanding along column 1

Cofactors of row 1

Cofactors of row 2

Cofactors of row 3

adj⁡A\operatorname{adj}AadjA and A−1A^{-1}A−1

Check: A(adj⁡A)=9IA(\operatorname{adj}A) = 9IA(adjA)=9I⁠, so AA−1=IAA^{-1} = IAA−1=I

Compute X=A−1BX = A^{-1}BX=A−1B

Check against the original sentences

Every fact used, and where it comes from

Write the system as $AX = B$

Compute $|A|$ ⁠, expanding along column 1

$\operatorname{adj}A$ and $A^{-1}$

Check: $A(\operatorname{adj}A) = 9I$ ⁠, so $AA^{-1} = I$

Compute $X = A^{-1}B$