CP1-Chp6-Matrices (2).pptx used for revision

CorePure1 Chapter 6 ::
Matrices
jfrost@tiffin.kingston.sch.uk
www.drfrostmaths.com
@DrFrostMaths
Last modified: 14th September 2018

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Chapter Overview
1:: Understand matrices and
perform basic operations
(adding, scalar multiplication)
4:: Solve simultaneous equations using matrices.
“Use matrices to solve the following
simultaneous equations:
𝑥 + 2𝑦 + 𝑧 = 4
𝑥 − 𝑦 + 3𝑧 = 1
2𝑥 + 5𝑦 − 𝑧 = 0
2:: Multiply Matrices
Those who have done either IGCSE Further Mathematics would have encountered some of this
content. Otherwise it will be completely new!
“Given that 𝐀 =
1 2
−3 0
and 𝐁 =
0 3
4 5
,
determine the matrix 𝐀𝐁.”
3:: Find the determinant or
inverse of a matrix.
“If 𝐀 =
1 2
−3 0
, determine
𝐀−1
.”
Teacher Notes: The matrices chapter from the old FP1 has largely been
split into two, with the latter half (Chapter 7) dedicated to transformations.
3 × 3 matrices from the old FP3 has been moved here. There is some new
discussion about the consistency of systems of equations.

Introduction
A matrix (plural: matrices) is simply an ‘array’ of numbers, e.g.
On a simple level, a matrix is just a way to organise values into rows and columns, and represent
these multiple values as a single structure.
Matrices are particularly useful in
3D graphics, as matrices can be
used to carry out
rotations/enlargements (useful for
changing the camera angle) or
project into a 2D ‘viewing’ plane.
But the power of matrices comes from them representing linear transformations/functions (which
we will particularly see in Chapter 7). We can
1. Represent linear transformations using matrices (e.g. rotations, reflections and enlargements)
2. Use them to solve linear simultaneous equations.
1 0 −2
3 3 0

(Just for Fun) Using matrices to represent data
This is a scene from the film Good Will Hunting.
Maths professor Lambeau poses a “difficult”* problem for his graduate students from
algebraic graph theory, the first part asking for a matrix representation of this graph.
Matt Damon anonymously solves the problem while on a cleaning shift.
* It really isn’t.
?
In an ‘adjacency matrix’, the
number in the 𝑖th row and
𝑗th column is the number of
edges directly connecting
node (i.e. dot) 𝑖 to dot 𝑗
?

(Just for Fun) Using matrices to represent data
In my 4th year undergraduate dissertation, I used matrices to help ‘learn’ mark schemes from
GCSE biology scripts*. Matrix algebra helped me to initially determine how words (and more
complex semantic information) tended to occur together with other words.
* Shameless Brag Opportunity: I won the “Best
Undergraduate Dissertation Prize” for this!

(Just for Fun) And matrices in Statistics…
In Stats Year 2, you have/will come across the Normal Distribution, where you need
to specify the variance. This can be extended to 2D (and beyond) by using a
“covariance matrix”, where each number in the matrix gives the extent to which each
axis varies with each other.
1D 2D

Matrix Fundamentals
#1 Dimensions of Matrices
The dimension of a matrix is its size, in terms of its number of
rows and columns (in that order).
Matrix Dimensions
2 × 3
3 × 1
1 × 3
?
?

Matrix Fundamentals
#2 Notation/Names for Matrices
A matrix can have square or curvy brackets*.
* The textbook only uses curvy.
Matrix Column Vector Row Vector
A matrix with one column is simply a vector in the usual sense!

Matrix Fundamentals
#3 Variables for Matrices
If the value of a variable is a matrix, we use bold, capital letters
(In contrast, vectors use bold, lowercase letters)
𝐀 =
1
6
−3
𝐂 = 𝐏−𝟏𝐓𝐏

Matrix Fundamentals
#4 Adding/Subtracting Matrices
Simply add/subtract the corresponding elements of each matrix.
They must be of the same dimension.
?
?

Matrix Fundamentals
#5 Scalar Multiplication
A scalar is a number which can ‘scale’ the elements inside a matrix.
?
?
?
1
2
3
Fro Side Note: You
first encountered
this at GCSE, in the
context of vectors.
3𝒂 is the vector 𝒂
‘scaled’ by the
scalar 3.

Matrix Fundamentals
#6 Special Matrices
A matrix is square if it has the same number of rows as columns.
1 2
3 4
3 1 4
2 2 5
−3 4 3
A zero matrix is one in which all its elements are 0. The dimensions are usually
clear from the context.
𝟎 =
0 0 0
0 0 0
0 0 0
A identity matrix 𝐈 is a square matrix which has 1’s in the ‘leading diagonal’
(starting top-left) and 0 elsewhere. Again, the dimensions depend on the context.
𝐈 =
1 0 0
0 1 0
0 0 1
, 𝐈 =
1 0
0 1
We will see the significance of
the identity matrix when we
cover matrix multiplication
imminently.

Exercise 6A
Pearson Pure Mathematics Year 1/AS
Pages 97-99
(Classes in a rush could probably get away with skipping this exercise)

Matrix Multiplication
1 0 3 -2
2 8 4 3
7 -1 0 2
5 1
1 7
0 3
8 -3
-11
We start with the first row and first column, and sum the products of each pair.
1 × 5 + 0 × 1 + 3 × 0 + −2 × 8 = −11
In Chapter 9 (Vectors), you will see that this is finding the “dot/scalar product”
of the two vectors.
16
Now repeat for the next row of the left matrix...
42 61
50 -6
Now repeat the process with the first row and second column.
=

Matrix Multiplication involving 𝐈
We earlier saw the identity matrix 𝐈 =
1 0
0 1
. What do you notice about…
1 0
0 1
3 2
1 0
=
3 2
1 0
3 2
1 0
1 0
0 1
=
3 2
1 0
?
?
In general 𝐀𝐈 = 𝐈𝐀 = 𝐀 for all matrices 𝐀.
So the identity matrix is a bit like the ‘1’ of matrix multiplication,
e.g. 1 × 3 = 3 × 1 = 3; multiplying by 1 has no effect, and multiplying by 𝐈 has
no effect.
For this reason, 1 is known as the ‘identity’ of multiplication over numbers.
And 0 is known as the ‘identity’ of addition over numbers, given that
𝑎 + 0 = 0 + 𝑎 = 𝑎 for all 𝑎.

1 2
3 4
2 0 −1
3 2 1
=
8 4 1
18 8 1
Test Your Understanding
?
?
1 2 3
1
2
3
= 14
1
2
3
1 2 3 =
1 2 3
2 4 6
3 6 9
?
?
a
b
c
N1
1 2
3 4
3
−1
=
1
5
?
1 2
3 4
2
=
7 10
15 22
?
N2
N3

Matrix Fundamentals
#7 Matrix Multiplication
Matrix multiplications are not always valid: the dimensions have to agree.
Note that only square matrices (i.e. same width as height) can be raised to a power.
Dimensions of 𝐀 Dimension of 𝐁 Dimensions of 𝐀𝐁 (if valid)
2  3 3  4 2  4
1  3 2  3 Not valid.
6  2 2  4 6  4
1  3 3  1 1  1
7  5 7  5 Not valid.
10  10 10  9 10  9
3  3 3  3 3  3
?
?
?
?
?
?
?

Exercise 6B
Pages 101-103

Determinant of a matrix
In Chapter 7, you will see that matrices can be thought of as a function
that can transform a point, eg:
1 2
3 4
1
1
=
3
7
1 2
3 4
(1,1)
(3,7)
A question might naturally be whether there is an ‘inverse
function/transformation’ that can retrieve the original point:
? ?
? ?
3
7
=
1
1 ? ?
? ?
(1,1)
(3,7)
This matrix would be known as the inverse of
1 2
3 4
.
With quadratics, we used to the word ‘discriminant’ for 𝑏2 − 4𝑎𝑐 because it
‘discriminates’ between the different cases of 0, 1, 2 roots.
Analogously, the ‘determinant’ 𝑨 or det(𝑨) for a matrix 𝑨 ‘determines’ whether it has
an inverse of not.

Determinants
 The determinant of a matrix 𝐀 =
𝑎 𝑏
𝑐 𝑑
is
det 𝐀 = |𝐀| = 𝑎𝑑 − 𝑏𝑐
 If 𝑑𝑒𝑡 𝐀 = 0, then 𝐀 is a singular matrix and it does not have an inverse.
 If det 𝐀 ≠ 0, then 𝐀 is a non-singular matrix and it has an inverse.
𝐀 det 𝐀
1 0
0 1
1
1 2
3 4
-2
0 3
−1 −4
3
10 −2
4 −1
-2
?
?
?
?
Quickfire Questions:

Test Your Understanding So Far
[Textbook]
𝐴 =
4 𝑝 + 2
−1 3 − 𝑝
Given that 𝐀 is singular, find the value of 𝑝.
Edexcel FP1(Old) Jan 2010 Q5
det 𝑨 = 4 3 − 𝑝 − −1 𝑝 + 2 = 0
14 − 3𝑝 = 0 ⇒ 𝑝 =
14
3
?
?
?

Determinants of 3 × 3 matrices
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
= 𝑎
𝑒 𝑓
ℎ 𝑖
− 𝑏
𝑑 𝑓
𝑔 𝑖
+ 𝑐
𝑑 𝑒
𝑔 ℎ
𝑎 𝑏 𝑐
3 1 4
2 2 5
−3 4 3
= 3
2 5
4 3
− 1
2 5
−3 3
+ 4
2 2
−3 4
= 3 −14 − 1 21 + 4 14
= −7
(note the minus for
the middle one)
?

𝑨 =
3 𝑘 0
−2 1 2
5 0 𝑘 + 3
where 𝑘 is a constant.
Given that 𝐴 is singular, find the possible values of 𝑘.
det 𝑨 = 2𝑘2
+ 19𝑘 + 9
If singular, 2𝑘2 + 19𝑘 + 9 = 0.
Solving:
𝑘 = −
1
2
, −9
?
𝑨 =
1 2 0
4 5 −6
−1 8 2
Determine det 𝐀 .
det 𝐴 = 1 10 + 48 − 2 8 − 6 + 0 32 + 5
= 54
?

Minors
The minor of an element in a 3 × 3 matrix is the determinant of the 2 × 2 matrix that
remains after the row and column containing that element have been crossed out.
1 2 0
4 5 −6
−1 8 2
Minor of 0:
𝟒 𝟓
−𝟏 𝟖
= 𝟑𝟕
Minor of -6:
𝟏 𝟐
−𝟏 𝟖
= 𝟏𝟎
Minor of 5:
𝟏 𝟎
−𝟏 𝟐
= 𝟐
?
?
?

Exercise 6C
Pages 106-108

Inverting a 2 × 2 matrix
We earlier saw that the inverse of a matrix 𝐌 (written 𝐌−1
) ‘undoes’ the effect of
the matrix. Thus:
𝐌𝐌−1
= 𝐌−1
𝐌 = 𝐈
as multiplying something by a matrix followed by its inverse has no overall effect
(i.e. the same as the identity matrix).
?
 If 𝐀 =
𝑎 𝑏
𝑐 𝑑
then 𝑨−1 =
1
det(𝐴)
𝑑 −𝑏
−𝑐 𝑎
• 𝐀−𝟏
is the ‘inverse’ of 𝐀, so that if 𝐀𝒙 = 𝒚, 𝐀−1
𝒚 = 𝒙
• 𝐀𝐀−𝟏 = 𝐀−𝟏𝐀 = 𝐈

Practising the Inverse
𝑎
1
𝑎𝑑 − 𝑏𝑐
𝑏
𝑐 𝑑
−
−
Divide by
determinant.
Swap NW-SE
elements.
Make SW-NE
elements negative.
Click to
Froinverse

2 0
0 2
−1
=
1
2
1 0
0 1
1 2
3 4
−1
=
1
2
−4 2
3 −1
7 2
1 −3
−1
=
1
23
3 2
1 −7
For what value of 𝑝 is
4 𝑝 + 2
−1 3 − 𝑝
singular? Given 𝑝 is not
this value, find the inverse.
𝒑 =
𝟏𝟒
𝟑
𝟏
𝟏𝟒−𝟑𝒑
𝟑 − 𝒑 − 𝒑 + 𝟐
𝟏 𝟒
?
?
?
? ?

Matrix Proofs Involving Inverse
If 𝐏 and 𝐐 are non-singular matrices, prove that 𝐏𝐐 −1 = 𝐐−1𝐏−1
Let 𝐶 = 𝑃𝑄 −1 then 𝑃𝑄 𝐶 = 𝐼
𝑃−1
𝑃𝑄𝐶 = 𝑃−1
𝐼
𝐼𝑄𝐶 = 𝑃−1
𝑄𝐶 = 𝑃−1
𝑄−1
𝑄𝐶 = 𝑄−1
𝑃−1
𝐼𝐶 = 𝑄−1𝑃−1
𝐶 = 𝑄−1𝑃−1
Fro Tip: You can rid of a matrix 𝐀 at the
front of the expression by multiplying
the front of each side of the equation
by 𝐀−1
(to get 𝐈). You can similarly
remove an 𝐀 at the end by multiplying
the end of each side by 𝐀−1
.
If 𝐴 and 𝐵 are non-singular matrices such that 𝐁𝐀𝐁 = 𝐈, prove that 𝐀 = 𝐁−1𝐁−𝟏
𝐵𝐴𝐵 = 𝐼
𝐵−1
𝐵𝐴𝐵 = 𝐵−1
𝐼
𝐼𝐴𝐵 = 𝐵−1
𝐴𝐵 = 𝐵−1
𝐴𝐵𝐵−1
= 𝐵−1
𝐵−1
𝐴𝐼 = 𝐵−1
𝐵−1
𝐴 = 𝐵−1
𝐵−1
?
?
Exam Note: I couldn’t find
any (old spec) FP1
questions of this type.

Exercise 6D
Pages 110-111

Matrix Transpose

𝐴𝑇 is the transpose of a matrix 𝐴, where the rows and columns are interchanged.
e.g.
1 2 3
4 5 6
𝑇
=
1 4
2 5
3 6
. An 𝑚 × 𝑛 matrix becomes 𝑛 × 𝑚.
Why transpose?
It is hard to have an exact conceptual sense of what the matrix transpose is.
But it allows a degree of algebraic manipulation:
When we multiply matrices we’re doing something called the ‘dot product’ (CP
Year 2) of each row of the first matrix and each column of the second.
Suppose we found the dot product of two vectors 𝒖 and 𝒗 and transformed the
first using a matrix 𝑨:
𝑨𝒖 ⋅ 𝒗
If we wanted to transform the second vector 𝑣 instead, we’d have to use the
transpose of 𝐴 instead to end up with the same dot product:
𝑨𝒖 ⋅ 𝒗 = 𝒖 ⋅ 𝑨𝑻
𝒗
This has a number of practical consequences.
(Far beyond understanding
required for exam)
e.g.
1 2
3 4
5
6
⋅
7
8
=
17
39
⋅
7
8
= 431
5
6
⋅
1 3
2 4
7
8
=
5
6
31
46
= 431

Inversing a 3 × 3 matrix
If 𝑨 =
1 2
3 4
, find 𝑨−1.
The method we previously used was a specific case of a more general method
which can be used for matrices of any size:
 Step 1: Find det(𝑨) det 𝑨 = 4 − 6 = −2
 Step 2: Form a matrix
of minors, 𝑴
Recap: The minor of each element
in a matrix is the determinant of
the remaining matrix when the
row and column are crossed out.
1 2
3 4
4
The minor of 1 is 4 because
the determinant of 4 is 4.
𝑴 =
4 3
2 1
of cofactors, 𝑪
A cofactor by definition is ‘a signed minor’.
We simply apply signs to each minor using
the following alternating pattern: (+ top left)
+ −
− +
𝑴 =
4 3
2 1
𝑪 =
4 −3
−2 1
 Step 4: 𝐴−1
=
1
det 𝑨
𝑪𝑇
𝑨−𝟏
= −
1
2
4 −2
−3 1
?
?

Inversing a 3 × 3 matrix
If 𝑨 =
1 3 1
0 4 1
2 −1 0
, find 𝑨−1.
 Step 1: Find det(𝑨)
of minors, 𝑴
of cofactors, 𝑪
 Step 4: 𝐴−1 =
1
det 𝑨
𝑪𝑇
det 𝑨 = 1 1 − 3 −2 + 1 −8 = −1
𝑴 =
1 −2 −8
1 −2 −7
−1 1 4
Fro Tip: Note we’ve
already found the top
row from above.
𝑪 =
1 2 −8
−1 −2 7
−1 −1 4
+ − +
− + −
+ − +
𝑨−1
=
1
−1
1 −1 −1
2 −2 −1
−8 7 4
=
−1 1 1
−2 2 1
8 −7 −4
?
?
?
?

Doing with your Classwiz
If 𝑨 =
1 3 1
0 4 1
2 −1 0
, find 𝑨−1.
1. Mode  Matrix.
2. Select 𝑀𝑎𝑡𝐴. This allows you to input your matrix, which will be
saved in a special variable ‘𝑀𝑎𝑡𝐴’.
3. Select 3 rows/cols and input each number, pressing = after each.
4. Press AC to start a calculation.
5. You want to write 𝑀𝑎𝑡𝐴−1. To get the 𝑀𝑎𝑡𝐴 in your expression:
OPTN for the matrix menu, then select 𝑀𝑎𝑡𝐴 to insert it into your
expression.
Use the special 𝒙−𝟏 key on your calculator, because the general
power button will not work in matrix mode.
6. Press =, and look appropriately smug.

Further Example
𝑨 =
−2 3 −3
0 1 0
1 −1 2
, and the matrix 𝐁 is such that 𝐀𝐁 −1 =
8 −17 9
−5 10 −6
−3 5 −4
.
(a) Show that 𝐀−1
= 𝐀.
(b) Find 𝑩−1.
If 𝐀−1 = 𝐀 then 𝐀𝐀−1 = 𝐀2 and hence 𝐀2 = 𝐈.
𝑨2
=
−2 3 −3
0 1 0
1 −1 2
−2 3 −3
0 1 0
1 −1 2
=
1 0 0
0 1 0
0 0 1
𝐴𝐵 −1
= 𝐵−1
𝐴−1
𝐴𝐵 −1𝐴 = 𝐵−1𝐴−1𝐴
𝐴𝐵 −1𝐴 = 𝐵−1
∴ 𝐵−1 =
8 −17 9
−5 10 −6
−3 5 −4
−2 3 −3
0 1 0
1 −1 2
=
−7 −2 −6
4 1 3
2 0 1
a
b
?
?

Exercise 6E
Pages 115-116

Frost Life StoriesTM
In the game Assassin’s Creed II, you encounter a variety of concentric ring picture
puzzles, which upon successfully completing, you unlock a segment of a secret video.
Rings are connected in pairs, and must be rotated together in their pairs. The aim is to
form a complete picture. Different possible pairs can be selected, for example, where
there just 3 rings, you could rotate A and B together, B and C together or C and A
together.
Only certain pairings are available.
Because I got stuck on one (this was back in
my uni days) and because I’m incredibly
uncool, I formed simultaneous equations
and used a matrix inverse to solve them,
which therefore told me how many times
to rotate each pair.
We’ll see how we can do this.

Using Matrices For Simultaneous Equations
 If 𝐀
𝑥
𝑦
𝑧
= 𝒗 then
𝑥
𝑦
𝑧
= 𝐀−1𝒗
[Textbook] Use an inverse matrix to solve the simultaneous equations:
−𝑥 + 6𝑦 − 2𝑧 = 21
6𝑥 − 2𝑦 − 𝑧 = −16
−2𝑥 + 3𝑦 + 5𝑧 = 24
We can write using a matrix multiplication:
−1 6 −2
6 −2 −1
−2 3 5
𝑥
𝑦
𝑧
=
21
−16
24
Find inverse of LHS matrix:
1
189
7 36 10
28 9 13
−14 9 34
∴
𝑥
𝑦
𝑧
=
1
189
7 36 10
28 9 13
−14 9 34
21
−16
24
=
−1
4
2
∴ 𝑥 = −1, 𝑦 = 4, 𝑧 = 2
If we multiplied out the LHS it’s easy
to see this gives us the equations in
the original question.
Use your calculator to find this directly.
Calculator Tip: You could check your answer
using the simultaneous equation solver.
?
?
?

Forming the equations yourself
[Textbook] A colony of 1000 mole-rats is made up of adult males, adult
females and youngsters. Originally there were 100 more adult females
than adult males.
After one year:
• The number of adult males had increased by 2%
• The number of adult females had increased by 3%
• The number of youngsters had decreased by 4%
• The total number of mole-rats had decreased by 20
Form and solve a matrix equation to find out how many of each type of
mole-rat were in the original colony.
Let 𝑥 = number of adult males
𝑦 = number of adult females
𝑧 = number of youngsters
𝑥 + 𝑦 + 𝑧 = 1000
𝑥 − 𝑦 = −100
1.02𝑥 + 1.03𝑦 + 0.96𝑧 = 980
1 1 1
1 −1 0
1.02 1.03 0.96
𝑥
𝑦
𝑧
=
1000
−100
980
→
“1000 mole-rats”
“Originally 100 more adult females than adult males”
“Total mole-rats after 1 year decreased by 20.”
𝑥
𝑦
𝑧
=
1
13
−96 7 100
−96 −6 100
205 −1 −200
1000
−100
980
=
100
200
700
100 adult males, 200 adult females, 700 youngsters in the
original colony.
?
?
?
?

Consistency of linear equations
𝑥 + 2𝑦 = 1
3𝑥 − 𝑦 = 0
System of equations
is consistent. It has
one solution.
The corresponding
matrix
1 2
3 −1
is
non-singular.
From Pure Year 1 you are already familiar with the idea that the solution of a system of two
equations (with two unknowns) can be visualised by finding the point of intersection of two lines.
A system of linear equations is known as consistent if there is at least one set of values that
satisfies all the equations simultaneously (i.e. at least one point of intersection).
𝑥 − 3𝑦 = 1
𝑥 − 3𝑦 = 2
System of equations
is inconsistent. It
has no solutions.
Matrix
1 −3
1 −3
is
singular.
𝑥 − 3𝑦 = 1
2𝑥 − 6𝑦 = 2
System of equations
is consistent. It has
infinitely many
solutions.
Matrix
1 −3
1 −3
is
singular.

Extending consistency to 3 variables
In Chapter 9 you will learn that just as 𝑎𝑥 + 𝑏𝑦 = 𝑐 gives the equation of a straight line,
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑 gives the equation of a plane.
Again, we get solutions to the system of linear equations when all of the planes intersect:
Scenario 1: Planes all
meet at a single point.
System of equations
consistent, and one
solution.
Scenario 2: Planes
form a sheaf.
They have a line of
intersection consisting
of infinitely many
points. System of
equations consistent
and infinitely many
solutions.
Scenario 3: Planes
form a prism.
While planes intersect
in pairs, they don’t all
intersect at any point.
System of equations is
inconsistent.
Scenario 4: Two of
more planes parallel
and non-identical.
Again, inconsistent, as
the parallel planes
never intersect, and
thus all equations
can’t be satisfied.
Any rows in the corresponding matrix which
are multiples of each other will be parallel.

Extending consistency to 3 variables
In Chapter 9 you will learn that just as 𝑎𝑥 + 𝑏𝑦 = 𝑐 gives the equation of a straight line,
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑 gives the equation of a plane.
Again, we get solutions to the system of linear equations when all of the planes intersect:
Scenario 5: Planes
represented by
equations are equivalent.
System of equations
consistent, and infinitely
many solutions.

Example
[Textbook] A system of equations is shown below:
3𝑥 − 𝑘𝑦 − 6𝑧 = 𝑘
𝑘𝑥 + 3𝑦 + 3𝑧 = 2
−3𝑥 − 𝑦 + 3𝑧 = −2
For each of the following values of 𝑘, determine whether the system of equations is
consistent or inconsistent. If the system is consistent, determine whether there is a
unique solution or an infinity of solutions. In each case, identify the geometric
configuration of the plane corresponding to each value of 𝑘.
(a) 𝑘 = 0 (b) 𝑘 = 1 (c) 𝑘 = −6 Remember that the system of equations is consistent if the
corresponding matrix is non-singular, i.e. its determinant is non-0.
𝑘 = 0:
3 0 −6
0 3 3
−3 −1 3
= −18
Matrix non-singular so a unique solution, i.e. planes meet at single point.
a
?

Example
[Textbook] A system of equations is shown below:
3𝑥 − 𝑘𝑦 − 6𝑧 = 𝑘
𝑘𝑥 + 3𝑦 + 3𝑧 = 2
−3𝑥 − 𝑦 + 3𝑧 = −2
For each of the following values of 𝑘, determine whether the system of equations is
consistent or inconsistent. If the system is consistent, determine whether there is a
unique solution or an infinity of solutions. In each case, identify the geometric
configuration of the plane corresponding to each value of 𝑘.
(a) 𝑘 = 0 (b) 𝑘 = 1 (c) 𝑘 = −6
𝑘 = 1:
3 −1 −6
1 3 3
−3 −1 3
= 0
3𝑥 − 𝑦 − 6𝑧 = 1 (1)
𝑥 + 3𝑦 + 3𝑧 = 2 (2)
−3𝑥 − 𝑦 + 3𝑧 = −2 (3)
1 + 2 × 2 : 5𝑥 + 5𝑦 = 5 (4)
2 − 3 : 4𝑥 + 4𝑦 = 4 (5)
Equations (4) and (5) are consistent so system is consistent and
has an infinity of solutions. Planes meet at a sheaf.
If the matrix is singular,
the system of equations
could still be consistent:
recall that we might have
a sheaf (i.e. planes
intersect at a line) or
equations represent same
plane.
Eliminate one of the
variables. If resulting two
equations are consistent,
then system will be
consistent.
b
?

The system of equations is consistent and has a single solution. Determine the
possible values of 𝑘.
2𝑥 + 3𝑦 − 𝑧 = 13
3𝑥 − 𝑦 + 𝑘𝑧 = 11
𝑥 + 𝑦 + 𝑧 = 7
2 3 −1
3 −1 𝑘
1 1 1
= 𝑘 − 15
To have a solution, we require that 𝑘 − 15 ≠ 0,
thus 𝑘 ≠ 15.
?

Exercise 6F
Pages 120-121

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