worksheet 21 matrices and applications name:

Maths Quest Maths C Year 12 for Queensland 2e 1

WorkSHEET 2.1 Matrices and applications Name: _________________________ 1 Evaluate the following:

2 If A = and B = calculate,

if possible, (a) AB (b) BA

(a)

(b) 𝐵𝐴 = $0 −5 −33 9 08 14 −6

.

3 If 𝐶 = 02 −13 4 2 calculate the determinant of C.

4 Calculate the inverse of C given in question 3.

5 Solve for X:

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2 1 1 0 3 2 2 2 1 3 3 41 1 3 0 0 2 1 2 3 3 0 4

4 131 7

AB´ + ´ - ´ ´- + ´ + - ´é ù

= ê ú´ + ´ + ´ ´- + ´ + ´ë û- -é ù

= ê úë û

111342

4312

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231411C

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22

6621

2002

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8623

20023

32

31

X

X

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The next 2 Questions refer to the following matrix:

6 Use your calculator to calculate the determinant of D.

Determinant of D = 21

7 Use your calculator to calculate the inverse of D.

8 Solve for X:

Becomes; 0−2 53 62 𝑋 = 034 41

57 602

−127 %

6 −5−3 −2) %

−2 53 6)𝑋 = −

127 %

6 −5−3 −2) %

34 4157 60)

𝑋 = −127 0

6 −5−3 −22 0

34 4157 602

𝑋 = −127 0

6 × 34 − 5 × 57 6 × 41 − 5 × 60−3 × 34 − 2 × 57 −3 × 41 − 2 × 602

𝑋 = −127 0

−81 −54−216 −2432

𝑋 = 03 2

8 92 make up your own 3x3 matrix equations and check them with your calculator.

9 Consider the matrix E =

For what values of x is the matrix singular?

If E is singular then the determinant of E = 0

Typeequationhere.

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10 Find 𝑎and𝑏 given 𝐴 = 01 𝑎𝑖1 𝑏 2

and |𝐴| = 6 + 2𝑖

|𝐴| = 𝑎𝑑 − 𝑏𝑐

so, 𝑏 − 𝑎𝑖 = 6 + 2𝑖

by correlating coefficients we have;

𝑏 = 6 and

𝑎 = −2 make up your own 3x3 matrix question similar to this and check it with your calculator.

11 If position vector ⟨1,1⟩ is rotated to ⟨−1,1⟩, use the rotational matrix to find what angle it has been rotated through. ** The rotational matrix is in your Formula Sheet. ** And yes you can already see the answer, but that’s not the point!

Use 𝑅 = 0cos 𝜃 − sin 𝜃sin 𝜃 cos 𝜃 2

Where;

𝑅𝑣 = 𝑣′ and

0cos 𝜃 − sin 𝜃sin 𝜃 cos 𝜃 2 × 0

112 = 0−11 2

0cos 𝜃 − sin 𝜃sin 𝜃 + cos 𝜃2 = 0−11 2

∴ we have;

cos 𝜃 − sin 𝜃 = −1𝑒𝑞. 1

sin 𝜃 + cos 𝜃 = 1𝑒𝑞. 2 from 𝑒𝑞. 1 cos 𝜃 = sin 𝜃 − 1 by sub into 𝑒𝑞. 2

sin 𝜃 + sin 𝜃 − 1 = 1

2 sin 𝜃 = 1

𝜃 =𝜋2


12 If vector 𝒂 = 𝑖 + 𝑗 is rotated to vector 𝒃 =]!"− √$

"^ 𝑖 + ]√$

"+ !

"^ 𝑗 . Use the rotational

matrix to find what angle vector 𝒂 has been rotated through to arrive at 𝒃.


Where;

𝑅𝑣 = 𝑣′ and

0cos 𝜃 − sin 𝜃sin 𝜃 cos 𝜃 2 × 0

112 =

⎣⎢⎢⎢⎡12−√32

√32 +

12⎦⎥⎥⎥⎤

0cos 𝜃 − sin 𝜃sin 𝜃 + cos 𝜃2 =

⎣⎢⎢⎢⎡12−√32

√32 +

12⎦⎥⎥⎥⎤

∴ we have;

cos 𝜃 − sin 𝜃 =12 −

√32 𝑒𝑞. 1

sin 𝜃 + cos 𝜃 =√32 +

12 𝑒𝑞. 2

from 𝑒𝑞. 1 cos 𝜃 = sin 𝜃 + !

"− √$

"

by sub into 𝑒𝑞. 2

sin 𝜃 + sin 𝜃 +12 −

√32 =

√32 +

12

sin 𝜃 =√32

𝜃 =𝜋3 𝑜𝑟

2𝜋3

check which angle, by checking solution in 𝑒𝑞. 2 because cos "%

$= − !

" … "%

$ does Not satisfy

the equation.

∴ 𝜃 =𝜋3


13 If vectors 𝒂 = 𝑖 + 𝑗 and 𝒃 = 3𝑖 + 4𝑗 are rotated using the Rotational Matrix to get

matrix h𝛼 − 𝛽 3𝛼 − 2√3𝛽 + 𝛼 3𝛽 + 4𝛼

k , then find 𝛼, 𝛽 and

the angle the vectors have been rotated.


Convert 𝒂 and 𝒃 to column matrix:01 3

1 42 Using

𝑅𝑣 = 𝑣′

!cos 𝜃 − sin 𝜃sin 𝜃 cos 𝜃 ) !

1 31 4) = .𝛼 − 𝛽 3𝛼 − 2√3

𝛽 + 𝛼 3𝛽 + 4𝛼4

!cos 𝜃 − sin 𝜃 3 cos 𝜃 − 4 sin 𝜃cos 𝜃 + sin 𝜃 3 cos 𝜃 + 4 sin 𝜃, = .𝛼 − 𝛽 3𝛼 − 2√3

𝛽 + 𝛼 3𝛽 + 4𝛼3

∴ cos 𝜃 − sin 𝜃 = 𝛼 − 𝛽𝑒𝑞. 1

3 cos 𝜃 − 4 sin 𝜃 = 3𝛼 − 2√3𝑒𝑞. 2

cos 𝜃 + sin 𝜃 = 𝛼 + 𝛽𝑒𝑞. 3

3 cos 𝜃 + 4 sin 𝜃 = 4𝛼 + 3𝛽𝑒𝑞. 4

Now, 𝑒𝑞. 1 + 𝑒𝑞. 32 cos 𝜃 = 2𝛼

and cos 𝜃 = 𝛼

𝑒𝑞. 1 − 𝑒𝑞. 3 − 2 sin 𝜃 = −2𝛽

and sin 𝜃 = 𝛽

sub in 𝑒𝑞. 23𝛼 − 4𝛽 = 3𝛼 − 2√3 and

𝛽 =√32

we can now say;

sin 𝜃 =√32 → 𝜃 =

𝜋3 𝑜𝑟

2𝜋3

because we because cos "%

$= − !

" … "%

$ does Not satisfy

the equation.

∴ 𝜃 =𝜋3


14 The next 3 Questions refer to the following information. The Leslie matrix describing the yearly changes in the number of female animals in a certain population is given below:

*** Leslie Matrices … external exam!

15 What does the number 2.5 in the first row of the matrix indicate?

This number indicates that each two-year-old female produces on average 2.5 offspring per year.

16 What does the number 0.3 in the second row of the matrix represent?

This number indicates that 30% of 1-year-old animals survive to two years of age.

17 If the initial population was Number of 1-year-olds = 220 Number of 2-year-olds = 180 Number of 3-year-olds = 260

Calculate the population in 3 years time.

The population in 3 years time, N3, is given by:

The population in 3 years time will be

Number of 1-year-olds = 793.5 Number of 2-year-olds = 103.5 Number of 3-year-olds = 145.5

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18 Technology active;

and Technology free

Describe how these planes intersect;

𝑥 + 2𝑦 + 𝑧 = 12

2𝑥 − 𝑦 + 𝑧 = 5

3𝑥 + 𝑦 − 2𝑧 = 1

Either by GC, or by lots of manual Gaussian work, get … RREF;

$1 0 00 1 00 0 1

|234.

Clearly, as there is ONE solution to the system of equations, the three planes intersect at ONE point, the point ⟨2, 3, 4⟩.

19 Technology active; and

Technology free Describe how these planes intersect;

3𝑥 + 2𝑦 + 𝑧 = 1

7𝑥 + 4𝑦 + 5𝑧 = 3

5𝑥 + 3𝑦 + 3𝑧 = 2


$1 0 30 1 −40 0 0

|1−10.

Clearly, as Row three gives us a Free Variable, there are an infinite number of solutions to the system of equations, therefore the three planes intersect at a LINE. (In the vectors worksheet you need to calculate the line of intersection as a vector function J )


20 Describe how these planes intersect;

6𝑥 − 15𝑦 − 3𝑧 = 11

2𝑥 − 5𝑦 − 𝑧 = 7

4𝑥 − 10𝑦 − 2𝑧 = 1


$2 −5 −10 0 00 0 0

|010.

Row 1 gives us a trivial solution where 𝑥, 𝑦𝑎𝑛𝑑𝑧 are all 0 Row 2 gives No solution, as zero times anything can’t equal 1 Row 3 gives us a free variable, but not much we can do with it. So, there is NO solution here! The three planes must be parallel. Lets evaluate the reasonableness of that. The three Normals to the planes are all scalars of each other, hence all planes must be parallel. Further, as there is not a consistent scalar for the Constant terms, the planes are at a different distance from the origin, hence parallel, but not coincident!



6𝑥 − 15𝑦 − 3𝑧 = 6

2𝑥 − 5𝑦 − 𝑧 = 2

4𝑥 − 10𝑦 − 2𝑧 = 4


$2 −5 −10 0 00 0 0

|100.

Row 1 now gives us an actual possible solution. We do not need to find it, but there has to be a value for x, y and z that will make that work! What ever that combination of x, y and z are, they will satisfy the other two equations too, as both Row 2 and Row 3 are free variables. Hence, these are Coincident planes Lets evaluate the reasonableness of that. The three Normals to the planes are all scalars of each other, hence all planes must be parallel. Further, as there is a consistent scalar for the Constant terms, the planes are at the same distance from the origin, hence they are coincident!



6𝑥 − 15𝑦 − 3𝑧 = −9

4𝑥 − 10𝑦 − 2𝑧 = 4

5𝑥 + 4𝑦 − 8𝑧 = −11


$3 0 −40 3 −10 0 0

|001.

OK … Really … ? The syllabus has this under Matrices, and in some cases, the description of the planes based on RREF is easy, but Wow, sometimes it’s easier just to take a look! Can you see the normal to planes 1 and 2 are scalars of each other … hence they are parallel. As the constant terms are not in the same scalar, we know they are different distances from the origin, hence planes 1 and 2 are parallel, but not coincident. Because the normal to plane 3 is not a scalar, it has to interest the other 2 planes … like:

worksheet 21 matrices and applications name:

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