Assignment 21 - Matrices II, SOLUTIONS

Mathematics Applications and Interpretations HL
The use of GDC is allowed and encouraged for all problems

Assignment 21 – Matrices II, SOLUTIONS
Problem 1
1 2
Let 𝐴 = ( ) . Find numbers 𝑎 and 𝑏 so that 𝐴2 = 𝑎𝐴 + 𝑏𝐼. [Hint: see example 9 on p. 313]
−1 −3
1 2 1 2 −1 −4
𝐴2 = ( )( )=( )
−1 −3 −1 −3 2 7
1 2 1 0 𝑎+𝑏 2𝑎
𝑎( )+𝑏( )=( )
−1 −3 0 1 −𝑎 −3𝑎 + 𝑏
So, 𝑎 + 𝑏 = −1, 2𝑎 = −4, − 𝑎 = 2, − 3𝑎 + 𝑏 = 7 .
Hence, 𝒂 = −𝟐 and 𝒃 = 𝟏
Problem 2
𝑎 𝑏 𝑑 −𝑏
Let 𝐴 = ( ) and 𝐵 = ( ).
𝑐 𝑑 −𝑐 𝑎
𝑎 𝑏 𝑑 −𝑏 𝒂𝒅 − 𝒃𝒄 𝟎 𝟏 𝟎
a. Calculate 𝐴𝐵 = ( )( )=( ) = (𝒂𝒅 − 𝒃𝒄) ( ).
𝑐 𝑑 −𝑐 𝑎 𝟎 −𝒃𝒄 + 𝒂𝒅 𝟎 𝟏
1
b. Hence write down a number 𝑘 so that 𝐴 × 𝑘𝐵 = 𝐼. 𝑘 = 𝑎𝑑−𝑏𝑑 = 𝟏/𝐝𝐞𝐭(𝑨)
1 𝑑 −𝑏
c. Have you now explained why 𝐴−1 = ( ) when det(𝐴) ≠ 0? YES!
det(𝐴) −𝑐 𝑎
Problem 3
In 3a below, just do what it asks
In 3b, the solution can of course be found by using the GDC app for simultaneous equations. However, please
solve with matrices (for the sake of practicing matrix algebra!) and then check your result by re-solving with the
GDC app.
a. 𝑥 = the cost of one football, 𝑦 = the cost of one baseball, 𝑧 = the cost of one basketball.
2 1 3
b. Let 𝐴 = [3 2 1] be the coefficient matrix for the system of simultaneous equations.
5 0 2
𝑥 135 𝑥 135 21
Then 𝐴 [𝑦] = [121.5], and hence [𝑦] = 𝐴−1 [121.5] = [16.5].
𝑧 156 𝑧 156 25.5
Now we know the prices of the different types of balls.
East Park International buys 4 footballs, 5 baseballs and 𝑁 basketballs for their $470 budget:
4 × 21 + 5 × 16.5 + 𝑁 × 25.5 = 470
𝑁 = 11.9
So, they can buy 11 basketballs.
Problem 4 A geneticist uses a Markov chain model to investigate changes in a specific gene in a cell as it divides.
Every time the cell divides, the gene may mutate between its normal state and other states.
𝑋 𝑋
The model is of the form ( 𝑛+1 ) = 𝐌 ( 𝑛 ) where 𝑋𝑛 is the probability of the gene being in its normal state after dividing
𝑍𝑛+1 𝑍𝑛
for the 𝑛th time, and 𝑍𝑛 is the probability of it being in another state after dividing for the 𝑛th time, where 𝑛 ∈ ℕ.
0.94 𝑏
Matrix 𝐌 is found to be ( ).
0.06 0.98
(a.i) Write down the value of 𝑏. 𝒃 = 𝟎. 𝟎𝟐 (since the numbers in each column must add up to 1) [1]
(a.ii) What does 𝑏 represent in this context?
There is a 2% probability that the gene turns from “other state” into “normal state” when the cells divides. [1]
(b) Find the eigenvalues of 𝐌. [3]
Well, we know from our exploration in class that 𝜆1 = 1 and 𝜆2 = 0.94 − 0.02 = 0.92, but we better find them
using the characteristic polynomial:
𝜆 − 0.94 −0.02
𝑃(𝜆) = det(𝜆𝐼 − 𝑀) = | | = (𝜆 − 0.94)(𝜆 − 0.98) − 0.06 × 0.02 = 𝜆2 − 1.92𝜆 + 0.92
−0.06 𝜆 − 0.98
Solve the quadratic equation 𝑃(𝜆) = 0 with PLYSMLT2: 𝝀 = 𝟎. 𝟗𝟐 or 𝝀 = 𝟏, as expected.
(c) Find the eigenvectors of 𝐌. [3]

𝑥 0.06 −0.02 𝑥 0.06𝑥 − 0.02𝑦 0
𝜆 = 1: (𝜆𝐼 − 𝑀) [𝑦] = [ ][ ] = [ ]=[ ],
−0.06 0.02 𝑦 −0.06𝑥 + 0.02𝑦 0
0.06𝑥 − 0.02𝑦 = 0
𝑦 = 3𝑥.
Let 𝑥 = 1, then 𝑦 = 3.
𝟏
Hence, the eigenvectors for 𝜆 = 1 are 𝒗𝟏 = 𝒕 [ ] , 𝒕 ≠ 𝟎
𝟑
𝑥 −0.02 −0.02 𝑥 −0.02𝑥 − 0.02𝑦 0
𝜆 = 0.92: (𝜆𝐼 − 𝑀) [𝑦] = [ ][ ] = [ ]=[ ],
−0.06 −0.06 𝑦 0.06𝑥 − 0.06𝑦 0
−0.02𝑥 − 0.02𝑦 = 0
𝑦 = −𝑥.
Let 𝑥 = 1, then 𝑦 = −1.

𝟏
Hence, the eigenvectors for 𝜆 = 0.92 are 𝒗𝟎.𝟗𝟐 = 𝒕 [ ] , 𝒕 ≠ 𝟎
−𝟏
The gene is in its normal state when 𝑛 = 0. Calculate the probability of it being in its normal state
(d.i) when 𝑛 = 5. [2]
0.94 0.02 5 1 0.744

[ ] [ ]=[ ]. Hence, after 5 divisions, the gene has a 74.4% probability of still being in normal state.
0.06 0.98 0 0.256
(d.ii) in the long term. [2]
0.94 0.02 100 1 0.250

Either: [ ] [ ]=[ ], so the probability of being in normal state in the long term is 25%.
0.06 0.98 0 0.750
1
0.25
Or: the normalized eigenvector for 𝜆 = 1 is [1+3
3 ] =[ ]. So, again, we see that the probability is 25%.
0.75
1+3
Two revision problems
R1: Differentiation
Let 𝑓(𝑥) = 3 cos(𝑥 2 + 1).
a. Find 𝑓 ′ (𝑥) = −3 sin(𝑥 2 + 1) × 2𝑥 = −𝟔𝒙 𝐬𝐢𝐧(𝒙𝟐 + 𝟏)
b. 𝑓 has a local extremum when 𝑥 = 𝑎, where 0 < 𝑎 < 2. Find the exact value of 𝑎.
Local extremum (minimum or maximum) means 𝑓 ′ (𝑎) = 0
−6𝑎 sin(𝑎2 + 1) = 0, so (by the null factor law) either 𝑎 = 0 or 𝑎2 + 1 = 0, ±𝜋, ±2𝜋, etc.
If 0 < 𝑎 < 2, then 0 < 𝑎2 + 1 < 5, and so we see that the only possible value for 𝑎2 + 1 is 𝜋.
𝑎2 + 1 = 𝜋,
𝑎 = ±√𝜋 − 1.
But 𝑎 is positive, so 𝒂 = √𝝅 − 𝟏
c. A plot of the graph for 𝑓 will immediately reveal that 𝑓(𝑎) is a minimum. However, for the sake of
practice, use the second derivative test to show that this indeed the case.
𝑑
𝑓 ′′ (𝑥) = (−6𝑥 sin(𝑥 2 + 1)) = −6 sin(𝑥 2 − 1) − 6𝑥 cos(𝑥 2 + 1) × 2𝑥
𝑑𝑥
𝑓 ′′ (𝑥) = −6 sin(𝑥 2 − 1) − 12𝑥 2 cos(𝑥 2 + 1)

When 𝑥 = 𝑎 = √𝜋 − 1, then 𝑥 2 + 1 = 𝜋, so
𝑓 ′′ (𝑎) = −6 sin(𝜋) − 12(𝜋 − 1) cos(𝜋) = 12(𝜋 − 1) > 0.
So, the graph of 𝑓 is concave up at 𝑥 = 𝑎, and hence there is a minimum there, QED.
R2: Solving nasty equations graphically
Use a graphical method the solve the equation 3 cos(𝑥 2 + 1) = 2𝑥 + 1.
Remember that this involves sketching the graphs of the left-hand side and of the right-hand side of the
equations.
Do not write any numbers on the axes other that the absolutely necessary ones.
Also remember that the sketch of the graphs is not the answer to the question…
There are three points of intersection as shown. So the solutions to the equation are
𝒙 = −𝟏. 𝟔𝟖 or 𝒙 = −𝟎. 𝟗𝟐𝟕 or 𝒙 = 𝟎. 𝟐𝟑𝟖

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Mathematics Applications and Interpretations HL

The use of GDC is allowed and encouraged for all problems

In 3a below, just do what it asks

(a.ii) What does 𝑏 represent in this context?

(b) Find the eigenvalues of 𝐌. [3]

(c) Find the eigenvectors of 𝐌. [3]

Let 𝑥 = 1, then 𝑦 = −1.

(d.i) when 𝑛 = 5. [2]

0.94 0.02 5 1 0.744

0.94 0.02 100 1 0.250

𝑓 ′′ (𝑥) = −6 sin(𝑥 2 − 1) − 12𝑥 2 cos(𝑥 2 + 1)

𝒙 = −𝟏. 𝟔𝟖 or 𝒙 = −𝟎. 𝟗𝟐𝟕 or 𝒙 = 𝟎. 𝟐𝟑𝟖

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