# IBDP Maths analysis and approaches Topic: AHL 1.16 :Solutions of systems of linear equations HL Paper 1

### Question:

Consider any three consecutive integers, n – 1, n and n + 1.
(a) Prove that the sum of these three integers is always divisible by 3.

Ans: (n – 1) + n + (n + 1)

= 3n

which is always divisible by 3

(b) Prove that the sum of the squares of these three integers is never divisible by 3.

Ans: (n -1)2 +n2 + (n+1)2     (=n2 – 2n + 1 + n2 + n2 + 2n + 1)
attempts to expand either  (n -1)2 or  (n+1)2  ( do not accept n2 – 1  or n2 + 1)
= 3n2 + 2

demonstrating recognition that 2 is not divisible by 3 or $$\frac{2}{3}$$   seen after correct
expression divided by 3
3n2  is divisible by 3 and so 3n2 + 2 is never divisible by 3
OR the first term is divisible by 3, the second is not

OR  $$3\left ( n^{2}+\frac{2}{3} \right ) OR \frac{3n^{2} + 2}{3} = n^{2} + \frac{2}{3}$$
hence the sum of the squares is never divisible by 3

### Question

Consider the following equations, where a , $$b \in \mathbb{R}:$$

$$x + 3y + (a – 1)z = 1$$

$$2x + 2y + (a – 2)z = 1$$

$$3x + y + (a – 3)z = b.$$

a. If each of these equations defines a plane, show that, for any value of a , the planes do not intersect at a unique point.[3]

Ans:

METHOD 1

$$\det \left( {\begin{array}{*{20}{c}} 1&3&{a – 1} \\ 2&2&{a – 2} \\ 3&1&{a – 3} \end{array}} \right)$$     M1

$$= 1\left( {2(a – 3) – (a – 2)} \right) – 3\left( {2(a – 3) – 3(a – 2)} \right) + (a – 1)(2 – 6)$$

(or equivalent)     A1

= 0 (therefore there is no unique solution)     A1

[3 marks]

METHOD 2

$$\left( {\begin{array}{*{20}{c}} 1&3&{a – 1}\\ 2&2&{a – 2}\\ 3&1&{a – 3} \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ 1\\ b. Find the value of b for which the intersection of the planes is a straight line.[4] ▶️Answer/Explanatio Ans: \end{array}} \right):\left( {\begin{array}{*{20}{c}} 1&3&{a – 1}\\ 0&{ – 4}&{ – a}\\ 0&{ – 8}&{ – 2a} \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ { – 1}\\ {b – 3} \end{array}} \right)$$     M1A1

$$:\left( {\begin{array}{*{20}{c}} 1&3&{a – 1}\\ 0&{ – 4}&{ – a}\\ 0&0&0 \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ { – 1}\\ {b – 1} \end{array}} \right)$$ (and 3 zeros imply no unique solution)     A1

[3 marks]

a.

METHOD 1

$$\left( {\begin{array}{*{20}{c}} 1&3&{a – 1}\\ 2&2&{a – 2}\\ 3&1&{a – 3} \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ 1\\ b \end{array}} \right):\left( {\begin{array}{*{20}{c}} 1&3&{a – 1}\\ 0&{ – 4}&{ – a}\\ 0&{ – 8}&{ – 2a} \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ { – 1}\\ {b – 3} \end{array}} \right)$$     M1A1

$$:\left( {\begin{array}{*{20}{c}} 1&3&{a – 1}\\ 0&{ – 4}&{ – a}\\ 0&0&0 \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ { – 1}\\ {b – 1} \end{array}} \right)$$     A1

b = 1     A1     N2

Note: Award M1 for an attempt to use row operations.

[4 marks]

METHOD 2

b = 1     A4

Note: Award A4 only if “ b −1 ” seen in (a).

[4 marks]

### Question

(a)     Show that the following system of equations has an infinite number of solutions.

$$x + y + 2z = – 2$$

$$3x – y + 14z = 6$$

$$x + 2y = – 5$$

The system of equations represents three planes in space.

Ans:

EITHER

$$\left( {\begin{array}{*{20}{c}}1&1&2\\3&{ – 1}&{14}\\1&2&0\end{array}\left| \begin{array}{c} – 2\\6\\ – 5\end{array} \right.} \right) \to \left( {\begin{array}{*{20}{c}}1&1&2\\0&1&{ – 2}\\0&0&0\end{array}\left| \begin{array}{c} – 2\\ – 3\\0\end{array} \right.} \right)$$     M1

row of zeroes implies infinite solutions, (or equivalent).     R1

Note:     Award M1 for any attempt at row reduction.

OR

$$\left| {\begin{array}{*{20}{c}}1&1&2\\3&{ – 1}&{14}\\1&2&0\end{array}} \right| = 0$$     M1

$$\left| {\begin{array}{*{20}{c}}1&1&2\\3&{ – 1}&{14}\\1&2&0\end{array}} \right| = 0$$ with one valid point     R1

OR

$$x + y + 2z = – 2$$

$$3x – y + 14z = 6$$

$$x + 2y = – 5$$   $$\Rightarrow x = – 5 – 2y$$

substitute $$x = – 5 – 2y$$ into the first two equations:

$$– 5 – 2y + y + 2z = – 2$$

$$3( – 5 – 2y) – y + 14z = 6$$     M1

$$– y + 2z = 3$$

$$– 7y + 14z = 21$$

the latter two equations are equivalent (by multiplying by 7) therefore an infinite number of solutions.     R1

OR

for example, $$7 \times {{\text{R}}_1} – {{\text{R}}_2}$$ gives $$4x + 8y = – 20$$     M1

this equation is a multiple of the third equation, therefore an infinite

number of solutions.     R1

[2 marks]

(b)     Find the parametric equations of the line of intersection of the three planes.

Ans:

let $$y = t$$     M1

then $$x = – 5 – 2t$$     A1

$$z = \frac{{t + 3}}{2}$$     A1

OR

let $$x = t$$     M1

then $$y = \frac{{ – 5 – t}}{2}$$     A1

$$z = \frac{{1 – t}}{4}$$     A1

OR

let $$z = t$$     M1

then $$x = 1 – 4t$$     A1

$$y = – 3 + 2t$$     A1

OR

attempt to find cross product of two normal vectors:

eg: $$\left| {\begin{array}{*{20}{c}}i&j&k\\1&1&2\\1&2&0\end{array}} \right| = – 4i + 2j + k$$     M1A1

$$x = 1 – 4t$$

$$y = – 3 + 2t$$

$$z = t$$     A1

(or equivalent)

[3 marks]

Total [5 marks]

### Question

The following system of equations represents three planes in space.

$x + 3y + z = – 1$

$x + 2y – 2z = 15$

$2x + y – z = 6$

Find the coordinates of the point of intersection of the three planes.

### Markscheme

EITHER

eliminating a variable, $$x$$, for example to obtain $$y + 3z = – 16$$ and $$– 5y – 3z = 8$$     M1A1

attempting to find the value of one variable     M1

point of intersection is $$( – 1,{\text{ }}2,{\text{ }} – 6)$$     A1A1A1

OR

attempting row reduction of relevant matrix, eg.      M1

correct matrix with two zeroes in a column, eg.      A1

further attempt at reduction     M1

point of intersection is $$( – 1,{\text{ }}2,{\text{ }} – 6)$$     A1A1A1

Note:     Allow solution expressed as $$x = – 1,{\text{ }}y = 2,{\text{ }}z = – 6$$ for final A marks.

[6 marks]

### Question

a. Consider the following system of equations:

$x + y + z = 1$

$2x + 3y + z = 3$

$x + 3y – z = \lambda$

where $$\lambda \in \mathbb{R}$$ .

Ans:

using row operations,     M1

to obtain 2 equations in the same 2 variables     A1A1

for example $$y – z = 1$$

$$2y – 2z = \lambda – 1$$

the fact that one of the left hand sides is a multiple of the other left hand side indicates that the equations do not have a unique solution, or equivalent     R1AG

[4 marks]

b. Show that this system does not have a unique solution for any value of $$\lambda$$ .[4]

(i)     Determine the value of $$\lambda$$ for which the system is consistent.

Ans: $$\lambda = 3$$     A1

(ii)     For this value of $$\lambda$$ , find the general solution of the system.[4]

Ans:

put $$z = \mu$$     M1

then $$y = 1 + \mu$$     A1

and $$x = – 2\mu$$ or equivalent     A1

[4 marks]

### Question

Solve the following system of equations:
3x – 2y + z = -4
x + y – z = -2
2x + 3y = 4

Ans
sing an elimination method,
3x – 2y +z = -4
x  –  y  –  z = -2
4x –  y = -6
4x + 6y = 8
7y = 14
Therefore x = -1, y = 2, z = 3

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