# IB DP Math AA :Topic : AHL 1.16 :Solutions of systems of linear equations: IB Style Questions- HL Paper 2

## Question

(a)     Find the set of values of k for which the following system of equations has

no solution.

x + 2y − 3z = k

3x + y + 2z = 4

5x + 7z = 5

(b)     Describe the geometrical relationship of the three planes represented by this system of equations.

## Markscheme

(a)

$$\left( {\begin{array}{*{20}{c}} 1&2&{ – 3}&k \\ 3&1&2&4 \\ 5&0&7&5 \end{array}} \right)$$     M1

$${R_1} – 2{R_2}$$

$$\left( {\begin{array}{*{20}{c}} { – 5}&0&{ – 7}&{k – 8} \\ 3&1&2&4 \\ 5&0&7&5 \end{array}} \right)$$     (A1)

$${R_1} + {R_3}$$

$$\left( {\begin{array}{*{20}{c}} 0&0&0&{k – 3} \\ 3&1&2&4 \\ 5&0&7&5 \end{array}} \right)$$     (A1)

Hence no solutions if $$k \in \mathbb{R}$$, $$k \ne 3$$     A1

(b)     Two planes meet in a line and the third plane is parallel to that line.

[5 marks]

## Examiners report

Most candidates realised that some form of row operations was appropriate here but arithmetic errors were fairly common. Many candidates whose arithmetic was correct gave their answer as k = 3 instead of $$k \ne 3$$ . Very few candidates gave a correct answer to (b) with most failing to realise that stating that there was no common point was not enough to answer the question.

## Question

The system of equations

$2x – y + 3z = 2$

$3x + y + 2z = – 2$

$– x + 2y + az = b$

is known to have more than one solution. Find the value of a and the value of b.

## Markscheme

EITHER

using row reduction (or attempting to eliminate a variable)     M1

$$\left( {\begin{array}{*{20}{ccc|c}} 2&{ – 1}&3&2 \\ 3&1&2&{ – 2} \\ { – 1}&2&a&b \end{array}} \right)\begin{array}{*{20}{l}} {} \\ { \to 2R2 – 3R1} \\ { \to 2R3 + R1} \end{array}$$

$$\left( {\begin{array}{*{20}{ccc|c}} 2&{ – 1}&3&2 \\ 0&5&{ – 5}&{ – 10} \\ 0&3&{2a + 3}&{2b + 2} \end{array}} \right)\begin{array}{*{20}{l}} {} \\ { \to R2/5} \\ {} \end{array}$$     A1

Note: For an algebraic solution award A1 for two correct equations in two variables.

$$\left( {\begin{array}{*{20}{ccc|c}} 2&{ – 1}&3&2 \\ 0&1&{ – 1}&{ – 2} \\ 0&3&{2a + 3}&{2b + 2} \end{array}} \right)\begin{array}{*{20}{l}} {} \\ {} \\ { \to R3 – 3R2} \end{array}$$

$$\left( {\begin{array}{*{20}{ccc|c}} 2&{ – 1}&3&2 \\ 0&1&{ – 1}&{ – 2} \\ 0&0&{2a + 6}&{2b + 8} \end{array}} \right)$$

Note: Accept alternative correct row reductions.

recognition of the need for 4 zeroes     M1

so for multiple solutions a = – 3 and b = – 4     A1A1

[5 marks]

OR

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

$$\Rightarrow 2(a – 4) + (3a + 2) + 3(6 + 1) = 0$$

$$\Rightarrow 5a + 15 = 0$$

$$\Rightarrow a = – 3$$     A1

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

$$\Rightarrow 2(b + 4) + (3b – 2) + 2(6 + 1) = 0$$     A1

$$\Rightarrow 5b + 20 = 0$$

$$\Rightarrow b = – 4$$     A1

[5 marks]

## Examiners report

Many candidates attempted an algebraic approach that used excessive time but still allowed few to arrive at a solution. Of those that recognised the question should be done by matrices, some were unaware that for more than one solution a complete line of zeros is necessary.

## Question

The three planes

$$2x – 2y – z = 3$$

$$4x + 5y – 2z = – 3$$

$$3x + 4y – 3z = – 7$$

intersect at the point with coordinates (a, b, c).

a.Find the value of each of a, b and c.[2]

b.

The equations of three planes are

$$2x – 4y – 3z = 4$$

$$– x + 3y + 5z = – 2$$

$$3x – 5y – z = 6$$.

Find a vector equation of the line of intersection of these three planes.[4]

## Markscheme

(a)     use GDC or manual method to find a, b and c     (M1)

obtain $$a = 2,{\text{ }}b = – 1,{\text{ }}c = 3$$ (in any identifiable form)     A1

[2 marks]

a.

use GDC or manual method to solve second set of equations     (M1)

obtain $$x = \frac{{4 – 11t}}{2};{\text{ }}y = \frac{{ – 7t}}{2};{\text{ }}z = t$$ (or equivalent)     (A1)

$$r = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 0 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { – 5.5} \\ { – 3.5} \\ 1 \end{array}} \right)$$ (accept equivalent vector forms)     M1A1

Note: Final A1 requires r = or equivalent.

[4 marks]

b.

## Examiners report

Generally well done.

a.

Moderate success here. Some forgot that an equation must have an = sign.

b.
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