## Question

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

no solution.

*x* + 2*y* − 3*z* = *k*

3*x* + *y* + 2*z* = 4

5*x* + 7*z* = 5

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

**▶️Answer/Explanation**

## 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*.

**▶️Answer/Explanation**

## 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]

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]

**▶️Answer/Explanation**

## 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]*

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]*

## Examiners report

Generally well done.

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