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

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]

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]