| Date | None Specimen | Marks available | 3 | Reference code | SPNone.1.hl.TZ0.9 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Determine | Question number | 9 | Adapted from | N/A |
Question
Consider the system of equations \[\left( {\begin{array}{*{20}{c}}
1&{ - 1}&2\\
2&2&{ - 1}\\
3&5&{ - 4}\\
3&1&1
\end{array}} \right)\left( \begin{array}{l}
x\\
y\\
z
\end{array} \right) = \left( \begin{array}{l}
5\\
3\\
1\\
k
\end{array} \right) .\]
By reducing the augmented matrix to row echelon form,
(i) find the rank of the coefficient matrix;
(ii) find the value of \(k\) for which the system has a solution.
For this value of \(k\) , determine the solution.
Markscheme
reducing to row echelon form
\(\begin{array}{*{20}{ccc|c}}
1&{ - 1}&2&5 \\
0&4&{ - 5}&{ - 7} \\
0&8&{ - 10}&{ - 14} \\
0&4&{ - 5}&{k - 15}
\end{array}\) (M1)(A1)
\(\begin{array}{*{20}{ccc|c}}
1&{ - 2}&2&5 \\
0&4&{ - 5}&{ - 7} \\
0&0&0&0 \\
0&0&0&{k - 8}
\end{array}\) A1
(i) this shows that the rank of the matrix is \(2\) A1
(ii) the equations can be solved if \(k = 8\) A1
[5 marks]
let \(z = \lambda \) A1
then \(y = \frac{{5\lambda - 7}}{4}\) A1
and \(x = \left( {5 - 2\lambda + \frac{{5\lambda - 7}}{4} = } \right)\frac{{13 - 3\lambda }}{4}\) A1
Note: Accept equivalent expressions.
[3 marks]
