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.

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]

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.