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.