Question 8:
8 | (a) | 2 1 3
3 − 1 − 2 [ = 0 ]
− 4 3 7
det = 0  planes do not meet at a point | M1
A1
[2] | 1.1a
1.1 | finding determinant of correct matrix
8 | (b) | Method 1: use 2 equations to find e.g. z and y in
terms of x, and substitute in 3rd equation e.g.
(1) + (2): 5x + z = 5,
 z = 5 – 5x, y = 13x – 12
substitute in (3): –4x + 39x – 36 + 35 – 35x = k
 k = −1 | M1
M1
M1
A1 | 2.1
2.1
2.1
2.2a | eliminating one variable
finding e.g. y and z in terms of x
substituting in 3rd equation
Method 2: eliminate one variable from 2 pairs of
eqns, and compare eqns in 2 remaining v’bles, e.g.
from (1) and (2): 5x + z = 5 (4)
from (1) and (3): 10x + 2z = 9 − k (5)
 9 − k = 10
 k = −1 | M1
M1
M1
A1 | eliminating one variable using two equations
eliminating same variable using two other equations
eliminating both v’bles using (4) and (5) equations (coeffs in
(4) and (5) must be correct)
Method 3: find linear combination by inspection
(3) = (1) − 2(2)
 k = 3 − 22 = −1 | M3
A1 | oe
for k = −1 unsupported, allow SCB3
Method 4: substitute a value of x, y or z , e.g.
substituting x = 0 y + 3z = 3, −y − 2z = 2
 z = 5, y = −12
substitute in (3): −36 + 35 = k
 k = −1 | M1
A1
M1
A1 | any value for x, y or z substituted
solve 2 equations for y, z
substitute values into third equation
[4]