Question 15:
15 | (a) | 1 𝑘 3
|3 4 2 |
1 3 −1
d e t M = 1  ( − 1 0 ) − k  ( − 5 ) + 3  5
= 5 k + 5
So planes meet at a point except when k = − 1 | M1
M1
A1
A1 | 2.1
2.1
1.1
2.2a | Considering correct determinant
A correct method to find the determinant, allow one slip.
Must contain a sum of three terms.
Could be implied by 𝑘 = −1 if working with det 𝐌 = 0
Must make it clear that the planes do meet at a point for
all values of 𝑘 other than −1. Do not accept “solution” for
“point”.
SC B2 if 5𝑘+5 found without working and a correct
conclusion given.
Alternative method | B2*
−2𝑘2+15𝑘+17 −7−7𝑘 3𝑘2−𝑘−4
𝑥 = ,𝑦 = ,𝑧 =
5𝑘+5 5𝑘+5 5𝑘+5 | Correctly finding 𝑥, 𝑦 or 𝑧 in terms of 𝑘 using
simultaneous equations
Undefined when 𝑘 = −1 | B1dep
So planes meet at a point except when k = − 1 | B1 | www. Must make it clear that the planes do meet at a
point for all values of 𝑘 other than −1. Do not accept
“solution” for “point”.
All three previous marks must have been awarded.
[4]
B2*
−10 𝑘+9 2𝑘−12
1
𝐌−1 = ( 𝟓 −𝟒 𝟕 )
5𝑘+5
5 𝑘−3 4−3𝑘
 x   − 1 0 k + 9 2 k − 1 2   1 
1
y = 5 − 4− 7− 3
5 k + 5
z 5 k 3 4 3 k − k
− 7 − 7 k 7
y = = − which is independent of k
5 k + 5 5 | M1
A1
M1
A1
M1
A1 | 3.1a
1.1
1.1
1.1
1.1
3.2a | Finding at least two correct cofactors (could be in matrix
of cofactors or adj 𝐌 or not in a matrix)
All cofactors in bold correct
1
Cofactor matrix transposed and multiplying by their
det𝐌
(could be seen in later calculation)
Correct inverse matrix (ignore first and third rows)
soi (may only see middle row). Dependent on at least M1
scored. Do not accept −1 substituted for 𝑘, this is not MR.
www, any values given in 𝐌−1 and any 𝑥 or 𝑧 coordinates
given must be correct. Statement of independence
required.
Alternative method 1 | NB This work may be seen in 15(a)
5x+10y=3−2k | 5x+10y=3−2k | M1* | M1* | Eliminating 𝑥 or 𝑧 using two equations. Allow one slip
only. Do not accept −1 substituted for 𝑘, this is not MR.
A1 | Correct elimination
4𝑥+(9+𝑘)𝑦 = 1−3𝑘 | 4𝑥+(9+𝑘)𝑦 = 1−3𝑘 | M1* | M1* | Eliminating same variable using different pair of
equations. Allow one slip only. Do not accept −1
substituted for 𝑘, this is not MR.
A1 | Correct elimination
(5+5𝑘)𝑦 = −7𝑘−7 | M1dep | Eliminating correctly to leave in terms of 𝑦 and 𝑘 only
− 7 − 7 k 7
y = = − which is independent of k
5 k + 5 5 | A1cao | www, any 𝑥 or 𝑧 coordinates given must be correct.
Statement of independence required.
Alternative method 2 | NB This work may be seen in 15(a)
5 x + 1 0 y = 3 − 2 k | M1* | Eliminating x, y or z. Allow one slip only. Do not accept
−1 substituted for 𝑘, this is not MR.
3−2𝑘−10𝑦
𝑥 =
5 | A1 | Finding (e.g.) x in terms of y and k.
(𝑘−3)𝑦+4𝑧 = 1+𝑘 | M1* | Eliminating another variable. Allow one slip only. Do not
accept −1 substituted for 𝑘, this is not MR.
1+𝑘−𝑘𝑦+3𝑦
𝑧 =
4 | A1 | Correctly
3 − 2 k − 1 0 y 3 (1 + k − k y + 3 y )
+ k y + = 1
5 4 | M1dep | Finding an equation for y in terms of k correctly
− 7 − 7 k 7
y = = −
which is independent of k
5 + 5 k 5 | A1cao | www, any 𝑥 or 𝑧 coordinates given must be correct.
Statement of independence required.
[6]
www, any 𝑥 or 𝑧 coordinates given must be correct.
Statement of independence required.