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System of equations question

I'm doing this question right now, and understand that a = -3 as the matrix must be singular, but I don't get why b = 13 as the matrix is singular so we can't inverse it, except that's what the mark scheme's done.

Question:
Consider this system of equations

2x+3y+z=6
−x+y+2z=7
ax+y+4z=b

Find the values of a
and b
for which the system has an infinite number of solutions.

Mark scheme (Q13): https://www.activeteachonline.com/default/player/document/id/721178/external/0/uid/357726
Reply 1
Original post by Amy.fallowfield
I'm doing this question right now, and understand that a = -3 as the matrix must be singular, but I don't get why b = 13 as the matrix is singular so we can't inverse it, except that's what the mark scheme's done.

Question:
Consider this system of equations

2x+3y+z=6
−x+y+2z=7
ax+y+4z=b

Find the values of a
and b
for which the system has an infinite number of solutions.

Mark scheme (Q13): https://www.activeteachonline.com/default/player/document/id/721178/external/0/uid/357726


If this matrix is singular, the third row is a linear combination of the first two rows. For the system of equations (including the right hand side) to be consistent, b must be the same linear combination of the first two elements of the right hand side. If not, the equations will be inconsistent and there will be no solutions.

So the first two planes/equations intersect in a line and you want the third plane/equation to include that line. You can do it using the adjugate (transpose of the cofactor matrix) - what does your textbook say about it?

Alternatively, if you consider the sub problem of finding a point on the solution line which is the intersection of the 3 planes (z=0, first two equations)
2x+3y=6
-x+y=7
fairly easily you get x=-3 y=4 which you can sub into the last equation to get b=13. You could similarly do x=0 and solve for y,z which would mean that youre ignoring the value of a.

Alternatively, if you can spot the multipliers for the first two rows that give the third, you can use that to get b (and a). So you solve (y and z coeffs)
3r + s = 1
r + 2s = 4
where r is the multiplier of equation 1 and s is the multiplier of equation 2. So r=-2/5 and s=11/5, which gives a=-3 and b=13. If you notice the terms in the adjugate matrix, theyre effectively these mutlipliers and if you calc the first column which becomes the first row of the adjugate, then it doesnt depend on a.

Edit - to come back to your original question (not sure what it says in the book) but if B = adj(A), so the adjugate matrix, then for every square matrix (including singular) you have
BA = AB = |A|I
Post multiplying by the vector x (= (x,y,z)^T.) and in this case its singular so |A|=0, so
BAx = 0
and the equations are consistent so Ax = c, where c is the vector (6,7,b)^T
Bc = 0
which is the condition you get to at the end of the model solution. Its worth noting that this only works if there is 1 equation which is a linear combination of the others. If there are two or more, the adjugate matrix is zero.
(edited 3 months ago)

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