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Matrices question

Screenshot 2024-04-05 at 16.17.54.png

I have done the first two parts of this question correctly (I don't have access to the mark scheme to this question but I checked my answers with a Wolfram Alpha widget and it worked out correctly).

How should I approach part iii of this question?

It clearly says that k can't be 5 at the start of the question and in part c, the k has been replaced with 5. This makes the determinant 0 so I'm having problems understanding how to work out the solutions to this.
Reply 1
Original post by vnayak
Screenshot 2024-04-05 at 16.17.54.png
I have done the first two parts of this question correctly (I don't have access to the mark scheme to this question but I checked my answers with a Wolfram Alpha widget and it worked out correctly).
How should I approach part iii of this question?
It clearly says that k can't be 5 at the start of the question and in part c, the k has been replaced with 5. This makes the determinant 0 so I'm having problems understanding how to work out the solutions to this.

Sometimes its easier to go back to the geometry, so each "row" represents one simulteaneous equation or a plane of solutions. When k=5, the normal of the first plane (the matrix coefficients) is a linear combination of the other two normals so theyre not independent. So the only way youll get a line of solutions is if you dont have a triangular prism of solutions and the distances from the origin (the right hand side values) match (so a sheaf) so youll get a line of solutions.

So whatever linear combination of the first and third row give the second, then you must have the same for the right hand side to get p. Then the solutions will be a line as you effectively have two equations in 3 variables (sheaf). So somehting like
https://pmt.physicsandmathstutor.com/download/Maths/A-level/Further/Core-Pure/AQA/Cheat-Sheets/Matrices%20III.pdf
(edited 3 months ago)
Reply 2
vnayak
..
To be honest, the simplest thing for the last part is probably "forget everything you know about matrices and just solve the problem using basic algebra".

To be a little more specific (if you've never solved 3 simultaneous equations in 3 unknowns): if you multiply the 2nd equation by 4, and then subtract 3 times the first equation from the 2nd equation (to get a new equation Eq1 that doesn't involve x), and similarly subtract twice the first equation from the 3rd equation to get an equation Eq2, then Eq1 & Eq2 form a pair of simultaneous equations in 2 unknowns (they only refer to y and z, not x). Then just solve these normally (you'll find your equations are inconsistent (have no solutions) except for a particular value of p).

[In fact, if you didn't already have the inverse of the matrix for (ii), it would probably be quicker to use the same approach ("just do the algebra") than to invert the matrix. It's certainly less arithmetic operations].
(edited 3 months ago)
Reply 3
Original post by DFranklin
To be honest, the simplest thing for the last part is probably "forget everything you know about matrices and just solve the problem using basic algebra".
To be a little more specific (if you've never solved 3 simultaneous equations in 3 unknowns): if you multiply the 2nd equation by 4, and then subtract 3 times the first equation from the 2nd equation (to get a new equation Eq1 that doesn't involve x), and similarly subtract twice the first equation from the 3rd equation to get an equation Eq2, then Eq1 & Eq2 form a pair of simultaneous equations in 2 unknowns (they only refer to y and z, not x). Then just solve these normally (you'll find your equations are inconsistent (have no solutions) except for a particular value of p).
[In fact, if you didn't already have the inverse of the matrix for (ii), it would probably be quicker to use the same approach ("just do the algebra") than to invert the matrix. It's certainly less arithmetic operations].

Thanks for clearing it up! Yeah the question had me stumped, I'll admit, and your method makes sense so thanks!
Reply 4
Original post by vnayak
Thanks for clearing it up! Yeah the question had me stumped, I'll admit, and your method makes sense so thanks!

A similar way to get the row multipliers so
row2 = alpha*row1 + beta*row3
would be just solve (x,y coefficients)
A [alpha, beta] = b
where A = [[4 8], [1 5]], b = [3, 2] and just invert/solve. Then check the z coefficient matches the same combination (so the equations are linearly dependent) and use them for the right hand side. In this case only alpha is necessary and you can get p with a (longish) one line.
(edited 3 months ago)
Reply 5
Original post by mqb2766
A similar way to get the row multipliers so
row2 = alpha*row1 + beta*row3
would be just solve (x,y coefficients)
A [alpha, beta] = b
where A = [[4 8], [1 5]], b = [3, 2] and just invert/solve. Then check the z coefficient matches the same combination (so the equations are linearly dependent) and use them for the right hand side.

Thanks!

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