davidjohn03
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Can someone please help me with part iii of this Q I have done the rest of the parts but unsure how to get to p
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davidjohn03
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SmorL
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I only had a glance but im pretty sure it has to do with substituting 5 for K and using it with the xyz matrix
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davidjohn03
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(Original post by SmorL)
I only had a glance but im pretty sure it has to do with substituting 5 for K and using it with the xyz matrix
Sorry Im unsure what u mean by this ?
Are you looking at part i
I am stuck on part iii I’ve completed i
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mqb2766
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(Original post by davidjohn03)
Sorry Im unsure what u mean by this ?
Are you looking at part i
I am stuck on part iii I’ve completed i
Ive not worked it through, but I guess the equations will be linearly dependent / consistent.
Try and find the two row multipliers (row reduction) which will make the left hand side of one equation zero . They'll be similar to how you solved i and ii. Then look at the right hand side of that equation and the value of p which makes that zero as well will mean the equations are consistent (not inconsistent) so there are solutions.
Last edited by mqb2766; 3 weeks ago
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davidjohn03
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(Original post by mqb2766)
Ive not worked it through, but I guess the equations will be linearly dependent / consistent.
Try and find the two row multipliers (row reduction) which will make the left hand side of one equation zero . They'll be similar to how you solved i and ii. Then look at the right hand side of that equation and the value of p which makes that zero as well will mean the equations are consistent (not inconsistent) so there are solutions.
Is there any other way of finding the solution as we haven't done anything like row reduction in class before
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mqb2766
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(Original post by davidjohn03)
Is there any other way of finding the solution as we haven't done anything like row reduction in class before
What did you do for the first two parts? This one shoud be similar.

Note you could solve for the two row multipliers by solving a 2*2 system in x and y. Then use the same multipliers to directly get p on the right hand side.
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davidjohn03
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(Original post by mqb2766)
What did you do for the first two parts? This one shoud be similar.

Note you could solve for the two row multipliers by solving a 2*2 system in x and y. Then use the same multipliers to directly get p on the right hand side.
Ive found the inverse in the first part and then used matrices multiplication in the 2nd part Name:  Screenshot 2021-09-19 at 09.13.35.jpg
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mqb2766
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(Original post by davidjohn03)
Ive found the inverse in the first part and then used matrices multiplication in the 2nd part Name:  Screenshot 2021-09-19 at 09.13.35.jpg
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Ok probably the easiest way is just to note that when k=5, the left hand sides are linearly dependent (det is zero in part i). This means that equation 2 can be expressed as a linear combination of equations 1 and 3, i.e.
3x + 2y +5z = a(4x + y + 5z) + b(8x + 5y + 13z)
So use the x and y coefficients to solve for a and b. The z coefficients should be matched by definition. In some cases you can determine a and b by inspection.

Then use them to combine the right hand sides to get the value of p to make the equations consistent.
Last edited by mqb2766; 3 weeks ago
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davidjohn03
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(Original post by mqb2766)
Ok probably the easiest way is just to note that when k=5, the left hand sides are linearly dependent (det is zero in part i). This means that equation 2 can be expressed as a linear combination of equations 1 and 3, i.e.
3x + 2y +5z = a(4x + y + 5z) + b(8x + 5y + 13z)
So use the x and y coefficients to solve for a and b. The z coefficients should be matched by definition. In some cases you can determine a and b by inspection.

Then use them to combine the right hand sides to get the value of p to make the equations consistent.
Thank you !
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