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further maths matrices help

Hi, I'm doing my homework on matrices and I'm struggling with two questions. The matrix (2 -3)
(a 6)
M is singular.
(a) Find the value of a.
(b) Show that M maps every point on the plane to a point on a straight line, and find the equation of this line
Ive got a as -4, but I'm unsure how to do part b?

Also, A transformation T is represented by the matrix M where (2 1 k)
(3 0 k+1)
(k-1 1 3)
The transformation T is applied to the unit cube.
(a) Given that k = 4, find the volume of the image of the unit cube.
(b) Find the values of k for which the image of the unit cube lies within a plane.
I got the volume as 8 but i have absolutely no idea what part b even means, so i don't know how to go about it?
Original post by Random11_206
Hi, I'm doing my homework on matrices and I'm struggling with two questions. The matrix (2 -3)
(a 6)
M is singular.
(a) Find the value of a.
(b) Show that M maps every point on the plane to a point on a straight line, and find the equation of this line
Ive got a as -4, but I'm unsure how to do part b?

Also, A transformation T is represented by the matrix M where (2 1 k)
(3 0 k+1)
(k-1 1 3)
The transformation T is applied to the unit cube.
(a) Given that k = 4, find the volume of the image of the unit cube.
(b) Find the values of k for which the image of the unit cube lies within a plane.
I got the volume as 8 but i have absolutely no idea what part b even means, so i don't know how to go about it?

First b) what happens when M multiplies a column vector (x,y). How would you interpret it?
Second b) similar to the first part, a 3D cube is projected onto a plane when the matrix is .... and what are the values of k?
Could you try and explain it in a different way, I’m still quite confused?
Original post by Random11_206
Could you try and explain it in a different way, I’m still quite confused?


First part b) what do you get when you mutiply an arbitrary point (x,y) by M? Can you intepret it as being a point on a line?
You get 2x - 3y = ? And -3x +6y = ? So could you get the line y= -2x? Or is that wrong?
Original post by Random11_206
You get 2x - 3y = ? And -3x +6y = ? So could you get the line y= -2x? Or is that wrong?

So as you say you get the point (X,Y)
X = 2x - 3y
Y = -4x + 6y = -2X
so apart from a typo (-4) and slightly "confusing" x,y and X,Y then yes.

second b) is similar idea but a projection from 3D space to 2D plane via a singular matrix.
(edited 1 year ago)
Okay, thanks so much, so for the second one how would you go from 3d to 2d, would you use x y z and eliminate the z? I’m not sure?
(edited 1 year ago)
Does the matrix have to be singular? So you use the detriment and get k=3 or k=-4?
Original post by Random11_206
Does the matrix have to be singular? So you use the detriment and get k=3 or k=-4?


Sounds about right. The unit cube has points (1,0,0), (0,1,0), (0,0,1) and when you multiply them by the matrix they give prouce the matrix columns as outputs. The only way these will lie on a plane is if they're linear combinations of each other so the matrix has to be singular.
(edited 1 year ago)
(a)The determinant of a 2x2 matrix is given by the formula:

det(M) = ad - bc

where a, b, c, and d are the elements of the matrix. In this case, we have:

det(M) = (2)(6) - (-3)(a) = 12 + 3a = 0

Solving for a we get a = -4
(b) the matrix transformation of a point (x,y) is given by:

(x' y') = M(x y)

(x' y') = (2 -3)(x) + (a 6)(y)

= (2x - 3y) + (ax + 6y)

= (2x - 3y + ay + 6y)

= (2x + (a-3)y + 6y)

so the equation of the line that is the image of the matrix transformation is:

y' = (2x + (a-3)y + 6y)

Now since a = -4
y' = (2x + (-4-3)y + 6y) = 2x - y + 6y = (2x+5y)

So, the equation of the straight line is 2x + 5y = 0, and this is the image of every point on the plane after the matrix transformation.
Original post by nephwriter
(a)The determinant of a 2x2 matrix is given by the formula:

det(M) = ad - bc

where a, b, c, and d are the elements of the matrix. In this case, we have:

det(M) = (2)(6) - (-3)(a) = 12 + 3a = 0

Solving for a we get a = -4
(b) the matrix transformation of a point (x,y) is given by:

(x' y') = M(x y)

(x' y') = (2 -3)(x) + (a 6)(y)

= (2x - 3y) + (ax + 6y)

= (2x - 3y + ay + 6y)

= (2x + (a-3)y + 6y)

so the equation of the line that is the image of the matrix transformation is:

y' = (2x + (a-3)y + 6y)

Now since a = -4
y' = (2x + (-4-3)y + 6y) = 2x - y + 6y = (2x+5y)

So, the equation of the straight line is 2x + 5y = 0, and this is the image of every point on the plane after the matrix transformation.

Best not to post solutions and let the OP try and work it out.
Looks incorrect anyway.
Reply 11
Original post by mqb2766
So as you say you get the point (X,Y)
X = 2x - 3y
Y = -4x + 6y = -2X
so apart from a typo (-4) and slightly "confusing" x,y and X,Y then yes.

second b) is similar idea but a projection from 3D space to 2D plane via a singular matrix.

so what actually is the answer?
Reply 12
Original post by Tomcm1485
so what actually is the answer?

Which part and what have you done?
Reply 13
Original post by mqb2766
Which part and what have you done?

part b of the first question posted and so far i got X=2x-3y and Y=-4x+6y i am unsure of how to proceed
Reply 14
Original post by Tomcm1485
part b of the first question posted and so far i got X=2x-3y and Y=-4x+6y i am unsure of how to proceed

If the projected point lies on a line, then you must have
Y = mX
for all points (x,y). Is this the case, can you determine m?

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