Statement on matrix and determinant

[Raghav Gupta]

Homework Statement

If A is a square matrix of order 3 then the true statement is
1. det(-A) = - det A
2.det A = 0
3.det ( A + I) = I + detA
4.det(2A) = 2detA

Homework Equations

NA

The Attempt at a Solution

2. option is obviously not true.
Making a random matrix A and verifying properties 1. , 3. , 4. would be lengthy.
What is the approach for this?

 

[Orodruin]

What is the approach for this?

Did you try simply using some basic rules for determinants? How would you write the determinant of a matrix in terms of its components?

 

[Raghav Gupta]

Did you try simply using some basic rules for determinants? How would you write the determinant of a matrix in terms of its components?

I know det(A^T) = det(A)
for equal size square matrices,
det(A)(B) = det(A)det(B),
What property should I apply as these I think are not applicable here?

 

[Orodruin]

det(A)(B) = det(A)det(B),

Can you write multiplication by a constant as a multiplication with a matrix? In that case, what matrix?

 

[Raghav Gupta]

Can you write multiplication by a constant as a multiplication with a matrix? In that case, what matrix?

Is it the identity matrix $I$ ?

 

[PeroK]

Is it the identity matrix $I$ ?

Why not take $A = I$, and calculate the determinants required? Just to see what happens. You are allowed to try things out with specific matrices!

 

[Orodruin]

Do you get 3A by multiplying A with the identity?

 

[Raghav Gupta]

Do you get 3A by multiplying A with the identity?

No. We get A only.

 

[Orodruin]

So how would you get 3A?

 

[Raghav Gupta]

So how would you get 3A?

By multiplying A with 3 or 3$I$

 

[Orodruin]

By multiplying A with 3 or 3$I$

Yes, try the latter and apply your determinant relations.

 

[Raghav Gupta]

Yes, try the latter and apply your determinant relations.

But by that I am getting options 1 and 4 true.
Among that one is supposed to be true.

 

[Orodruin]

You are then doing it wrong. What is the determinant of 3I?

 

[Raghav Gupta]

You are then doing it wrong. What is the determinant of 3I?

27

 

[Orodruin]

27

Yes, so what is then the determinant of 2A expressed in det(A)?

 

[Raghav Gupta]

Yes, so what is then the determinant of 2A expressed in det(A)?

8det(A).Option 4 rejected.
Option 1 is okay.
What about option 3 ?

 

[Ray Vickson]

8det(A).Option 4 rejected.
Option 1 is okay.
What about option 3 ?

You tell us.

 

[Raghav Gupta]

You tell us.

But we can't apply the property there of
det(A)det(B) = det(A)(B) as there is no use of it?

 

[Orodruin]

Why don't you try insering a well chosen matrix and see what comes out? I suggest one which will make the determinants easy to compute.

 

[Raghav Gupta]

Why don't you try insering a well chosen matrix and see what comes out? I suggest one which will make the determinants easy to compute.

Oh, thanks that option 3 is also false.
But I have verified that by taking any random matrix.
Is there any general proof of that or we learn it in higher sections?

 

[Orodruin]

To prove that something is false you only need a counter example.

 

[Ray Vickson]

Oh, thanks that option 3 is also false.
But I have verified that by taking any random matrix.
Is there any general proof of that or we learn it in higher sections?

Right: not only is (3) false, it does not even have any meaning. The left-hand-side det(A+I) is a number, while the right-hand-side det(A) + I is a number plus a matrix---which does not exist in any reasonable way.

 

[Orodruin]

Right: not only is (3) false, it does not even have any meaning. The left-hand-side det(A+I) is a number, while the right-hand-side det(A) + I is a number plus a matrix---which does not exist in any reasonable way.

I always took the I there to mean 1, precisely since it does not have any meaning otherwise ...

 

[Fredrik]

In some contexts (especially in books on functional analysis) it's considered acceptable to write down equations of the type "operator = number" and sums of the type "operator + number". The number is then interpreted as that number times the identity operator. So (3) could be interpreted as $\det(A+I)I=I+(\det A)I$, but it seems very unlikely that this is the intended interpretation. I'm inclined to go with Ray Vickson's interpretation first (the equality is nonsense), and Orodruin's interpretation second (the I on the right is supposed to be 1).