# Homework Help: Statement on matrix and determinant

1. Apr 27, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data
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
2. Relevant equations
NA

3. 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?

2. Apr 27, 2015

### Orodruin

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

3. Apr 27, 2015

### Raghav Gupta

I know det(AT) = 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?

4. Apr 27, 2015

### Orodruin

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

5. Apr 27, 2015

### Raghav Gupta

Is it the identity matrix $I$ ?

6. Apr 27, 2015

### PeroK

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!

7. Apr 27, 2015

### Orodruin

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

8. Apr 27, 2015

### Raghav Gupta

No. We get A only.

9. Apr 27, 2015

### Orodruin

Staff Emeritus
So how would you get 3A?

10. Apr 28, 2015

### Raghav Gupta

By multiplying A with 3 or 3$I$

11. Apr 28, 2015

### Orodruin

Staff Emeritus
Yes, try the latter and apply your determinant relations.

12. Apr 28, 2015

### Raghav Gupta

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

13. Apr 28, 2015

### Orodruin

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

14. Apr 28, 2015

### Raghav Gupta

27

15. Apr 28, 2015

### Orodruin

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

16. Apr 28, 2015

### Raghav Gupta

8det(A).Option 4 rejected.
Option 1 is okay.

17. Apr 28, 2015

### Ray Vickson

You tell us.

18. Apr 28, 2015

### Raghav Gupta

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

19. Apr 28, 2015

### Orodruin

Staff Emeritus
Why dont you try insering a well chosen matrix and see what comes out? I suggest one which will make the determinants easy to compute.

20. Apr 28, 2015

### Raghav Gupta

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?

21. Apr 28, 2015

### Orodruin

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

22. Apr 28, 2015

### Ray Vickson

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.

23. Apr 28, 2015

### Orodruin

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

24. Apr 28, 2015

### Fredrik

Staff Emeritus
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).