## DLS one year ago Matrix question!

1. DLS

2. DLS

Is it the original matrix?

3. shahara12

The matrix formed by taking the transpose of the cofactor matrix of a given original matrix. The adjoint of matrix A is often written adj A

4. cap_n_crunch

It depends on your matrix for a $$2 \times 2$$ matrix, it's the original matrix.

5. DLS

How and why?What about 3x3 one?why not?

6. DLS

@shubhamsrg That solves the IIT Main question,we had to take the adjoint of that ajdoint matrix then finally the determinant then equate it to the given value If i'm not wrong/

7. cap_n_crunch

For any other $$n \times n$$ matrix it's $$\det(A)^{n-2} A$$

8. DLS

Its 3x3

9. DLS

I am given the adjoint of a matrix I want to find the original matrix.

10. cap_n_crunch

So if it's $$3 \times 3$$ i could get slightly nasty :)

11. cap_n_crunch

But we know that $$\det(\text{adj}(A)) = \det(A)^{n-1}$$

12. cap_n_crunch

Which, for your case would mean: $\det(A) = \sqrt{\det(\text{adj}(A))}$

13. DLS

:/

14. cap_n_crunch

Okay lets call your matrix $\text{adj}(A) = G$

15. cap_n_crunch

We know: $\underbrace{\text{adj}(G)}_{\text{adj}(\text{adj}(A))} = \det(A)^{n-2}A$ and $\det(G) = \det(A)^{n-1}$ So in your case it's $$3 \times 3$$. Therefore we can write: $\det(G) = \det(A)^{3-1} = \det(A)^2$ Which makes$\det(A) = \sqrt{\det(G)}$ Now we're coming back to our first formula: $\text{adj}(G) = \det(A)^{n-2} A$ We plug in the formula we got for $$\det(A):$$$\text{adj}(G) = \sqrt{\det(G)}^{n-2} A$ With $$n = 3$$ it simplifies to: $\text{adj}(G) =\sqrt{\det(G)} A$ Finally, to get $$A$$ we multiply $$\frac{1}{\sqrt{\det(G)}}$$ on both sides: $\text{adj}(G) \frac{1}{\sqrt{\det(G)}} =\sqrt{\det(G)}\frac{1}{\sqrt{\det(G)}} A$ We can do some cancellations and end up with: $A = \frac{1}{\sqrt{\det(G)}} \text{adj}(G) = \frac{1}{\sqrt{\det(G)}} \text{adj}(\text{adj}(A))$

16. DLS

oh..well.....

17. cap_n_crunch

It looks complicated but try to understand it step by step. And if you struggle with anythink let me know.