Question

Linear Algebra Question

Answer 1

\[A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right] has determinant -3\]

Answer 2

\[B = \left[\begin{matrix}x & y \\ c & d \end{matrix}\right] has determinant 2\]

Answer 3

\[C = \left[\begin{matrix}3c & 3d \\ a + 2x & b + 2y\end{matrix}\right]\]

Answer 4

Find the determinant of \[2adj(C ^{-1})\]

Answer 5

@Hero @dpaInc @cwrw238

Answer 6

Btw I know what the answer is supposed to be and I've gone over this question 3 times. I want to know what you guys get to check if perhaps the key is wrong

Answer 7

is it just 2 ?

Answer 8

im using a property which adj(A) = det(A) * A^-1

Answer 9

No, 2 isn't even one of the choices and I got something different

Answer 10

@estudier @robtobey

Answer 11

so what is wrong here ?
adj(C) = det(C) * C^-1
so adj(C^-1) = det(C^-1) * C
so your question becomes :
det(2adj(C^-1)) = det(2*det(C^-1) *C) =
2 * Det(C^-1) * Det(C) = 2

Answer 12

mm wait

Answer 13

The purpose is to find the determinant of the adjoint and the formula for that is
det(adj(A)) = det(A)^(n-1)
You do not know the inverse of C, nor the determinant of either, that is what makes the question challenging.

Answer 14

The determinant is a number. The answer is supposed to be (-4/3) although me and my friend get (-4/5)

Answer 15

ops finding determinant

Answer 16

The matrix resulting from A+2B is only one row operation away from C and I think that's how the question has to be started to solve

Answer 17

@Agent_Sniffles @kymber

Answer 18

Anyone?

Answer 19

Calculating 2Adj(C^(-1) for a general matrix Z, we get that it is equal to 2/Det(Z).
For that C you gave us, its determinant is:\[Det(C)=3cb+6cy-3da-6dx=-3(da-cb)-6(dx-cy)=-3\].Therefore, \[2Adj(C^{-1})=-\frac{ 2 }{ 3 }\]

Answer 20

Is that the answer you had in your key?

Answer 21

The answer in the key is -4/3

Answer 22

You did it in an interesting way though

Answer 23

I think you missed the part where 2 has to be squared

Answer 24

Yeah
When you take the 2 out, it has to be squared

Answer 25

But thanks that was a very good answer

Answer 26

Your welcome, and thank you for the medal. I did it again and I didn't find the part where the 2 is squared.

Answer 27

because we are trying to find the determinant of 2adj(C^-1)
so det(2adjC-1) = 2^2 * det(adjC-1)

Answer 28

Oh, got it. I had put the 2 out of the determinant. Nice problem.