Finding the determinant of adjoint matrix

[amninder15]

Any help would be appreciated.
I tried to solve this problem by first the adjoint of A but then that get really complicated I have no clue how to do this.

 

[DryRun]

Evaluate the determinant of A (work it out, even though you are given the numerical answer). Equate the expression with 4. After you've found the adjoint of A, find the determinant. Replace first expression into second one. Expand. Terms cancel out. I just worked it out, but i'll withhold on the answer, until you've shown some work.

 

[amninder15]

Ok so this is what I am doing I found the determinant of A its something like this
a(ei-fh)-b(di-gh)+c(dh-eg) = 4
Then I found the adjoint A which has terms like this ei-fh, ch-bi and so on. But now the problem I am facing is that with these type of terms finding determinant is getting really complicated I don't think I can use the row reduced method coz i can't see that there is any cancellations going on. So i tried to use the cofactor way but that's getting really long. I am not sure I am doing it right.

 

[DryRun]

For starters, your determinant of A is wrong. But there is a simpler way to solve this, using the properties of determinants. You are already given that the determinant of matrix A is 4.

From the definition of $$A^{-1}=\frac{1}{det(A)}\times adj(A)$$
Multiply by matrix A throughout and cross-multiply:
$$A\times Adj(A)=det(A)\times I$$
where I is the 3x3 identity matrix.

Now taking determinant on both sides we get,

$$det(A)\times det(Adj(A))=det(det(A)\times I)$$

Recall, for an n x n square matrix,
$$det(kB)=k^n\times det(B)$$
where k is a numerical constant.

$$det(A)\times det(Adj(A))=(det(A))^n \times det(I)$$

Since determinant of an identity matrix is 1,

$$det(A)\times det(Adj(A))=(det(A))^n$$
Therefore,
$$det(Adj(A))=\frac{(det(A))^n}{det(A)}=(det(A))^{n-1}$$

From there, you can easily find the answer.

 

[amninder15]

Thank You So much
I have exam tommorow and this was question in practise exam
and I wasnt able to do it. But you explained it real nice

Thanks!

 

[DryRun]

It's preferable to memorize the formula, which is much easier and faster than deriving it, especially for multiple choice questions:
$$det(Adj(A))=(det(A))^{n-1}$$