If $ A=\begin{bmatrix} 1 & a & b\\ -a & -1 & c \\ -b & -c & 1 \end{bmatrix}$, show that $detA=1+a^2+b^2+c^2$. Hence, find $A^{-1}$ for any a,b, and c.
Any help/guidance would be appreciated!
Thanks in advance!
Using Rule of Sarrus I can confirm: $detA=b^2+c^2+a^2+1-acb+bac=1+a^2+b^2+c^2$
I am going to let you find the determinant of $A$ it is pretty straight forward to calculate the determinant of a $3\times 3$ matrix, and if you don't know it yet, it is worth looking up.
$A^{-1}$
One way to do it is to take $A$ put the indentity matrix next to it. Perform row operations on $A$ and perform the identical operations on $I$ When you have reduced $A$ to the identity, what was the identity has become $A^{-1}$
Another idea. The second and third row vectors of $A$ are both orthogonal to the first column vector of $A^{-1}$
if you find $(-a, -1, c)\times (-b,-c, 1)$ it will give you a vector that is parallel to the necessary first column vector of $A^{-1}$ then you multiply (1,a,b) by that vector to size it correctly.
$\frac{1}{1+a^2 + b^2+c^2}\begin {bmatrix} c^2-1\\a-bc\\ ac-b \end{bmatrix}$ is the first column vector.
By the way, are you sure it is a $-1$ in the middle there? If it is then $\det A = -1+a^2 - b^2 + c^2$