Question

If A=$\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]$ then det $\left(A^{-1}\right)$ =

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

By property $det\left(A\right)$$=\frac{1}{det\left(A^{-1}\right)}$
so
det (A) $=\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]=-1[1-1]-1(-1-1)+1(1+1)$ $=4$
$\therefore |A^{-1}|=\frac{1}{|A|}=\frac{1}{4}$.