If $(2I-A)^{-1} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\-1 & 1 & 0 \end{pmatrix}.$
Evaluate for matrix A.
what i did is i got the adj( $(2I-A)^{-1}$ ) but I'm not sure if this was how the approach was supposed to be..
On
I haven't done linear algebra for a year and a half, but I would like to give this a go.
I worked out the inverse of $2I-A$ as follows. Gauss reduce the right hand side of
$$\left(\begin{array}{rrr|rrr} 1&0&0 &1&1&1\\ 0&1&0 &1&0&1\\ 0&0&1 &-1&1&0 \end{array}\right)$$
to get
$$\left(\begin{array}{rrr|rrr} 1&-1&-1 &1&0&0\\ 1&-1&0 &0&1&0\\ -1&2&1 &0&0&1 \end{array}\right).$$
Then calculate
$$\left(\begin{array}{rrr} 2&0&0\\ 0&2&0\\ 0&0&2 \end{array}\right)- \left(\begin{array}{rrr} 1&-1&-1\\ 1&-1&0\\ -1&2&1 \end{array}\right)= \left(\begin{array}{rrr} 1&1&1\\ -1&3&0\\ 1&-2&1 \end{array}\right).$$
If $(2I - A)^{-1} = B$, then $A = 2I - B^{-1}$.