Question

if $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right],6A^{-1}=A^2+cA+dI$, then $(c,d)$ is

• A. $(-6,11)$
• B. $(-11,6)$
• C. $(11,6)$
• D. $(6,11)$

Answer 1

Answer: (A)

Given, $6A^1=A^2+cA+dI$

$\Rightarrow 6A^1-A-A^{2}A+cA+dI\cdot A$

$\Rightarrow 6I=A^3+c{A}^2+dA$

$\Rightarrow A^3+c{A}^2+dA-6I=0$

Now, $A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 5\\ 0& -10& 14\end{array}\right]$

$A^3\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 5\\ 0& -10& 14\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 5\\ 0& -10& 14\end{array}\right]$

$\therefore$ From equation$\left(i\right)$, we have

$\left[\begin{array}{ccc}1& 0& 0\\ 0& -11& 19\\ 0& -38& 46\end{array}\right]+c\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 5\\ 0& -10& 14\end{array}\right]+d\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]-6\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=0$

$\Rightarrow \left[\begin{array}{ccc}1+c+d-6& 0& 0\\ 0& -11-c+d-6& 19+5c+d\\ 0& -38-10c-2d& 46+14c+4d-6\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

$\Rightarrow 1+c+d-6=0,-11-c+d-6=0$

$\Rightarrow c+d=5,-c+d=17$

$\Rightarrow c=-6,d=11$