The matrix A satisfies the equation 6A^-1=A²+cA+dl where c and d are scalars and I is the identity matrix. Then (c+d) is equal to

The correct option is 5.

From the given image, we identify the matrix $A$ as follows:
$A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& -2\\ 0& 1& 1\end{array}\right]$

To find the characteristic equation of matrix $A$, we compute the determinant $\det (A-\lambda I)=0$:
$\left|\begin{array}{ccc}1-\lambda & 0& 0\\ 0& 4-\lambda & -2\\ 0& 1& 1-\lambda \end{array}\right|=0$

Expanding the determinant along the first row:
$(1-\lambda )\left[\right(4-\lambda \left)\right(1-\lambda )-(-2\left)\right(1\left)\right]=0$
$(1-\lambda )[{\lambda }^2-5\lambda +4+2]=0$
$(1-\lambda )({\lambda }^2-5\lambda +6)=0$
${\lambda }^2-5\lambda +6-{\lambda }^3+5{\lambda }^2-6\lambda =0$
$-{\lambda }^3+6{\lambda }^2-11\lambda +6=0$

Multiplying the equation by $-1$ gives the characteristic equation:
${\lambda }^3-6{\lambda }^2+11\lambda -6=0$

By the Cayley-Hamilton Theorem, matrix $A$ satisfies its own characteristic equation:
$A^3-6A^2+11A-6I=O$
where $I$ is the identity matrix and $O$ is the zero matrix.

Rearranging the equation to solve for the identity term:
$6I=A^3-6A^2+11A$

Since the determinant of $A$ is non-zero, $A$ is invertible. Multiplying both sides by $A^{-1}$:
$6A^{-1}=A^2-6A+11I$

Comparing this with the given equation:
$6A^{-1}=A^2+cA+dI$

By comparing the coefficients, we obtain:
$c=-6$
$d=11$

Thus, we compute the sum $c+d$:
$c+d=-6+11=5$