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.

Step-by-step Explanation:

First, we refer to the provided image which defines the matrix 3 × 3 matrix $A$ as:
$$A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& -2\\ 0& 1& 1\end{array}\right]$$

To find the equation satisfied by $A$, we can determine its characteristic equation using:
$$det(A-\lambda I)=0$$
Expanding along the first row, we get:
$$det\left[\begin{array}{ccc}1-\lambda & 0& 0\\ 0& 4-\lambda & -2\\ 0& 1& 1-\lambda \end{array}\right]=0$$
$$(1-\lambda )\left[\right(4-\lambda \left)\right(1-\lambda )-(-2\left)\right(1\left)\right]=0$$
Simplifying the terms inside the brackets:
$$(1-\lambda )[{\lambda }^2-5\lambda +4+2]=0$$
$$(1-\lambda )({\lambda }^2-5\lambda +6)=0$$
Expanding this product:
$${\lambda }^2-5\lambda +6-{\lambda }^3+5{\lambda }^2-6\lambda =0$$
$$-{\lambda }^3+6{\lambda }^2-11\lambda +6=0$$
Multiplying by -1, we obtain the characteristic equation:
$${\lambda }^3-6{\lambda }^2+11\lambda -6=0$$

According to the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation. Therefore:
$$A^3-6A^2+11A-6I=0$$

Since the determinant of $A$ is $6\ne 0$, the matrix is invertible. Multiplying the entire equation by the inverse matrix $A^{-1}$:
$$A^2-6A+11I-6A^{-1}=0$$
Rearranging the equation to isolate $6A^{-1}$:
$$6A^{-1}=A^2-6A+11I$$

Now, we compare this derived equation with the given equation:
$$6A^{-1}=A^2+cA+dI$$
By equating the corresponding coefficients:
$$c=-6$$
$$d=11$$

We need to calculate the value of $(c+d)$:
$$c+d=-6+11=5$$