Matrix

From (1), we get ${ A }^{ 3 }-I=O$ . (O is zero matrix) This is equalvent to

$(A-I)({ A }^{ 2 }+A+I)=O$

Multiply inverse of A-I, we get $({ A }^{ 2 }+A+I)=O$

Since ${ (A-I) }^{ 60 }={ (A-I) }^{ { 2 }^{ 30 } }$ , ${ (A-I) }^{ 2 }={ A }^{ 2 }-2A+I$ , ${ A }^{ 2 }=-A-I,\quad \therefore -A-I-2A+I=-3A$ .

${ (-3A) }^{ 30 }={ 3 }^{ 30 }\cdot { A }^{ 30 }={ 3 }^{ 30 }\cdot A^{ { 3 }^{ 10 } }={ 3 }^{ 30 }\cdot I^{ 10 }={ 3 }^{ 30 }I=\begin{pmatrix} { 3 }^{ 30 } & 0 \\ 0 & { 3 }^{ 30 } \end{pmatrix}$

Therefore sum of all components of matrix ${ (A-I) }^{ 60 }$ is ${ 2 }^{ 1 }\cdot { 3 }^{ 30 }$ . a=1 and b=30 . Therefore a+b = $\boxed{31}$

Since $A^3 = I$ , the matrix $A$ is diagonalisable, so there exists a nonsingular matrix $P$ such that $PAP^{-1}$ is diagonal, and hence such that $PAP^{-1} \; = \; \left(\begin{array}{cc} u & 0 \\ 0 & v \end{array}\right)$ where $u^3 = v^3 = 1$ . Since $A-I$ is nonsingular, neither $u$ nor $v$ is equal to $1$ , and so each of $u,v$ is equal to $\omega$ or $\omega^2$ . where $\omega = e^{\frac{2\pi i}{3}}$ is the primitive cube root of unity.

Now $\omega-1 = -i\sqrt{3}\omega^2$ and $\omega^2-1 = i\sqrt{3}\omega$ , so that $(u-1)^{60} = (v-1)^{60} = (\omega-1)^{60} = (\omega^2-1)^{60} = 3^{30}$ , and hence $P(A-I)^{60}P^{-1} \; = \; \left(\begin{array}{cc} (u-1)^{60} & 0 \\ 0 & (v-1)^{60}\end{array}\right) \; = \; \left(\begin{array}{cc} 3^{30} & 0 \\ 0 & 3^{30}\end{array}\right) \; = \; 3^{30}I$ and hence $(A-I)^{60} = 3^{30}I$ , making the sum of coefficients equal to $2 \times 3^{30}$ . This makes the answer $\boxed{31}$ .