Question

If $A$ is a non-singular matrix of order $n$ satisfying the matrix equation $I+A+A^{2}+A^{3}+...+A^{10}=0$, where $I$ and $0$ are, respectively, unit and null matrices of order $n$, then $A^{10}=$

• A. $A^{-1}$
• B. $I$
• C. $A$
• D. $I+A$
• E. $0$

Hint:

Logic Tip: A geometric series of matrices $I + A + A^2 + \dots + A^k = 0$ behaves identically to a geometric series of numbers. Multiplying by $(I - A)$ yields $I - A^{k+1} = 0$, meaning $A^{11} = I$. Multiplying by $A^{-1}$ immediately gives $A^{10} = A^{-1}$.

Answer 1

The Correct Option is A

Concept:
For matrix equations, multiplying the entire equation by the inverse of a matrix (if it exists) is a standard technique to isolate terms or lower the powers of the matrices. Since $A$ is given as non-singular, its inverse $A^{-1}$ exists.
Step 1: Multiply the given equation by $A^{-1}$.
We are given: $$I + A + A^2 + A^3 + \dots + A^{10} = 0$$ Pre-multiply (or post-multiply) both sides by $A^{-1}$: $$A^{-1}(I + A + A^2 + A^3 + \dots + A^{10}) = A^{-1}(0)$$
Step 2: Distribute $A^{-1}$ across the terms.
$$A^{-1}I + A^{-1}A + A^{-1}A^2 + \dots + A^{-1}A^{10} = 0$$ Using the properties $A^{-1}I = A^{-1}$ and $A^{-1}A^k = A^{k-1}$: $$A^{-1} + I + A + A^2 + \dots + A^9 = 0$$
Step 3: Relate the new equation back to the original equation.
From our original equation, we can isolate the sum of terms up to $A^9$: $$(I + A + A^2 + \dots + A^9) + A^{10} = 0$$ $$(I + A + A^2 + \dots + A^9) = -A^{10}$$
Step 4: Substitute and solve for $A^{10}$.
Substitute this group into the equation we derived in Step 2: $$A^{-1} + (I + A + A^2 + \dots + A^9) = 0$$ $$A^{-1} + (-A^{10}) = 0$$ $$A^{-1} = A^{10}$$