Let A = [[α, -1], [6, β]], α 0, such that det(A) = 0 and α + β = 1. If I denotes 2 × 2 identity ma

Q: Wait, why can't we just multiply the matrix by itself 8 times? It's just a 2x2 matrix, right?

While technically possible, it is a trap! In JEE Advanced, time is your most precious resource. Calculating $A^2, A^3, \dots, A^8$ manually is a recipe for arithmetic errors. The examiners want to see if you can spot the pattern of idempotency ($A^2 = A$). Always look for structural shortcuts before diving into brute-force calculation.

Q: How do I know for sure that $I$ and $A$ commute? Can I just use the Binomial Theorem blindly?

That is a fantastic question. The Binomial Theorem $(A+B)^n = \sum \binom{n}{k} A^{n-k} B^k$ is only valid if $AB = BA$. In this case, $B = I$. Since the identity matrix $I$ commutes with every square matrix ($IA = AI = A$), the condition is satisfied. You are safe to use the theorem!

Q: Why did the sum of binomial coefficients start from $k=1$ in the final step?

Great observation. The expansion is $(I+A)^8 = \binom{8}{0}I^8 + \binom{8}{1}I^7A + \binom{8}{2}I^6A^2 + \dots + \binom{8}{8}A^8$. The first term is $\binom{8}{0}I = I$. Every subsequent term contains at least one power of $A$. Since $A^k = A$ for all $k \ge 1$, we factor out $A$ from all terms *except* the first one. That is why the sum is $\binom{8}{1} + \binom{8}{2} + \dots + \binom{8}{8}$.

Question

Let $A = \begin{bmatrix} \alpha & -1 \ 6 & \beta \end{bmatrix}$, $\alpha > 0$, such that $\det(A) = 0$ and $\alpha + \beta = 1$. If $I$ denotes $2 	imes 2$ identity matrix, then the matrix $(I + A)^8$ is:

Options:

(A) $\begin{bmatrix} 4 & -1 \ 6 & -1 \end{bmatrix}$

(B) $\begin{bmatrix} 257 & -64 \ 514 & -127 \end{bmatrix}$

(C) $\begin{bmatrix} 1025 & -511 \ 2024 & -1024 \end{bmatrix}$

(D) $\begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Matrices

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a tedious exercise in matrix multiplication. You see a matrix $A = \begin{bmatrix} \alpha & -1 \ 6 & \beta \end{bmatrix}$ and a power of 8, and your instinct might be to start multiplying.

But hold on—in the world of JEE Advanced, brute force is rarely the intended path. There is a hidden elegance here, a mathematical rhythm waiting to be discovered. Let us embark on this journey together.

### Decoding the Matrix

We are given two constraints: $\det(A) = 0$ and $\alpha + \beta = 1$. Let us translate these into the language of algebra.

The determinant of a $2 	imes 2$ matrix is defined as the product of the main diagonal minus the product of the off-diagonal. Thus:

$$\det(A) = (\alpha)(\beta) - (-1)(6) = 0$$

This simplifies beautifully to $\alpha\beta + 6 = 0$, or $\alpha\beta = -6$. Now, we have a system of two equations:

1) $\alpha + \beta = 1$
2) $\alpha\beta = -6$

If you have spent enough time with quadratic equations, you might recognize these as the sum and product of roots. We can substitute $\beta = 1 - \alpha$ into the second equation: $\alpha(1 - \alpha) = -6$.

Expanding this gives us $\alpha - \alpha^2 = -6$, which rearranges into the standard quadratic form: $\alpha^2 - \alpha - 6 = 0$. Factoring this, we get $(\alpha - 3)(\alpha + 2) = 0$.

Since the problem explicitly states $\alpha > 0$, we must reject $\alpha = -2$ and accept $\alpha = 3$. Consequently, $\beta = 1 - 3 = -2$. Our matrix $A$ is revealed:

$$A = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix}$$

### The Idempotent Revelation

Now, here is where the magic happens. Before we even think about raising this to the power of 8, let us test the behavior of $A$ by squaring it. Calculate $A^2 = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix}$.

Performing the multiplication:
- Top-left: $(3)(3) + (-1)(6) = 9 - 6 = 3$
- Top-right: $(3)(-1) + (-1)(-2) = -3 + 2 = -1$
- Bottom-left: $(6)(3) + (-2)(6) = 18 - 12 = 6$
- Bottom-right: $(6)(-1) + (-2)(-2) = -6 + 4 = -2$

Look at that result! $A^2 = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} = A$. This is not a coincidence; this is an **idempotent matrix**.

Because $A^2 = A$, it follows that $A^3 = A^2 \cdot A = A \cdot A = A$. By induction, $A^n = A$ for any positive integer $n$. This is the "Aha!" moment that separates the top rankers from the rest.

### The Binomial Shortcut

We need to evaluate $(I + A)^8$. Since the identity matrix $I$ commutes with $A$, we can apply the Binomial Theorem:

$$(I + A)^8 = \binom{8}{0}I^8 + \binom{8}{1}I^7A + \binom{8}{2}I^6A^2 + \dots + \binom{8}{8}A^8$$

Since $I^n = I$ and $A^k = A$ for all $k \ge 1$, this simplifies to:

$$(I + A)^8 = I + \left( \binom{8}{1} + \binom{8}{2} + \dots + \binom{8}{8} ight) A$$

Recall the identity for the sum of binomial coefficients: $\sum_{k=0}^{n} \binom{n}{k} = 2^n$. Our sum is missing the $\binom{8}{0}$ term, which is 1.

Therefore, the sum of the coefficients is $2^8 - 1 = 256 - 1 = 255$. Our expression is now simply $I + 255A$.

### Final Calculation

We are at the finish line. Let us perform the final matrix addition:

$$(I + A)^8 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + 255 \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix}$$

Multiplying the scalar 255 into $A$:

$$255A = \begin{bmatrix} 765 & -255 \ 1530 & -510 \end{bmatrix}$$

Adding the identity matrix:

$$(I + A)^8 = \begin{bmatrix} 1 + 765 & 0 - 255 \ 0 + 1530 & 1 - 510 \end{bmatrix} = \begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$$

And there it is. We have navigated the complexity, identified the structural pattern, and arrived at the solution with precision. The final result is:

$$\begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$$

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [α 6 - 1 β]$ , $α > 0$ , such that $det (A) = 0$ and $α + β = 1$ . If $I$ denotes $2 \times 2$ identity matrix, then the matrix $(I + A)^{8}$ is:

Select Answer:

Visualized Solution

Analyze the Determinant Condition

Use the Sum Condition $α + β = 1$

Formulate the Quadratic Equation

Solve for $α$ and $β$

Construct Matrix $A$

Check the Idempotent Property

Apply Binomial Theorem

Simplify the Binomial Sum

Perform Final Matrix Addition

Final Result

The Sigma Insight: Types of Matrices

The Hidden Symmetry of Matrices

Decoding the Matrix

The Idempotent Revelation

The Binomial Shortcut

Final Calculation

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