Let B = [[1, 3], [1, 5]] and A be a 2 × 2 matrix such that AB⁻¹ = A⁻¹. If BCB⁻¹ = A and C⁴ + αC² + β

Q: Wait, why can't I just calculate A by finding the inverse of B and solving the system? Isn't that safer?

That is a valid instinct, but in JEE Advanced, time is your most precious currency. Calculating inverses and solving systems of linear equations is prone to arithmetic errors and is computationally heavy. By using matrix properties to derive A^2 = B and C^2 = B, we bypass the tedious arithmetic entirely. Always look for the algebraic shortcut before diving into the calculation!

Q: How did we know that C^2 = B? Is that a general property?

It is not a general property, but a specific consequence of the relationships given in this problem. When we calculated C^2 = (B^{-1}AB)(B^{-1}AB), the middle terms B(B^{-1}) became the identity matrix I. This 'telescoping' effect is common in matrix problems. Always look for those internal cancellations!

Q: Why are we using the Cayley-Hamilton Theorem here? Can't we just compute B^2 directly?

You absolutely could compute B^2 directly, but the Cayley-Hamilton Theorem is the 'elegant' path. It connects the matrix B to its characteristic polynomial \lambda^2 - \text{Tr}(B)\lambda + |B| = 0. By comparing the coefficients of B^2 + \alpha B + \beta I = O with the characteristic equation, we solve for \alpha and \beta in one swift step. It is about working smarter, not harder.

Question

Let $B = \begin{bmatrix} 1 & 3 \ 1 & 5 \end{bmatrix}$ and $A$ be a $2 	imes 2$ matrix such that $AB^{-1} = A^{-1}$. If $BCB^{-1} = A$ and $C^4 + \alpha C^2 + \beta I = O$, then $2\beta - \alpha$ is equal to

Options:

(A) 16

(B) 2

(C) 8

(D) 10

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are not just solving a matrix problem; we are uncovering a hidden symmetry. Many students look at a matrix equation like $AB^{-1} = A^{-1}$ and immediately reach for the inverse formula, ready to grind through rows and columns.

But stop. Take a breath. In JEE Advanced, the most elegant path is rarely the one that requires the most arithmetic. Let us walk through this together.

### The Algebraic Anchor

We start with the equation $AB^{-1} = A^{-1}$. Our goal is to simplify this into something meaningful by isolating the relationship between $A$ and $B$.

By pre-multiplying both sides by $A$, we get $A(AB^{-1}) = A(A^{-1})$. Since $A(A^{-1}) = I$, the right side simplifies to the identity matrix $I$.

Now we have $A^2 B^{-1} = I$. To clear that $B^{-1}$, we post-multiply by $B$. The result is a beautiful, clean anchor: 
$$A^2 = B$$

### The Transformation of $C$

Now, we turn our attention to the matrix $C$, defined by $BCB^{-1} = A$. We need to understand $C^2$ to solve the polynomial equation.

Let us isolate $C$. By pre-multiplying by $B^{-1}$ and post-multiplying by $B$, we find $C = B^{-1}AB$.

Now, let us square it: 
$$C^2 = (B^{-1}AB)(B^{-1}AB)$$

Look closely at the center of this expression. We have $B(B^{-1})$, which is just $I$. The expression collapses to $B^{-1}A^2B$.

And here is the magic—we already know that $A^2 = B$! Substituting this in, we get:
$$C^2 = B^{-1}BB = B$$
The complexity vanishes, leaving us with the simple truth that $C^2 = B$.

### The Cayley-Hamilton Revelation

We are left with the polynomial $C^4 + \alpha C^2 + \beta I = O$. Since $C^2 = B$, then $C^4 = B^2$.

Our equation becomes $B^2 + \alpha B + \beta I = O$. This is the moment to invoke the **Cayley-Hamilton Theorem**.

This theorem is the bridge between the abstract matrix and its characteristic polynomial. For our matrix $B = \begin{bmatrix} 1 & 3 \ 1 & 5 \end{bmatrix}$, the characteristic equation is $\lambda^2 - 	ext{Tr}(B)\lambda + |B| = 0$.

Calculating the trace, $	ext{Tr}(B) = 1 + 5 = 6$, and the determinant, $|B| = (1)(5) - (3)(1) = 2$, we get:
$$\lambda^2 - 6\lambda + 2 = 0$$

By the Cayley-Hamilton theorem, it follows that:
$$B^2 - 6B + 2I = O$$

### Final Calculation

Comparing $B^2 + \alpha B + \beta I = O$ with $B^2 - 6B + 2I = O$, we see that $\alpha = -6$ and $\beta = 2$.

The question asks for $2\beta - \alpha$. Substituting our values:
$$2(2) - (-6) = 4 + 6 = 10$$

We did not just calculate; we navigated the structure of the matrices. The final answer is **10**.

Animated Solution for Mathematics - Matrices and Determinants: Let $B = [1135]$ and $A$ be a $2 \times 2$ matrix such that $A B^{- 1} = A^{- 1}$ . If $B C B^{- 1} = A$ and $C^{4} + α C^{2} + β I = O$ , then $2 β - α$ is equal to

Select Answer:

Visualized Solution

Analyzing the First Equation

Pre-multiplying by $A$

Simplifying the Equation

Isolating $A^{2}$

Analyzing the Second Equation

Isolating Matrix $C$

Calculating $C^{2}$

The Polynomial Equation

Substituting $C^{2} = B$

Characteristic Equation Formula

Trace and Determinant of $B$

Cayley-Hamilton Theorem

Final Calculation

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Algebraic Anchor

The Transformation of $C$

The Cayley-Hamilton Revelation

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Analyzing the First Equation

Pre-multiplying by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Simplifying the Equation

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Analyzing the Second Equation

Isolating Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2

The Polynomial Equation

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2=B

Characteristic Equation Formula

Trace and Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Cayley-Hamilton Theorem

Final Calculation

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Algebraic Anchor

The Transformation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The Cayley-Hamilton Revelation

Final Calculation

Similar Questions

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Pre-multiplying by $A$

Isolating $A^{2}$

Isolating Matrix $C$

Calculating $C^{2}$

Substituting $C^{2} = B$

Trace and Determinant of $B$

The Transformation of $C$

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