If A = [[1, 5], [λ, 10]], A⁻¹ = αA + βI and α + β = -2, then 4α² + β² + λ² is equal to : (1) 12 (2)

Q: Wait, why can't I just calculate $A^{-1}$ using the adjoint method? Isn't that safer?

You absolutely could! But in JEE Advanced, time is your most precious currency. The Cayley-Hamilton approach bypasses the tedious calculation of the adjoint and determinant, turning a messy algebraic problem into a elegant comparison of coefficients.

Q: How do I know for sure that the Cayley-Hamilton Theorem applies here?

The theorem states that every square matrix satisfies its own characteristic equation. Since $A$ is a $2 \times 2$ matrix, it is guaranteed to satisfy the equation derived from $\det(A - kI) = 0$. It is a fundamental property, not a shortcut!

Q: Why did we multiply by $A^{-1}$ instead of just rearranging the terms?

Multiplying by $A^{-1}$ is the standard 'operator' technique to shift powers of $A$. It turns $A^2$ into $A$, $A$ into $I$, and $I$ into $A^{-1}$, which perfectly matches the form $A^{-1} = \alpha A + \beta I$ given in the problem.

Q: What if the determinant of $A$ was zero? Would the inverse exist?

Great catch! If $\det(A) = 0$, the matrix is singular and $A^{-1}$ does not exist. In this problem, the existence of the relation $A^{-1} = \alpha A + \beta I$ implicitly assumes $A$ is invertible. If you ever find $\det(A) = 0$, you know the premise of the question would be invalid.

Q: Is there any risk in comparing coefficients $\alpha$ and $\beta$ directly?

As long as the identity $A^{-1} = \alpha A + \beta I$ holds for all entries, the coefficients of $A$ and $I$ must be unique. Comparing them is mathematically sound because $A$ and $I$ are linearly independent matrices.

Question

If $A = \begin{bmatrix} 1 & 5 \ \lambda & 10 \end{bmatrix}$, $A^{-1} = \alpha A + \beta I$ and $\alpha + \beta = -2$, then $4\alpha^2 + \beta^2 + \lambda^2$ is equal to :

Options:

(A) 12

(B) 19

(C) 14

(D) 10

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the road to JEE excellence. Today, we aren't just solving a matrix problem; we are uncovering the hidden symmetry of linear transformations.

You are presented with a matrix $A = \begin{bmatrix} 1 & 5 \ \lambda & 10 \end{bmatrix}$ and a mysterious relationship $A^{-1} = \alpha A + \beta I$.

At first glance, you might be tempted to dive into the standard formula for the inverse, calculating the determinant and the adjoint. But stop—take a breath. In the world of JEE Advanced, we look for the path of least resistance, the path that reveals the soul of the matrix.

### The Characteristic Soul

Every square matrix carries a 'DNA' known as its characteristic equation. This equation, defined by $\det(A - kI) = 0$, tells us how the matrix behaves under transformation.

Let us compute this for our matrix $A$:

$$\begin{vmatrix} 1-k & 5 \ \lambda & 10-k \end{vmatrix} = 0$$

Expanding this, we get $(1-k)(10-k) - 5\lambda = 0$, which simplifies beautifully to $k^2 - 11k + (10 - 5\lambda) = 0$. This is the heartbeat of our matrix.

### The Power of Cayley-Hamilton

Here is where the magic happens. The **Cayley-Hamilton Theorem** tells us that a matrix satisfies its own characteristic equation. So, we replace $k$ with $A$:

$$A^2 - 11A + (10 - 5\lambda)I = 0$$

This equation is a bridge. We need to reach the form $A^{-1} = \alpha A + \beta I$. We simply multiply the entire equation by $A^{-1}$:

$$A^{-1}(A^2 - 11A + (10 - 5\lambda)I) = 0$$

Distributing $A^{-1}$, we get $A - 11I + (10 - 5\lambda)A^{-1} = 0$. Now, watch as we isolate $A^{-1}$ with the precision of a surgeon:

$$(10 - 5\lambda)A^{-1} = -A + 11I$$

$$A^{-1} = \frac{-1}{10 - 5\lambda}A + \frac{11}{10 - 5\lambda}I$$

### The Final Unveiling

By comparing this to our given relation $A^{-1} = \alpha A + \beta I$, we identify our coefficients: $\alpha = \frac{-1}{10 - 5\lambda}$ and $\beta = \frac{11}{10 - 5\lambda}$.

We are given the constraint $\alpha + \beta = -2$. Substituting our expressions, we find:

$$\frac{-1 + 11}{10 - 5\lambda} = -2 \implies \frac{10}{10 - 5\lambda} = -2$$

Solving this simple linear equation, we find $10 = -20 + 10\lambda$, which leads us to $\lambda = 3$.

With $\lambda$ in hand, the rest is a victory lap. We calculate $\alpha = \frac{1}{5}$ and $\beta = -\frac{11}{5}$. Finally, we evaluate the expression $4\alpha^2 + \beta^2 + \lambda^2$:

$$4\left(\frac{1}{5}ight)^2 + \left(-\frac{11}{5}ight)^2 + 3^2 = 4\left(\frac{1}{25}ight) + \frac{121}{25} + 9 = \frac{125}{25} + 9 = 5 + 9 = 14$$

And there it is—**14**. The complexity dissolves into simplicity when you understand the underlying structure. Keep this perspective, and no matrix will ever intimidate you again. Onward!

Animated Solution for Mathematics - Matrices and Determinants: If $A = [1 λ 510]$ , $A^{- 1} = α A + β I$ and $α + β = - 2$ , then $4 α^{2} + β^{2} + λ^{2}$ is equal to :

Select Answer:

Visualized Solution

Matrix $A$ and the Given Equation

The Characteristic Equation

Expanding the Determinant

Applying Cayley-Hamilton Theorem

Multiplying by $A^{- 1}$

Isolating $A^{- 1}$

Comparing Coefficients

Using the Condition $α + β = - 2$

Solving for $λ$

Calculating $α$ and $β$

The Final Calculation

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Characteristic Soul

The Power of Cayley-Hamilton

The Final Unveiling

Similar Questions

Let $A = [1 - 1 24]$ . If $A^{- 1} = α I + β A, α, β \in R, I$ is a $2 \times 2$ identity matrix, then $4 (α - β)$ is equal to :

Let $A = [1 - 2 2 - 5]$ . Let $α, β \in R$ be such that $α A^{2} + β A = 2 I$ . Then $α + β$ is equal to -

Let $A$ be a $2 \times 2$ symmetric matrix such that $A [11] = [37]$ and the determinant of $A$ be 1. If $A^{- 1} = α A + β I$ , where $I$ is an identity matrix of order $2 \times 2$ , then $α + β$ equals _______

Let $A = 121 - 1 11 1 - 3 1$ and $10 B = 4 - 5 1 20 - 2 2 α 3$ . If $B$ is the inverse of matrix $A$ , then $α$ is

If $A = (2924)$ and $I = (1001)$ , then $10 A^{- 1}$ is equal to :

$A = 100 01 - 2 014$ and $I = 100010001$ and $A^{- 1} = [\frac{1}{6} (A^{2} + c A + d I)]$ , then the value of $c$ and $d$ are

If $A = [2924]$ and $I = [1001]$ , then $10 A^{- 1}$ is equal to:

If $A = [0 tan (θ /2) - tan (θ /2) 0]$ and $(I_{2} + A) (I_{2} - A)^{- 1} = [a b - b a]$ , then $13 (a^{2} + b^{2})$ is equal to

Let $X = 000100010, Y = α I + β X + γ X^{2}$ and $Z = α^{2} I - α β X + (β^{2} - α γ) X^{2}, α, β, γ \in R$ . If $Y^{- 1} = 1/5 00 - 2/5 1/5 0 1/5 - 2/5 1/5$ , then $(α - β + γ)^{2}$ is equal to

Let A be a $3 \times 3$ matrix such that $A + A^{T} = O$ . If $A 1 - 1 0 = 332, A^{2} 1 - 1 0 = - 3 19 - 24$ and $det (adj (2 adj (A + I))) = (2)^{α} \cdot (3)^{β} \cdot (11)^{γ}, α, β, γ$ are non-negative integers, then $α + β + γ$ is equal to .........

Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A=[1λ​510​], A−1=αA+βI and α+β=−2, then 4α2+β2+λ2 is equal to :

Select Answer:

Visualized Solution

Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and the Given Equation

The Characteristic Equation

Expanding the Determinant

Applying Cayley-Hamilton Theorem

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

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

Comparing Coefficients

Using the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α+β=−2

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

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