Let A be a 2 × 2 symmetric matrix such that A [1, 1]ᵀ = [3, 7]ᵀ and the determinant of A be 1. If A⁻

Q: Wait, why can't I just calculate $A^{-1}$ using the formula $\frac{1}{|A|} \text{adj}(A)$? Isn't that easier?

You absolutely can! That is a perfectly valid path. However, the Cayley-Hamilton approach is a powerful tool in your JEE arsenal. It allows you to express the inverse as a linear combination of the matrix itself without needing to perform the full adjugate calculation. It is about choosing the most elegant path, not just the first one you see. Mastering this technique will save you precious time in the exam hall.

Q: Why does the Cayley-Hamilton theorem allow us to replace $\lambda$ with $A$?

That is a profound question. The characteristic equation $p(\lambda) = \det(A - \lambda I) = 0$ is a polynomial in $\lambda$. Cayley-Hamilton asserts that the matrix $A$ itself satisfies this polynomial equation. It is not just a substitution; it is a fundamental property of linear operators. Think of it as the matrix 'obeying' its own internal structure. It is a deep, beautiful result that bridges the gap between scalar algebra and matrix algebra.

Q: How do I know that $\alpha$ and $\beta$ are unique?

Great observation. In this $2 \times 2$ case, $A$ and $I$ are linearly independent unless $A$ is a scalar multiple of $I$. Since our matrix $A = \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix}$ is not a scalar multiple of the identity, the representation $A^{-1} = \alpha A + \beta I$ is unique. You can safely equate the coefficients.

Question

Let $A$ be a $2 	imes 2$ symmetric matrix such that $A \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 3 \ 7 \end{bmatrix}$ and the determinant of $A$ be 1. If $A^{-1} = \alpha A + \beta I$, where $I$ is an identity matrix of order $2 	imes 2$, then $\alpha + \beta$ equals _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with a symmetric $2 	imes 2$ matrix $A$. By the definition of symmetry, $A = A^T$, which implies the off-diagonal elements must be equal. We define the matrix as:

$$A = \begin{bmatrix} a & b \ b & d \end{bmatrix}$$

We are given the transformation condition $A \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 3 \ 7 \end{bmatrix}$. Performing the matrix multiplication, we obtain:

$$\begin{bmatrix} a+b \ b+d \end{bmatrix} = \begin{bmatrix} 3 \ 7 \end{bmatrix}$$

This yields the system of linear equations:
1) $a + b = 3$
2) $b + d = 7$

From these, we express $a$ and $d$ in terms of $b$:
$a = 3 - b$
$d = 7 - b$

### Applying the Determinant Constraint

We are given that the determinant of the matrix is $|A| = 1$. The determinant of a $2 	imes 2$ matrix is calculated as:

$$|A| = ad - b^2 = 1$$

Substituting our expressions for $a$ and $d$ into this equation, we get:

$$(3 - b)(7 - b) - b^2 = 1$$

Expanding the product, we have:

$$21 - 3b - 7b + b^2 - b^2 = 1$$

The $b^2$ terms cancel out, simplifying the equation to:

$$21 - 10b = 1 \Rightarrow 10b = 20 \Rightarrow b = 2$$

Substituting $b=2$ back into our expressions for $a$ and $d$, we find $a = 3 - 2 = 1$ and $d = 7 - 2 = 5$. Thus, the matrix is:

$$A = \begin{bmatrix} 1 & 2 \ 2 & 5 \end{bmatrix}$$

### Utilizing the Cayley-Hamilton Theorem

To find $A^{-1}$ in the form $\alpha A + \beta I$, we utilize the **Cayley-Hamilton Theorem**, which states that a matrix satisfies its own characteristic equation:

$$\lambda^2 - 	ext{Tr}(A)\lambda + |A| = 0$$

The trace of $A$ is $	ext{Tr}(A) = 1 + 5 = 6$, and the determinant is $|A| = 1$. The characteristic equation is:

$$\lambda^2 - 6\lambda + 1 = 0$$

Replacing $\lambda$ with $A$, we obtain:

$$A^2 - 6A + I = 0$$

Multiplying the entire equation by $A^{-1}$, we get:

$$A - 6I + A^{-1} = 0 \Rightarrow A^{-1} = -A + 6I$$

### Final Calculation

Comparing $A^{-1} = -A + 6I$ with the required form $A^{-1} = \alpha A + \beta I$, we identify the coefficients:

$$\alpha = -1, \quad \beta = 6$$

The final sum $\alpha + \beta$ is:

$$-1 + 6 = \mathbf{5}$$

Animated Solution for Mathematics - Matrices and Determinants: 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 _______

Enter Numerical Value:

Visualized Solution

Defining the Symmetric Matrix $A$

Applying the Matrix Multiplication Condition

Expressing $a$ and $d$ in terms of $b$

Using the Determinant Condition $∣ A ∣ = 1$

Solving for the Variable $b$

Finding the Elements $a$ and $d$

The Characteristic Equation of $A$

Applying the Cayley-Hamilton Theorem

Finding the Expression for $A^{- 1}$

Comparing and Finding $α$ and $β$

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

Applying the Determinant Constraint

Utilizing the Cayley-Hamilton Theorem

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

Applying the Matrix Multiplication Condition

Using the Determinant Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣=1

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

Finding the Elements .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

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