A = [1 0 0; 0 1 1; 0 -2 4] and I = [1 0 0; 0 1 0; 0 0 1] and A⁻¹ = [1/6(A² + cA + dI)], then the val

Q: Why can't I just use the adjoint method? It's a standard formula.

The adjoint method is mathematically sound, but in the heat of a JEE Advanced exam, it is a time-sink. Calculating the determinant and the matrix of cofactors for a $3 \times 3$ matrix is prone to arithmetic errors. The Cayley-Hamilton approach is elegant, faster, and demonstrates a deeper mastery of linear algebra.

Q: Is it always true that $A^3 - 6A^2 + 11A - 6I = 0$?

Yes! This is the beauty of the Cayley-Hamilton Theorem. It states that every square matrix satisfies its own characteristic equation. It is not a coincidence; it is a fundamental property of matrices that bridges the gap between scalar polynomials and matrix algebra.

Q: When I multiply by $A^{-1}$, what happens to the identity matrix $I$?

Great question. Remember that $A^{-1} \cdot I = A^{-1}$. When you multiply the constant term (which is $6I$) by $A^{-1}$, you get $6A^{-1}$. This is exactly the term you need to isolate to find the inverse.

Question

$A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 0 & -2 & 4 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ and $A^{-1} = \left[ \frac{1}{6}(A^2 + cA + dI) ight]$, then the value of $c$ and $d$ are

Options:

(A) (a) $(-6, -11)$

(B) (b) $(6, 11)$

(C) (c) $(-6, 11)$

(D) (d) $(6, -11)$

Sigma Percentile · Accepted Answer

### The Art of the Shortcut: Mastering Cayley-Hamilton

Welcome, future engineers. Today, we stand before a problem that separates the calculators from the thinkers.

You see a matrix $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 0 & -2 & 4 \end{bmatrix}$ and an expression for its inverse, and your instinct might be to dive into the abyss of adjoints and cofactors. Stop. Take a breath.

In the JEE Advanced, speed is not just about moving your hand faster; it is about choosing the path of least resistance. We are going to use the **Cayley-Hamilton Theorem**, a tool that turns a tedious calculation into a moment of mathematical elegance.

### Phase 1: The DNA of the Matrix

First, we define the characteristic equation. We set the determinant of $A - \lambda I$ to zero. This isn't just algebra; it is finding the 'DNA' of the matrix.

We set up the determinant:

$$\begin{vmatrix} 1-\lambda & 0 & 0 \ 0 & 1-\lambda & 1 \ 0 & -2 & 4-\lambda \end{vmatrix} = 0$$

Look at the first row. It is a gift, given those two zeros. Expanding along the first row, we quickly arrive at the cubic polynomial.

We take $(1-\lambda)$ and multiply it by the determinant of the remaining $2 	imes 2$ matrix:

$$(1-\lambda) \left[ (1-\lambda)(4-\lambda) - (-2)(1) ight] = 0$$

Simplifying the terms inside the square brackets, we get $(1-\lambda)(\lambda^2 - 5\lambda + 6) = 0$. Multiplying this out, we arrive at the characteristic polynomial:

$$\lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0$$

### Phase 2: The Transformation

Now, the magic happens. We invoke the **Cayley-Hamilton Theorem**. This theorem tells us that we can replace the scalar variable $\lambda$ directly with our matrix $A$.

The constant term, $6$, becomes $6I$. Our equation transforms into:

$$A^3 - 6A^2 + 11A - 6I = O$$

This is the bridge between scalar polynomials and matrix algebra. We have successfully encoded the properties of the matrix into a single equation.

### Phase 3: The Final Reveal

To find $A^{-1}$, we don't need to invert anything. We simply multiply the entire equation by $A^{-1}$.

Remember, $A^{-1} \cdot A = I$. So, the power of every $A$ term reduces by one:

$$A^2 - 6A + 11I - 6A^{-1} = O$$

Rearranging to isolate $A^{-1}$, we get $6A^{-1} = A^2 - 6A + 11I$. Dividing by $6$, we find:

$$A^{-1} = \frac{1}{6}(A^2 - 6A + 11I)$$

Comparing this to the given form $A^{-1} = \frac{1}{6}(A^2 + cA + dI)$, we can clearly see that **$c = -6$** and **$d = 11$**.

It is clean, it is fast, and it is beautiful. You have just solved a complex matrix problem without ever calculating a single cofactor. Keep this mindset, and you will conquer the JEE.

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

Select Answer:

Visualized Solution

Introduction to the Problem

The Characteristic Equation

Setting up $∣ A - λ I ∣ = 0$

Choosing the Expansion Row

Expanding the Determinant

Simplifying the Quadratic

Forming the Cubic Polynomial

Final Characteristic Polynomial

Applying Cayley-Hamilton Theorem

Introducing $A^{- 1}$

Simplifying the Matrix Equation

Isolating $A^{- 1}$

Comparing Coefficients

The Sigma Insight: Adjoint and Inverse of a Matrix

The Art of the Shortcut

Phase 1

Phase 2

Phase 3

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