If A = [[2, 2], [9, 4]] and I = [[1, 0], [0, 1]], then 10A⁻¹ is equal to :
(A) 4I – A
(B) 6I - A
(C)

Q: Wait, why can't we just use the standard formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$?

You absolutely can! The adjoint method is a perfectly valid and reliable way to find the inverse. However, in competitive exams like the JEE, time is your most precious resource. The Cayley-Hamilton approach allows you to express the inverse as a linear combination of the matrix $A$ and the identity matrix $I$ without calculating the adjoint or the inverse explicitly. It is a powerful shortcut that demonstrates a deeper understanding of matrix properties.

Q: What exactly is the Cayley-Hamilton Theorem, and why does it work?

The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. If the characteristic polynomial is $p(\lambda) = \det(A - \lambda I)$, then $p(A) = 0$. It is a fundamental result in linear algebra that bridges the gap between scalar polynomials and matrix algebra. It is incredibly useful for finding higher powers of matrices and their inverses.

Q: Why do we replace the constant term $10$ with $10I$ in the final equation?

This is a common point of confusion! In the characteristic equation, we are dealing with scalars (like $\lambda^2 - 6\lambda - 10 = 0$). When we transition to matrix algebra, every term must be a matrix. Since $A^2$ and $A$ are matrices, the constant term must also be a matrix to maintain dimensional consistency. Therefore, we multiply the scalar constant by the identity matrix $I$ to make it a matrix.

Question

If $A = \begin{pmatrix} 2 & 2 \ 9 & 4 \end{pmatrix}$ and $I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$, then $10A^{-1}$ is equal to :

Options:

(A) $4I - A$

(B) $6I - A$

(C) $A - 4I$

(D) $A - 6I$

Sigma Percentile · Accepted Answer

### The Elegance of Matrix Algebra: Beyond the Adjoint Method

Welcome, aspiring engineers! Today, we are going to look at a classic matrix problem that often appears in the JEE Advanced. We are given a matrix $A = \begin{pmatrix} 2 & 2 \ 9 & 4 \end{pmatrix}$ and asked to find the expression for $10A^{-1}$.

When you first see this, your instinct might be to jump straight into the standard formula: $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$. While that is a perfectly correct path, it is the path of the novice. Today, we are going to walk the path of the master by using the **Cayley-Hamilton Theorem**.

### The Gateway: The Characteristic Equation

To unlock the power of Cayley-Hamilton, we first need the characteristic equation of matrix $A$. This equation is defined as $|A - \lambda I| = 0$. Think of this as the 'DNA' of the matrix—it encapsulates the essential properties of the transformation $A$ represents.

Let's set up the determinant:

$$\begin{vmatrix} 2-\lambda & 2 \ 9 & 4-\lambda \end{vmatrix} = 0$$

Expanding this $2 	imes 2$ determinant is straightforward, but this is where precision matters. We cross-multiply the diagonal elements and subtract the product of the off-diagonal elements:

$$(2-\lambda)(4-\lambda) - (9 	imes 2) = 0$$

Expanding the brackets, we get:

$$8 - 2\lambda - 4\lambda + \lambda^2 - 18 = 0$$

Simplifying this, we arrive at our characteristic equation:

$$\lambda^2 - 6\lambda - 10 = 0$$

### The Power of Cayley-Hamilton

Now, here is the magic. The Cayley-Hamilton Theorem tells us that every square matrix satisfies its own characteristic equation. This means we can replace the scalar $\lambda$ with the matrix $A$ itself!

But remember, we must treat the constant term as a matrix, so we multiply it by the identity matrix $I$:

$$A^2 - 6A - 10I = 0$$

This equation is a beautiful relationship between the matrix $A$, its square, and the identity matrix. It is the key to solving our problem without ever calculating the adjoint.

### The Final Leap: Isolating the Inverse

We need to find $10A^{-1}$. Look at our equation: $A^2 - 6A - 10I = 0$. If we multiply the entire equation by $A^{-1}$, we can isolate the inverse term:

$$A^{-1}(A^2) - 6A^{-1}(A) - 10A^{-1}(I) = 0$$

Since $A^{-1}A = I$ and $A^{-1}I = A^{-1}$, this simplifies elegantly to:

$$A - 6I - 10A^{-1} = 0$$

Now, simply rearrange the terms to solve for $10A^{-1}$:

$$10A^{-1} = A - 6I$$

And there you have it! We have bypassed the tedious calculation of the adjoint and arrived at the solution using the structural properties of the matrix. This is the kind of mathematical elegance that will not only save you time in the exam hall but also deepen your appreciation for the underlying beauty of linear algebra. Keep practicing these shortcuts—they are the tools that separate the good from the elite!

Animated Solution for Mathematics - Matrices and Determinants: If $A = (2924)$ and $I = (1001)$ , then $10 A^{- 1}$ is equal to :

Select Answer:

Visualized Solution

Introduction to Matrix $A$

The Characteristic Equation

Setting up the Determinant

Expanding the Determinant

Solving the Quadratic

Applying Cayley-Hamilton Theorem

Checking for Invertibility

Multiplying by $A^{- 1}$

Isolating $10 A^{- 1}$

Final Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

The Elegance of Matrix Algebra

The Gateway

The Power of Cayley-Hamilton

The Final Leap

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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=(29​24​) and I=(10​01​), then 10A−1 is equal to :