Suppose A is any 3 × 3 non-singular matrix and (A-3I)(A-5I) = 0, where I = I3 and O = O3. If αA + βA

Q: Wait, why can we just multiply by $A^{-1}$? Is that always allowed?

Great question! We can only multiply by $A^{-1}$ because the problem explicitly states that $A$ is non-singular. If $\det(A)$ were zero, $A^{-1}$ wouldn't exist, and this entire path would be blocked. Always check for that 'non-singular' keyword!

Q: Do I need to find the actual matrix $A$ to solve this?

Absolutely not! In JEE problems, you are rarely asked to find the specific elements of $A$ unless it's a very simple case. Here, we are dealing with the algebraic properties of the matrix. We are working with the characteristic equation, not the matrix entries themselves.

Q: Why does $(A-3I)(A-5I) = O$ simplify to $A^2 - 8A + 15I = O$?

It's just like expanding $(x-3)(x-5) = 0$ in scalar algebra! Because $A$ and $I$ commute (since $AI = IA = A$), we can treat them like variables. Just be careful: if you had two different matrices $A$ and $B$, you couldn't just expand them like that unless they commuted.

Question

Suppose A is any $3 	imes 3$ non-singular matrix and $(A-3I)(A-5I) = 0$, where $I = I_3$ and $O = O_3$. If $\alpha A + \beta A^{-1} = 4I$, then $\alpha + \beta$ is equal to :-

Options:

(A) 13

(B) 7

(C) 12

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dive into a problem that might look like a daunting wall of symbols, but is actually a beautiful, structured dance of logic.

We are dealing with a $3 	imes 3$ matrix $A$, and we are given a polynomial equation: 
$$(A-3I)(A-5I) = O$$

Our mission is to find the sum $\alpha + \beta$ such that $\alpha A + \beta A^{-1} = 4I$. Let's break this down step-by-step.

### Decoding the Polynomial

First, let's look at the given equation: $(A-3I)(A-5I) = O$. In the world of matrices, we must be careful with multiplication, but here, the identity matrix $I$ commutes with everything.

This means we can expand this just like a standard quadratic equation in scalar algebra. Let's distribute the terms:
$$A^2 - 5A - 3A + 15I = O$$

Since $I^2 = I$, our equation simplifies to:
$$A^2 - 8A + 15I = O$$

This is the characteristic-like behavior of our matrix $A$. It tells us exactly how $A^2$ relates to $A$ and $I$.

### The Power of the Inverse

Now, look at our target: $\alpha A + \beta A^{-1} = 4I$. We have an $A^{-1}$ term in our target, but our current equation only has $A^2$, $A$, and $I$.

The problem guarantees that $A$ is non-singular, meaning $\det(A) 
eq 0$, which ensures that $A^{-1}$ exists. This is our green light to multiply the entire equation $A^2 - 8A + 15I = O$ by $A^{-1}$.

When we multiply by $A^{-1}$, we get:
$$A^{-1}(A^2 - 8A + 15I) = A^{-1}O$$

Distributing $A^{-1}$ gives us $A^{-1}A^2 - 8A^{-1}A + 15A^{-1}I = O$. Since $A^{-1}A = I$ and $A^{-1}I = A^{-1}$, this simplifies to:
$$A - 8I + 15A^{-1} = O$$

### The Final Alignment

We are almost there! Let's rearrange this to look like our target form. Moving the $8I$ to the other side, we get:
$$A + 15A^{-1} = 8I$$

Now, compare this to $\alpha A + \beta A^{-1} = 4I$. The right-hand side of our equation is $8I$, but the target is $4I$. To fix this, we simply divide the entire equation by $2$:
$$\frac{1}{2}A + \frac{15}{2}A^{-1} = 4I$$

By comparing $\frac{1}{2}A + \frac{15}{2}A^{-1} = 4I$ with $\alpha A + \beta A^{-1} = 4I$, we can clearly see that $\alpha = \frac{1}{2}$ and $\beta = \frac{15}{2}$.

The question asks for the sum $\alpha + \beta$. Adding these together, we get:
$$\alpha + \beta = \frac{1}{2} + \frac{15}{2} = \frac{16}{2} = 8$$

And there you have it! The final answer is **8**. A truly elegant solution to a problem that initially seemed so complex.

Animated Solution for Mathematics - Matrices and Determinants: Suppose A is any $3 \times 3$ non-singular matrix and $(A - 3 I) (A - 5 I) = 0$ , where $I = I_{3}$ and $O = O_{3}$ . If $α A + β A^{- 1} = 4 I$ , then $α + β$ is equal to :-

Select Answer:

Visualized Solution

Introduction to the Problem

Expanding the Matrix Polynomial

Simplifying the Expansion

The Strategy: Introducing $A^{- 1}$

Multiplying by $A^{- 1}$

Simplifying the New Equation

Rearranging to Match the Target Form

Scaling the Equation

Comparing Coefficients

Final Calculation

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

Decoding the Polynomial

The Power of the Inverse

The Final Alignment

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