If A = [α 0; 1 1] and B = [1 0; 5 1], then value of α for which A² = B, is (a) 1 (b) -1 (c) 4 (d) no

Q: Wait, I got $\alpha = 1$ from the top-left element. Why is that not the answer?

That is a classic trap! In matrix equality, $\alpha$ must satisfy *all* equations simultaneously. While $\alpha = 1$ or $\alpha = -1$ satisfies the top-left element, it must also satisfy the bottom-left element ($\alpha + 1 = 5$). Since $1+1 \neq 5$ and $-1+1 \neq 5$, $\alpha = 1$ and $\alpha = -1$ are not solutions to the entire matrix equation.

Q: Why can't I just solve one element and call it a day?

Because a matrix is a single mathematical object representing a collection of data. For two matrices to be equal, they must be identical in every position. If you only solve one element, you are ignoring the other constraints that the matrix structure imposes. Always check for consistency across all elements!

Q: Is it possible for a matrix equation to have no solution even if the individual equations have solutions?

Absolutely! This is a perfect example of an inconsistent system. Each individual equation (like $\alpha^2 = 1$ or $\alpha + 1 = 5$) has a solution, but there is no single value of $\alpha$ that satisfies both at the same time. In JEE problems, always look for this 'consistency check'—it's where the real test lies.

Question

If $A = \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$, then value of $\alpha$ for which $A^2 = B$, is

Options:

(A) (a) 1

(B) (b) $-1$

(C) (c) 4

(D) (d) no real values

Sigma Percentile · Accepted Answer

### The Matrix as a Universe of Constraints

Welcome, future engineers! Today, we are going to peel back the layers of a deceptively simple matrix problem. In the world of JEE Advanced, matrices are not just grids of numbers; they are containers of information, and when we equate them, we are essentially setting up a system of logical constraints.

Let's dive into the problem of finding $\alpha$ such that $A^2 = B$.

### The Setup: Defining the Battlefield

We are given two matrices:

$$A = \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$$

$$B = \begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$$

Our mission is to find the value of $\alpha$ that satisfies the condition $A^2 = B$. The first step is to understand that $A^2$ represents the matrix product $A 	imes A$. This is where the "row-by-column" dance begins.

### The Multiplication: The Dance of Rows and Columns

To compute $A^2$, we perform the multiplication:

$$A^2 = \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} 	imes \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$$

Let's break this down element by element. For the top-left element, we take the first row of the first $A$ and the first column of the second $A$:

$$(\alpha \cdot \alpha) + (0 \cdot 1) = \alpha^2$$

For the top-right element, we take the first row of the first $A$ and the second column of the second $A$:

$$(\alpha \cdot 0) + (0 \cdot 1) = 0$$

Moving to the second row, the bottom-left element is the second row of the first $A$ multiplied by the first column of the second $A$:

$$(1 \cdot \alpha) + (1 \cdot 1) = \alpha + 1$$

Finally, the bottom-right element is the second row of the first $A$ multiplied by the second column of the second $A$:

$$(1 \cdot 0) + (1 \cdot 1) = 1$$

So, our resulting matrix $A^2$ is:

$$A^2 = \begin{bmatrix} \alpha^2 & 0 \ \alpha + 1 & 1 \end{bmatrix}$$

### The Comparison: The Moment of Truth

Now, we equate this to matrix $B$:

$$\begin{bmatrix} \alpha^2 & 0 \ \alpha + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$$

For two matrices to be equal, every corresponding element must be identical. This gives us a system of equations:

1. Top-left: $\alpha^2 = 1$
2. Bottom-left: $\alpha + 1 = 5$

### The Reality Check: Why Consistency Matters

Here is where the JEE trap is set. If you only solve the first equation, you get $\alpha = \pm 1$. If you only solve the second, you get $\alpha = 4$.

A matrix equation is a single, unified statement. The variable $\alpha$ must satisfy **all** conditions simultaneously.

If $\alpha = 1$, then $\alpha + 1 = 2$, which is not $5$. If $\alpha = -1$, then $\alpha + 1 = 0$, which is not $5$. If $\alpha = 4$, then $\alpha^2 = 16$, which is not $1$.

There is no single value of $\alpha$ that makes both equations true. Therefore, the system is inconsistent. There is **no real value of $\alpha$** that satisfies the condition $A^2 = B$.

### Conclusion: The JEE Mindset

This problem is a beautiful reminder that in mathematics, especially in competitive exams, the "obvious" path is often a trap. We didn't just calculate; we analyzed the consistency of the system.

Always remember: when you are dealing with matrices, you are dealing with a system of constraints. If those constraints contradict each other, the answer is simply that no such value exists. Keep practicing, stay rigorous, and never stop questioning the logic behind the numbers!

Animated Solution for Mathematics - Matrices and Determinants: If $A = [α 1 01]$ and $B = [1501]$ , then value of $α$ for which $A^{2} = B$ , is

Select Answer:

Visualized Solution

Identify Matrices $A$ and $B$

Defining the Condition $A^{2} = B$

Matrix Multiplication: First Row

Matrix Multiplication: Second Row

Equating $A^{2}$ to $B$

Extracting Equations for $α$

Solving the First Equation

Solving the Second Equation

Checking for Consistency

Final Answer

The Sigma Insight: Algebraic Operations on Matrices

The Matrix as a Universe of Constraints

The Setup

The Multiplication

The Comparison

The Reality Check

Conclusion

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Visualized Solution

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