Let A = [[1, -1], [2, α]] and B = [[β, 1], [1, 0]], α, β ∈ R. Let α₁ be the value of α which satisfi

Q: Wait, why can't we just use the formula $(A+B)^2 = A^2 + 2AB + B^2$?

That is the most common trap in linear algebra! That formula only holds if $A$ and $B$ commute, meaning $AB = BA$. In general, matrix multiplication is not commutative. Always expand $(A+B)(A+B)$ as $A^2 + AB + BA + B^2$.

Q: How did you know to pick the (1,2) entry to solve for $\alpha$?

Great observation! In matrix equations, every entry must satisfy the equality. I chose the $(1,2)$ entry because it resulted in a simple linear equation. You could have chosen any entry, but why do extra work? Always look for the 'path of least resistance' in the matrix.

Q: Does the value of $\beta$ matter for $\alpha_1$?

In Case 1, notice that the $(1,2)$ entry of $(A+B)^2$ is $0$. When we compare it to the $(1,2)$ entry of $A^2 + K$, the $\beta$ terms cancel out or are absent, allowing us to solve for $\alpha_1$ directly. It is a beautiful simplification!

Q: Why did we get $\beta = 0$ in the second case?

By equating the $(1,2)$ entry of $(A+B)^2$ (which is $0$) to the $(1,2)$ entry of $B^2$ (which is $\beta$), the equality forces $\beta$ to be $0$. It is not an assumption; it is a mathematical necessity imposed by the problem's condition.

Q: Is it possible that $\alpha_1$ or $\alpha_2$ could be complex numbers?

The problem states $\alpha, \beta \in \mathbb{R}$. If our calculations had led to a quadratic with a negative discriminant, we would have to re-evaluate. Here, the equations were linear, ensuring real solutions.

Question

Let $A = \begin{bmatrix} 1 & -1 \ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \ 1 & 0 \end{bmatrix}, \alpha, \beta \in \mathbb{R}$. Let $\alpha_1$ be the value of $\alpha$ which satisfies $(A+B)^2 = A^2 + \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix}$ and $\alpha_2$ be the value of $\alpha$ which satisfies $(A+B)^2 = B^2$. Then $|\alpha_1 - \alpha_2|$ is equal to

Sigma Percentile · Accepted Answer

### The Dance of Matrices: A Journey into Algebraic Elegance

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a matrix problem; we are choreographing a dance between two matrices, $A$ and $B$.

Many students look at matrix algebra and see a wall of numbers. I want you to see a structure, a system that obeys its own unique laws—laws that are often more subtle than the arithmetic we learned in primary school.

### The Setup: Defining Our Players

We begin with two matrices:

$$A = \begin{bmatrix} 1 & -1 \ 2 & \alpha \end{bmatrix}, \quad B = \begin{bmatrix} \beta & 1 \ 1 & 0 \end{bmatrix}$$

Our first task is to find the sum $A+B$. This is the easy part, the warm-up. We add the corresponding elements to obtain:

$$A + B = \begin{bmatrix} \beta+1 & 0 \ 3 & \alpha \end{bmatrix}$$

Notice that zero in the top-right corner? That is a gift. In mathematics, whenever you see a zero, you should smile, as it is a simplification waiting to happen.

### The Expansion: Squaring the Sum

Now, we calculate $(A+B)^2$. Remember, matrix multiplication is row-by-column.

When we perform the multiplication, we get:

$$(A+B)^2 = \begin{bmatrix} \beta+1 & 0 \ 3 & \alpha \end{bmatrix} \begin{bmatrix} \beta+1 & 0 \ 3 & \alpha \end{bmatrix} = \begin{bmatrix} (\beta+1)^2 & 0 \ 3(\beta+1) + 3\alpha & \alpha^2 \end{bmatrix}$$

Look at that! The top-right entry remains zero. This is the 'anchor' of our calculation, and no matter what $\alpha$ or $\beta$ are, that entry is locked at zero.

### Case 1: The First Constraint

We are told that $(A+B)^2 = A^2 + \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix}$. First, let us find $A^2$:

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

Now, we compare the $(1,2)$ entry of both sides. On the left, we have $0$. On the right, we have the $(1,2)$ entry of $A^2$ plus the $(1,2)$ entry of the constant matrix, which is $2$.

So, we have:

$$0 = (-1 - \alpha) + 2 \implies 1 - \alpha = 0 \implies \alpha_1 = 1$$

### Case 2: The Second Constraint

Now, we face the condition $(A+B)^2 = B^2$. First, we compute $B^2$:

$$B^2 = \begin{bmatrix} \beta & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} \beta & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} \beta^2+1 & \beta \ \beta & 1 \end{bmatrix}$$

Comparing the $(1,2)$ entries again: on the left, we have $0$. On the right, we have $\beta$. Thus, $\beta = 0$.

Now, we look at the $(2,1)$ entry to find $\alpha_2$. From our previous expansion of $(A+B)^2$, the $(2,1)$ entry is $3(\beta+1) + 3\alpha$. Setting this equal to the $(2,1)$ entry of $B^2$, which is $\beta$:

$$3(\beta+1) + 3\alpha = \beta$$

Substituting $\beta = 0$ into the equation:

$$3(1) + 3\alpha = 0 \implies 3\alpha = -3 \implies \alpha_2 = -1$$

### The Final Victory

We have arrived at our destination. We found $\alpha_1 = 1$ and $\alpha_2 = -1$. The problem asks for $|\alpha_1 - \alpha_2|$:

$$|1 - (-1)| = |1 + 1| = \mathbf{2}$$

It is a beautiful result. We navigated the non-commutative nature of matrices, used the simplicity of the zero entry to our advantage, and arrived at a clean, integer answer.

Enter Numerical Value:

Visualized Solution

Define Matrices $A$ and $B$

Compute the Sum Matrix $(A + B)$

Calculate $(A + B)^{2}$

Calculate $A^{2}$ for Case 1

Solve for $α_{1}$ in Case 1

Calculate $B^{2}$ for Case 2

Solve for $α_{2}$ in Case 2

Final Calculation

The Sigma Insight: Algebraic Operations on Matrices

The Dance of Matrices

The Setup

The Expansion

Case 1

Case 2

The Final Victory

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Enter Numerical Value:

Visualized Solution

Define Matrices .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; } B

Compute the Sum Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A+B)

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A+B)2

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2 for Case 1

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α1​ in Case 1

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2 for Case 2

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2​ in Case 2