Let A be a 2 x 2 matrix with real entries. Let I be the 2 x 2 identity matrix. Denote by tr(A), the

Q: Wait, why do we reject the case where $a+d \neq 0$?

Great question! If $a+d \neq 0$, then the equations $b(a+d)=0$ and $c(a+d)=0$ force $b=0$ and $c=0$. This makes $A$ a diagonal matrix $\begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$. Since $A^2 = I$, we get $a^2=1$ and $d^2=1$, meaning $a, d \in \{1, -1\}$. This would mean $A$ is either $I$, $-I$, or $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ or $\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$. The problem explicitly excludes $A=I$ and $A=-I$. While the other two diagonal matrices are possible, the condition $A \neq \pm I$ is a general constraint. By proving $a+d=0$ is necessary, we are exploring the structure of all such matrices that satisfy $A^2=I$.

Q: Is it possible for $tr(A)$ to be non-zero?

In this specific problem, no. We derived that $tr(A) = a+d = 0$ is a necessary condition for $A^2=I$ given $A \neq \pm I$. If the trace were non-zero, the matrix would have to be one of the trivial cases we already excluded. So, for any matrix $A$ satisfying the given conditions, the trace must be zero.

Q: How does $det(A) = -1$ relate to the eigenvalues?

That is a brilliant connection! If $A^2 = I$, then the eigenvalues $\lambda$ of $A$ must satisfy $\lambda^2 = 1$, so $\lambda = \pm 1$. The determinant of a matrix is the product of its eigenvalues. If the eigenvalues are $1$ and $-1$, the determinant is $(1)(-1) = -1$. If the eigenvalues were $1, 1$ or $-1, -1$, the determinant would be $1$, but that would correspond to $A=I$ or $A=-I$. Since we are not in those cases, the eigenvalues must be $1$ and $-1$, confirming $det(A) = -1$.

Question

Let $A$ be a $2 	imes 2$ matrix with real entries. Let $I$ be the $2 	imes 2$ identity matrix. Denote by $tr(A)$, the sum of diagonal entries of $A$. Assume that $A^2 = I$. \ Statement-1: If $A 
eq I$ and $A 
eq -I$, then $det(A) = -1$ \ Statement-2: If $A 
eq I$ and $A 
eq -I$, then $tr(A) 
eq 0$.

Options:

(A) Statement-1 is false, Statement-2 is true

(B) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

(C) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

(D) Statement-1 is true, Statement-2 is false

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a transformation, a matrix $A$ that acts on a two-dimensional space. You are told that applying this transformation twice brings you right back to where you started, the identity transformation $I$.

This is the essence of the condition $A^2 = I$. Let us define our matrix $A$ as:

$$A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$$

The identity matrix is $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$. The trace, $tr(A)$, is the sum of the diagonal elements, $a+d$. This trace is a fundamental invariant of the matrix.

### The Algebraic Unfolding

To understand $A^2 = I$, we must compute the product $A \cdot A$. Performing the multiplication, we get:

$$A^2 = \begin{bmatrix} a^2 + bc & b(a+d) \ c(a+d) & bc + d^2 \end{bmatrix}$$

Look closely at the off-diagonal terms. We can factor them as $b(a+d)$ and $c(a+d)$. Since $a+d$ is the trace, we equate the elements of $A^2$ to $I$:

$$a^2 + bc = 1$$
$$b(a+d) = 0$$
$$c(a+d) = 0$$
$$d^2 + bc = 1$$

### The Constraint of Non-Triviality

We are given that $A 
eq I$ and $A 
eq -I$. What happens if $a+d 
eq 0$?

If the trace is not zero, then the equations $b(a+d) = 0$ and $c(a+d) = 0$ force $b=0$ and $c=0$. If $b=0$ and $c=0$, then $A$ is a diagonal matrix $\begin{bmatrix} a & 0 \ 0 & d \end{bmatrix}$.

With $a^2=1$ and $d^2=1$, $A$ would be one of the following:

$$\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}, 	ext{ or } \begin{bmatrix} -1 & 0 \ 0 & 1 \end{bmatrix}$$

The first two are $I$ and $-I$, which are excluded. This forces us to conclude that for any non-trivial solution, the trace $a+d$ must be exactly $0$. This immediately tells us that Statement-2, which claims $tr(A) 
eq 0$, is **false**.

### The Determinant's Elegance

Now, let us find the determinant, $det(A) = ad - bc$. We know $a+d = 0$, so $d = -a$.

From our first equation, $a^2 + bc = 1$, we can write $bc = 1 - a^2$. Substituting these into the determinant formula:

$$det(A) = a(-a) - (1 - a^2) = -a^2 - 1 + a^2$$

The $a^2$ terms cancel out, leaving us with $det(A) = -1$. This confirms that Statement-1 is **true**.

We have navigated the algebraic landscape and found the truth: **Statement-1 is true, and Statement-2 is false.** It is a beautiful reminder that in mathematics, constraints are not just limitations; they are the keys that unlock the solution.

Select Answer:

Visualized Solution

Defining the Matrix $A$

Computing $A^{2}$

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

Analyzing the Constraint $A \neq = \pm I$

Evaluating Statement-2

Evaluating Statement-1

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Algebraic Unfolding

The Constraint of Non-Triviality

The Determinant's Elegance

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