Let A be a 2 × 2 matrix with real entries such that Aᵀ = αA + I, where α ∈ ℝ -{-1,1}. If det (A² – A

Q: Wait, why did we assume $A$ is a scalar matrix? Can't it be any matrix?

Great question! We didn't assume it; we derived it. By equating the elements of $A^T$ and $\alpha A + I$, we found that $b=c=0$ and $a=d$. This forces the matrix to be a scalar multiple of the identity matrix. It's a consequence of the constraints, not an assumption.

Q: Why do we reject the case where the discriminant is negative?

The problem explicitly states that $\alpha$ is a real number. Since $a = \frac{1}{1-\alpha}$, if $\alpha$ is real, $a$ must also be real. Therefore, we only consider the real roots of the quadratic equation.

Q: Is there a faster way to solve $\det(A^2 - A) = 4$?

You could find the eigenvalues of $A$. Since $A = aI$, the eigenvalues are both $a$. The eigenvalues of $A^2 - A$ are $a^2 - a$. The determinant is the product of eigenvalues, so $(a^2 - a)^2 = 4$. It's the same logic, just a different perspective!

Question

Let $A$ be a $2 	imes 2$ matrix with real entries such that $A^T = \alpha A + I$, where $\alpha \in \mathbb{R} -\{-1,1\}$. If $\det (A^2 – A)= 4$, the sum of all possible values of $\alpha$ is equal to

Options:

(A) 0

(B) $\frac{1}{2}$

(C) 2

(D) $\frac{5}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Matrix Structure

We begin with the given matrix equation:
$$A^T = \alpha A + I$$

Let us represent the matrix $A$ as:
$$A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$$

Taking the transpose of $A$, the off-diagonal elements swap positions. The equation becomes:
$$\begin{pmatrix} a & c \ b & d \end{pmatrix} = \begin{pmatrix} \alpha a + 1 & \alpha b \ \alpha c & \alpha d + 1 \end{pmatrix}$$

### Equating Elements

By equating the corresponding elements of the matrices, we obtain the following system of equations:
1. $a = \alpha a + 1$
2. $d = \alpha d + 1$
3. $c = \alpha b$
4. $b = \alpha c$

From equations (1) and (2), we observe that $a = d = \frac{1}{1 - \alpha}$. Substituting equation (4) into (3), we get $c = \alpha^2 c$, which implies $c(1 - \alpha^2) = 0$.

Assuming $\alpha^2 
eq 1$, we find $c = 0$ and consequently $b = 0$. Thus, $A$ simplifies to a scalar matrix of the form $A = aI$.

### Solving the Determinant Equation

We are given the condition $\det(A^2 - A) = 4$. Since $A = aI$, we can write:
$$A^2 - A = a^2I - aI = (a^2 - a)I$$

The determinant of a $2 	imes 2$ scalar matrix $kI$ is $k^2$. Therefore:
$$(a^2 - a)^2 = 4$$

This yields two possible cases for the quadratic expression:
$$a^2 - a = 2 \quad 	ext{or} \quad a^2 - a = -2$$

### Final Calculation

For the first case, $a^2 - a - 2 = 0$, which factors as $(a - 2)(a + 1) = 0$, giving $a = 2$ or $a = -1$. For the second case, $a^2 - a + 2 = 0$, the discriminant is $D = 1 - 8 = -7$, which yields no real solutions for $a$.

Using the relation $a = \frac{1}{1 - \alpha}$, we solve for $\alpha$:
$$\alpha = 1 - \frac{1}{a}$$

For $a = 2$, $\alpha = 1 - \frac{1}{2} = \frac{1}{2}$. For $a = -1$, $\alpha = 1 - \frac{1}{-1} = 2$.

The sum of all possible real values of $\alpha$ is:
$$\frac{1}{2} + 2 = \mathbf{\frac{5}{2}}$$

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = α A + I$ , where $α \in R - {- 1, 1}$ . If $det (A^{2} - A) = 4$ , the sum of all possible values of $α$ is equal to

Select Answer:

Visualized Solution

Define Matrix $A$

Transpose Equation Setup

Equating Diagonal Elements

Equating Off-Diagonal Elements

Deducing $b$ and $c$

Simplified Matrix $A$

Determinant Condition

Substituting $A = a I$

Solving for $a$

Finding Values of $α$

Final Sum of $α$

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Matrix Structure

Equating Elements

Solving the Determinant Equation

Final Calculation

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Select Answer:

Visualized Solution

Define Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Transpose Equation Setup

Equating Diagonal Elements

Equating Off-Diagonal Elements

Deducing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Simplified Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Determinant Condition

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=aI

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

Finding Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Final Sum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α