Q.If $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that $A^2 = I$, then (A) $1 + \alpha^2 + \beta\gamma = 0$ (B) $1 - \alpha^2 + \beta\gamma = 0$ (C) $1 - \alpha^2 - \beta\gamma = 0$ (D) $1 + \alpha^2 - \beta\gamma = 0$

Concept & Intuition

When a matrix satisfies a polynomial equation like $A^2 = 4A + 3I$, it means $A$ behaves like a root of that polynomial. We can manipulate this equation algebraically just like we would with numbers — but with matrices, we must be careful about commutativity (here, $A$ commutes with itself and with $I$, so we're safe).

The trick: if we multiply both sides by $A^{-1}$ (which exists, as we'll see), we get a linear expression for $A$ in terms of $I$. Then we can substitute that back into the given form $A^{-1} = xA + yI$ to find $x$ and $y$.

Step-by-step solution

1. Start with the given equation

   $A^2 = 4A + 3I$

   This is a matrix equation — every term is a $2\times2$ (or $n\times n$) matrix.

2. Multiply both sides by $A^{-1}$ on the left

   Since $A^{-1}A = I$, we get:

   $A^{-1}A^2 = A^{-1}(4A + 3I)$

   $\Rightarrow (A^{-1}A)A = 4A^{-1}A + 3A^{-1}I$

   $\Rightarrow IA = 4I + 3A^{-1}$

   So:

   $A = 4I + 3A^{-1}$

3. Rearrange to express $A^{-1}$ in terms of $A$ and $I$

   $3A^{-1} = A - 4I$

   $\Rightarrow A^{-1} = \frac{1}{3}A - \frac{4}{3}I$

4. Compare with the given form

   We are told $A^{-1} = xA + yI$.

   Matching coefficients:

   $x = \frac{1}{3}$,

   $y = -\frac{4}{3}$ …

Concept and Intuition

In scalar algebra, we are accustomed to the identity $a^2 - b^2 = (a - b)(a + b)$. This identity relies on the commutative property of multiplication, where $ab = ba$. For example, when we expand $(a - b)(a + b)$, we get $a^2 + ab - ba - b^2$, and since $ab = ba$, the middle terms cancel out, leaving $a^2 - b^2$.

However, matrix multiplication is generally not commutative. That is, for two matrices $A$ and $B$, it is usually the case that $AB \neq BA$. This non-commutativity is the crucial difference that this problem tests. If we blindly apply scalar algebra rules to matrices without considering the order of multiplication, we might make an error. The problem statement essentially gives us a condition under which the scalar identity does hold for matrices, and we need to find out what that condition implies about $A$ and $B$.

Step-by-step Derivation

1. Start with the given equation: We are given that $A$ and $B$ are square matrices of order $m$ such that:

2. Expand the right-hand side carefully: When multiplying matrices, we must maintain the order of multiplication. We distribute $(A - B)$ over $(A + B)$:

Now, distribute $A$ and $B$ into their respective parentheses:

This simplifies to:

> [!WARNING]
> A common mistake is to assume $AB = BA$ from the start, which would incorrectly simplify $AB - BA$ to $0$. Remember that matrix multiplication is not generally commutative.

3. Substitute the expanded form back into the original equation:

Now we equate the left-hand side of the given equation with our expanded right-hand side:

4. Simplify the equation: We can subtract $A^2$ from both sides of the equation:

Next, add $B^2$ to both sides:

Rearranging this equation, we get:

5. Interpret the result:

   The derivation shows that for the identity $A^2 - B^2 = (A - B)(A + B)$ to hold true for matrices $A$ and $B$, it is necessary that $AB = BA$. This means that matrices $A$ and $B$ must commute.

6. Evaluate the given options: …

The key insight here is to use the given relations to simplify powers of $A$ and $B$. We're told that $AB = A$ and $BA = B$, which means each matrix "absorbs" the other in a specific order. These relations are our tools to reduce any product back to simpler forms.

Let's compute $A^2$ and $B^2$ separately, then add them.

Finding $A^2$:

1. Write $A^2 = A \cdot A$.

2. We need to express this using our given relations. Notice that from $AB = A$, we can write $A = AB$.

3. Substitute this into $A^2$:

4. But we know $BA = B$, so:

Finding $B^2$:

1. Write $B^2 = B \cdot B$.

2. From $BA = B$, we have $B = BA$.

3. Substitute:

4. Using $AB = A$: $B^2 = B(AB) = BA = B$ …

Concept and Intuition

When you square a sum of matrices, you get the same expansion as with numbers: $(A+B)^2 = A^2 + AB + BA + B^2$. The only difference is that matrix multiplication is not commutative — $AB$ and $BA$ are generally different. The given condition says this sum equals $A^2 + B^2$, so the two middle terms $AB$ and $BA$ must add up to the zero matrix. That means $AB + BA = O$, which rearranges to $AB = -BA$. This is the definition of anti-commuting matrices.

Step-by-Step Solution

1. Expand the square Since matrix multiplication is distributive, we have:

This is exactly like the binomial expansion for numbers, but we must keep the order of multiplication.

2. Apply the given condition The problem states:

Substituting the expansion:

3. Cancel the common terms Subtract $A^2 + B^2$ from both sides:

4. Rearrange to the required form The equation $AB + BA = O$ is equivalent to:

This is the defining relation for matrices that anti-commute. …

We start with the matrix $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$. Notice that $A$ is a special matrix — it behaves like the imaginary unit $i$ in complex numbers because $A^2 = -I$. Let’s verify:

$A^2 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$.

This property is the heart of the problem. When we multiply two linear combinations of $I$ and $A$, the result will be a combination of $I$ and $A$ again, but because $A^2 = -I$, any $A^2$ term collapses back to a multiple of $I$. So the product $(3I+4A)(3I-4A)$ should simplify to something like $(\text{number})I + (\text{number})A$. Let’s find out exactly.

1. Expand the product carefully — but treat $I$ and $A$ as commuting matrices (they do, since $I$ commutes with everything).

   $(3I+4A)(3I-4A) = 3I \cdot 3I + 3I \cdot (-4A) + 4A \cdot 3I + 4A \cdot (-4A)$

   $= 9I^2 -12IA + 12AI -16A^2$.

   Since $I^2 = I$, $IA = A$, and $AI = A$, the middle terms $-12A + 12A$ cancel exactly. So we get:

   $= 9I - 16A^2$.

2. Now use $A^2 = -I$ to replace $A^2$:

   $9I - 16(-I) = 9I + 16I = 25I$.

   So the product simplifies to $25I$, a pure scalar multiple of the identity matrix.

3. The problem states that this product equals $x^2 I$. Therefore: …

When asked to compute a high power of a matrix, such as $A^{2023}$, the most efficient approach is to calculate the first few powers ($A^2, A^3, A^4, \dots$) and look for a pattern. Direct multiplication 2023 times is not feasible.

Matrices often exhibit predictable patterns in their powers. Common patterns include:

• Cyclic behavior: Powers repeat after a certain number of steps (e.g., $A^k = I$, where $I$ is the identity matrix).
• Nilpotency: A certain power of the matrix becomes the zero matrix. Once a matrix power is the zero matrix, all subsequent higher powers will also be the zero matrix. This is a very common scenario for matrices with many zero entries.
• Idempotency: $A^2 = A$. In this case, $A^n = A$ for all $n \ge 1$.

The matrix $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ has a simple structure with many zeros, which strongly suggests that it might be nilpotent. Let's calculate its powers to find the pattern.

1. Calculate $A^2$: We multiply $A$ by itself:

To perform matrix multiplication, we take the dot product of the rows of the first matrix with the columns of the second matrix.
*   The element in the first row, first column of $A^2$ is $(0)(0) + (1)(0) = 0$.
*   The element in the first row, second column of $A^2$ is $(0)(1) + (1)(0) = 0$.
*   The element in the second row, first column of $A^2$ is $(0)(0) + (0)(0) = 0$.
*   The element in the second row, second column of $A^2$ is $(0)(1) + (0)(0) = 0$.

Thus, we find:

2. Identify the pattern:

   We have found that $A^2$ is the zero matrix. A matrix $M$ is called nilpotent if $M^k = \mathbf{0}$ for some positive integer $k$, where $\mathbf{0}$ denotes the zero matrix. In this case, $A$ is a nilpotent matrix with an index of 2.

3. Generalize for $A^{2023}$:

   Since $A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, let's consider any higher power, say $A^n$ where $n \ge 2$. We can write $A^n$ as $A^2 \cdot A^{n-2}$. …