Question

If $A$ is a matrix such that $A(A + I) + 2I = 0$, then which of the following is/are true?

• A. A is non-singular
• B. A is symmetric
• C. A cannot be skew-symmetric
• D. $A^{-1} = -\frac{1}{2}(A+I)$

Answer 1

Answer: (A), (C), (D)

(A), (C), (D)

Answer 2

The given matrix equation is $A(A + I) + 2I = 0$. Expanding this, we get $A^2 + AI + 2I = 0$, which simplifies to $A^2 + A + 2I = 0$.

Let's analyze each option:

(A) $A$ is non-singular

From the equation $A^2 + A + 2I = 0$, we can write $A(A+I) = -2I$. Taking the determinant of both sides:

$\det(A(A+I)) = \det(-2I)$

$\det(A) \det(A+I) = \det(-2I)$

If $A$ is an $n \times n$ matrix, then $\det(-2I) = (-2)^n \det(I) = (-2)^n \cdot 1 = (-2)^n$.

Since $(-2)^n \neq 0$ for any integer $n$, it implies that $\det(A) \det(A+I) \neq 0$.

Therefore, $\det(A) \neq 0$, which means $A$ is non-singular. This statement is true.

(B) $A$ is symmetric

A matrix $A$ is symmetric if $A^T = A$.

The matrix $A$ satisfies the polynomial equation $x^2 + x + 2 = 0$. The eigenvalues of $A$ are the roots of this characteristic equation.

The roots are $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(2)}}{2} = \frac{-1 \pm \sqrt{1 - 8}}{2} = \frac{-1 \pm \sqrt{-7}}{2} = \frac{-1 \pm i\sqrt{7}}{2}$.

These are complex eigenvalues.

If $A$ is a real symmetric matrix, its eigenvalues must be real. Since the eigenvalues are complex, $A$ cannot be a real symmetric matrix. In the context of JEE/NEET, matrices are generally assumed to be real unless specified otherwise.

Even if $A$ is a complex matrix, the statement "A is symmetric" implies it is a necessary property. However, one can construct non-symmetric complex matrices that satisfy this equation. Therefore, this statement is not universally true for any matrix $A$ satisfying the condition. This statement is false.

(C) $A$ cannot be skew-symmetric

A matrix $A$ is skew-symmetric if $A^T = -A$.

Assume, for contradiction, that $A$ is skew-symmetric.

We are given $A^2 + A + 2I = 0$.

Taking the transpose of the entire equation:

$(A^2 + A + 2I)^T = 0^T$

$(A^2)^T + A^T + (2I)^T = 0$

$(A^T)^2 + A^T + 2I = 0$

Since we assumed $A^T = -A$, substitute this into the equation:

$(-A)^2 + (-A) + 2I = 0$

$A^2 - A + 2I = 0$ (Equation 1)

We also have the original equation:

$A^2 + A + 2I = 0$ (Equation 2)

Subtract Equation 1 from Equation 2:

$(A^2 + A + 2I) - (A^2 - A + 2I) = 0 - 0$

$A^2 + A + 2I - A^2 + A - 2I = 0$

$2A = 0$

$A = 0$

Now, substitute $A=0$ back into the original equation $A^2 + A + 2I = 0$:

$0^2 + 0 + 2I = 0$

$2I = 0$

This implies that the identity matrix $I$ is a zero matrix, which is false for any identity matrix of order $n \ge 1$.

Therefore, our initial assumption that $A$ is skew-symmetric must be false. Hence, $A$ cannot be skew-symmetric. This logic holds for both real and complex matrices. This statement is true.

(D) $A^{-1} = -\frac{1}{2}(A+I)$

From the given equation $A^2 + A + 2I = 0$.

Since $A$ is non-singular (as established in option A), $A^{-1}$ exists.

Multiply the entire equation by $A^{-1}$ from the left (or right):

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

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

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

$IA + I + 2A^{-1} = O$

$A + I + 2A^{-1} = O$

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

$A^{-1} = -\frac{1}{2}(A+I)$

This statement is true.

Therefore, the true statements are (A), (C), and (D).