Question

If A is matrix such that $A^2+A+2I=0$, then which of the following is INCORRECT?

• A. A is non-singular
• B. A ≠ 0
• C. A is symmetric
• D. $A^{-1} = -\frac{1}{2}(A+I)$

Answer 1

Answer: (C)

(C)

Answer 2

The given equation is $A^2+A+2I=0$, where A is a matrix, I is the unit matrix of order 2, and O is the null matrix of order 2.

Let's analyze each option:

(A) A is non-singular.

From the given equation, we can write $A^2+A = -2I$. Factoring A on the left, we get $A(A+I) = -2I$. Taking the determinant of both sides: $\det(A(A+I)) = \det(-2I)$ $\det(A)\det(A+I) = \det(-2I)$ Since I is a 2x2 matrix, $-2I = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}$. $\det(-2I) = (-2)(-2) - (0)(0) = 4$. So, $\det(A)\det(A+I) = 4$. Since the product is 4 (which is non-zero), neither $\det(A)$ nor $\det(A+I)$ can be zero. Thus, $\det(A) \neq 0$, which means A is non-singular. Alternatively, from $A(A+I) = -2I$, we can write $A \left(-\frac{1}{2}(A+I)\right) = I$. This shows that the inverse of A is $A^{-1} = -\frac{1}{2}(A+I)$. Since the inverse exists, A is non-singular. Statement (A) is correct.

(B) A ≠ 0.

If A were the null matrix O, the equation $A^2+A+2I=0$ would become $O^2+O+2I=0$, which simplifies to $O+O+2I=O$, or $2I=O$. For a 2x2 matrix, $2I = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ and $O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. $\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ is false. Thus, A cannot be the null matrix. Statement (B) is correct.

(C) A is symmetric.

A matrix A is symmetric if $A^T = A$. The equation $A^2+A+2I=0$ implies that A satisfies the polynomial $p(x) = x^2+x+2=0$. By the Cayley-Hamilton theorem, the characteristic polynomial of A must be $\lambda^2+\lambda+2=0$. The eigenvalues of A are the roots of this characteristic equation: $\lambda = \frac{-1 \pm \sqrt{1^2 - 4(1)(2)}}{2} = \frac{-1 \pm \sqrt{-7}}{2} = \frac{-1 \pm i\sqrt{7}}{2}$. The eigenvalues are complex. If A were 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. Assuming A is a real matrix, it cannot be symmetric. If A were allowed to be a complex matrix, it is possible for a complex symmetric matrix to satisfy the equation (e.g., a diagonal matrix with entries being the roots of $x^2+x+2=0$). However, the statement "A is symmetric" typically implies it is a necessary property given the condition. Since we can find non-symmetric matrices satisfying the equation (as shown in the thought process, constructing a non-symmetric matrix with complex entries satisfying the system of equations derived from $A^2+A+2I=0$ and $A^T \neq A$ is possible), the statement "A is symmetric" is not necessarily true for any matrix A satisfying the condition. Therefore, it is an incorrect statement about A in general.

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

From the given equation $A^2+A+2I=0$. We know from (A) that A is non-singular, so $A^{-1}$ exists. Multiply the equation by $A^{-1}$ on the left: $A^{-1}(A^2+A+2I) = A^{-1}O$ $A^{-1}A^2 + A^{-1}A + A^{-1}(2I) = O$ $(A^{-1}A)A + I + 2A^{-1}I = O$ $IA + I + 2A^{-1} = O$ $A + I + 2A^{-1} = O$ $2A^{-1} = -(A+I)$ $A^{-1} = -\frac{1}{2}(A+I)$. Statement (D) is correct.

The incorrect statement is (C).