Matrices & Determinants
Orthogonal and Skew-Symmetric Matrices
GRB_1000_MCQ
Grade Class 12

Question:

If both $A - \dfrac{I}{2}$ and $A + \dfrac{I}{2}$ are orthogonal matrix, then which of the following statements are <b>incorrect</b>? (where $I$ is an identity matrix order same as that of $A$.)
$A$ is skew-symmetric matrix of odd order.
$A^2 = \dfrac{3}{4}I$
$A$ is skew-symmetric matrix of even order.
$A$ is orthogonal

Step-by-Step Solution

Key Concept: The key idea is to use the definition of an orthogonal matrix ($P^T P = I$) for both given matrices, then combine the resulting matrix equations through addition and subtraction to deduce fundamental properties of matrix A.
Step 1: Let $P = A - \frac{I}{2}$ and $Q = A + \frac{I}{2}$. Since $P$ and $Q$ are orthogonal matrices, we have $P^T P = I$ and $Q^T Q = I$. From $P^T P = I$: $$ \left(A - \frac{I}{2}\right)^T \left(A - \frac{I}{2}\right) = I $$ $$ \left(A^T - \frac{I}{2}\right) \left(A - \frac{I}{2}\right) = I $$ $$ A^T A - \frac{1}{2}A^T - \frac{1}{2}A + \frac{1}{4}I = I \quad \cdots (1) $$ From $Q^T Q = I$: $$ \left(A + \frac{I}{2}\right)^T \left(A + \frac{I}{2}\right) = I $$ $$ \left(A^T + \frac{I}{2}\right) \left(A + \frac{I}{2}\right) = I $$ $$ A^T A + \frac{1}{2}A^T + \frac{1}{2}A + \frac{1}{4}I = I \quad \cdots (2) $$ Step 2: Add equations (1) and (2): $$ \left(A^T A - \frac{1}{2}A^T - \frac{1}{2}A + \frac{1}{4}I\right) + \left(A^T A + \frac{1}{2}A^T + \frac{1}{2}A + \frac{1}{4}I\right) = I + I $$ $$ 2A^T A + \frac{1}{2}I = 2I $$ $$ 2A^T A = \frac{3}{2}I $$ $$ A^T A = \frac{3}{4}I $$ Step 3: Subtract equation (1) from equation (2): $$ \left(A^T A + \frac{1}{2}A^T + \frac{1}{2}A + \frac{1}{4}I\right) - \left(A^T A - \frac{1}{2}A^T - \frac{1}{2}A + \frac{1}{4}I\right) = I - I $$ $$ A^T + A = 0 $$ $$ A^T = -A $$ Thus, $A$ is a skew-symmetric matrix. Step 4: Substitute $A^T = -A$ into $A^T A = \frac{3}{4}I$: $$ (-A)A = \frac{3}{4}I $$ $$ -A^2 = \frac{3}{4}I $$ $$ A^2 = -\frac{3}{4}I $$ Step 5: Evaluate the given statements: 1. **$A$ is skew-symmetric matrix of odd order.** If $A$ is a skew-symmetric matrix of odd order, then $\det(A) = 0$. From $A^T A = \frac{3}{4}I$, we have $\det(A^T A) = \det\left(\frac{3}{4}I\right)$. $\det(A^T)\det(A) = \left(\frac{3}{4}\right)^n \det(I)$, where $n$ is the order of $A$. $(\det(A))^2 = \left(\frac{3}{4}\right)^n$. If $n$ is odd, $\det(A)=0$, which implies $0 = (\frac{3}{4})^n$, a contradiction. Therefore, $A$ cannot be of odd order. This statement is incorrect. 2. **$A^2 = \frac{3}{4}I$.** From Step 4, we found $A^2 = -\frac{3}{4}I$. This statement is incorrect. 3. **$A$ is skew-symmetric matrix of even order.** From Step 3, $A$ is skew-symmetric. From the analysis in statement 1, $A$ must be of even order. This statement is correct. 4. **$A$ is orthogonal.** For $A$ to be orthogonal, $A^T A = I$. From Step 2, we found $A^T A = \frac{3}{4}I$. Since $\frac{3}{4}I \neq I$, $A$ is not orthogonal. This statement is incorrect. The incorrect statements are 1, 2, and 4.
Correct Answer: 1, 4

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