MCQ
Assertion $(A)$ : If $A=\frac{1}{3}\left[\begin{array}{ccc}1 & -2 & 2 \\ -2 & 1 & 2 \\ -2 & -2 & -1\end{array}\right]$, then $A\left(A^T\right)=I$
Reason $(R)$ : For any square matrix $A,\left(A^T\right)^T=A$
  • A
    Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.
  • Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A)$.
  • C
    $(A)$ is true but $(R)$ is false.
  • D
    $(A)$ is false but $(R)$ is true.

Answer

Correct option: B.
Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A)$.
$\because A A^T=\frac{1}{3}\left[\begin{array}{ccc}1 & -2 & 2 \\ -2 & 1 & 2 \\ -2 & -2 & -1\end{array}\right] \cdot $
$\frac{1}{3}\left[\begin{array}{ccc}1 & -2 & -2 \\ -2 & 1 & -2 \\ 2 & 2 & -1\end{array}\right] $
$=\frac{1}{9}\left[\begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Consider the function $f(x)=[\sin x], x \in[0, \pi]$. Assertion $(A) : f(x)$ is not continuous at $x=\frac{\pi}{2}$.
Reason $(R) : \lim _{x \rightarrow \frac{\pi}{2}} f(x)$ does not exist.
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as :
Assertion : For two matrices $A$ and $B$ of order $3, \mid\text{A}\mid=2\mid\text{B}\mid=-3$ then if $\mid2\text{AB}\mid$ is $-48.$
Reason : For a square matrix $A, \text{A}(\text{adj}\ \text{A})=(\text{adj}\ \text{A})\text{A}=\mid\text{A}\mid.$
Directions: In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as:
Assertion: The solution of $\sin^{-1}(6\text{x})+\sin^{-1}(6\sqrt{3}\text{x})=\frac{-\pi}{2}$ is $\text{x}=\frac{1}{12}.$
Reason: $\cot^{-1}\text{x}$ is increasing function for $0\leq\text{x}\leq1.$
Directions : In the following questions, the Assertions $(A)$ and Reason $(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : Maximum value of $z = 11x + 7y,$ subjectto constraints $2\text{x}+\text{y}\leq6,\text{x}\leq2,\text{x}\geq0,\text{y}\geq0$ will be obtained at $(0, 6).$
Reason : In a bounded feasible region, it always exist a maximum and minimum value.
Directions : In the following questions, the Assertions $(A)$ and Reason $(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) : \text{f(x)}=\tan^2\text{x}$ is continuous at $\text{x}=\frac{\pi}{2}$
Reason $(R) :\ ?^2$ is continuous at $\text{x}=\frac{\pi}{2}$
Directions: In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R)$. Mark the correct choice as:
Assertion: If $\text{A}=\begin{pmatrix}1 & 2\\ 2& 3 \end{pmatrix}$ and $\text{B}=\begin{pmatrix}-1&4\\0&5\end{pmatrix}.$ $(A + B)^2 = A^2 + 2AB + B^2.$
Reason: $\text{AB}\neq\text{BA}.$
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):$ The curve $y = x^2$ represents a parabola with vertex at origin.
Reason $(R):$ For a curve Tangent and Normal lines are always perpendicular at thepoint of contact.
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following :
Assertion $(A) :$ The function $f(x) = x^3 - 3x^2 + 6x - 100$ is strictly increasing on $R$
Reason $(R) :$ A strictly increasing functions is an injective function.
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) f(x) = x - 1 + x - 2$ is continuous but not differentiable at $x = 1, 2.$
Reason $(R)$ Every differentiable function is continuous
Assertion (A) : If $A$ is skew-symmetric of order 3, then its determinant should be zero.
Reason (R) : If $A$ is square matrix, then $\operatorname{det} A=\operatorname{det} A^{\prime}=\operatorname{det}\left(-A^{\prime}\right)$.