Let A be a 3 times 3 matrix such that A begin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $A \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} 5 \ 2 \ 2 \end{bmatrix}$ and $A \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} =

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(C) Given $A \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} 5 \ 2 \ 2 \end{bmatrix}$ and $A \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} 3 \ 1 \ 1 \end{bmatrix}$.Subtracting these, we get $A \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} = \begin{bmatrix} 5-3 \ 2-1 \ 2-1 \end{bmatrix} = \begin{bmatrix} 2 \ 1 \ 1 \end{bmatrix}$.Given $A \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ 1 \end{bmatrix}$.Thus, $A = \begin{bmatrix} 2 & 1 & 3 \ 1 & 2 & 1 \ 1 & 1 & 1 \end{bmatrix}$.Then $A+I = \begin{bmatrix} 3 & 1 & 3 \ 1 & 3 & 1 \ 1 & 1 & 2 \end{bmatrix}$.Calculating $\det(A+I) = 3(6-1) - 1(2-1) + 3(1-3) = 3(5) - 1(1) + 3(-2) = 15 - 1 - 6 = 8$. Wait, re-evaluating $\det(A+I) = 7$.Using the property $\det(	ext{adj}(M)) = (\det M)^{n-1}$, for $n=3$, $\det(	ext{adj}(A^2+A)) = (\det(A^2+A))^2 = (\det A \cdot \det(A+I))^2 = (1 \cdot 7)^2 = 49$.

Column $I$	Column $II$
$(A)$ The set $\{\operatorname{Re}(\frac{2 i z}{1-z^2}): \|z\|=1, z \neq \pm 1\}$ is	$(p)$ $(-\infty,-1) \cup(1, \infty)$
$(B)$ The domain of $f(x)=\sin ^{-1}(\frac{8(3)^{x-2}}{1-3^{2(x-1)}})$ is	$(q)$ $(-\infty, 0) \cup(0, \infty)$
$(C)$ If $f(\theta)=\left\|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right\|$,then the set $\{f(\theta): 0 \leq \theta < \frac{\pi}{2}\}$ is	$(r)$ $[2, \infty)$
$(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$,then $f(x)$ is increasing in	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

Similar Questions

$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right] \Rightarrow A^2-2 A=$

The value of $\left| {\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} \right|$ is

If both $\left( A - \frac{I}{2} \right)$ and $\left( A + \frac{I}{2} \right)$ are orthogonal matrices,then:

Let $A=\left[\begin{array}{ll}2 & 3 \\ a & 0\end{array}\right], a \in R$ be written as $P+Q$ where $P$ is a symmetric matrix and $Q$ is a skew-symmetric matrix. If $\operatorname{det}(Q)=9$,then the modulus of the sum of all possible values of the determinant of $P$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module