MCQ
Let $A=\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|, $ then
  • A
    $A^2=I$
  • B
    $A ^2=4$
  • $A ^2= A$
  • D
    $A ^2=0$

Answer

Correct option: C.
$A ^2= A$
 $A^2=\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|=\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|=A$

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

Let $A = \left[ {\begin{array}{*{20}{c}}1&0&0\\5&2&0\\{ - 1}&6&1\end{array}} \right]$, then the adjoint of $ A $ is
$\vec{a}+\vec{b}+\vec{c}=0$ such that $|\vec{a}|=3,|\vec{b}|=5$ and $|\vec{c}|=7$.
What is the angle between $\vec{a}$ and $\vec{b}$ ?
If $\left( {\frac{1}{2},\frac{1}{3},n} \right)$ are the direction cosines of a line, then the value of $n$ is
What is the length of the longer diagonal of the parallelogram constructed on $5\vec{\text{a}}+2\vec{\text{b}}$ and $\vec{\text{a}}-3\vec{\text{b}}$ if it is given that $|\vec{\text{a}}|=2\sqrt{2},\big|\vec{\text{b}}\big|=3$ and the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{4}$?
Point $A$ lies at a distance of $6$ units from the point $(1 ,0, 1)$ , on the line $\frac{{x - 1}}{2} = \frac{y}{2} = \frac{{z - 1}}{1}$ , in $-ve\ z$ direction, then co-ordinates of $A$ are
Let $f:R \to R,f\left( x \right) = a{x^3} + b{x^2} + cx + d,$ has no extreme value , then which of the following is always correct ?
If the points whose position, vectors are $3i - 2j - k,$ $2i + 3j - 4k,$ $ - i + j + 2k$and $4i + 5j + \lambda k$ lie on a plane, then $\lambda = $
The corner points of the feasible region determined by the system of linear constraints are $(0,0),(0,40),(20,40),(60,20),(60,0)$. The objective function is $Z=4 x+3 y$.
Compare the quantity in Column A and Column B
Column AColumn B
Maximum of Z325
Find values of $\mathrm{k}$ if area of triangle is $4$ square units and vertices are  $(\mathrm{k}, 0),(4,0),(0,2)$
If $2f(x) - 3f\left( {\frac{1}{x}} \right) = x$, then $\int_1^2 {f(x)} \;dx$ is equal to