Tag: maths

Questions Related to maths

If $A$ and $B$ are two non singular matrices of the same order such that ${ B }^{ r }=I$, for some positive integer $r>1$, then ${ A }^{ -1 }{ B }^{ r-1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A=$

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $I$

  2. $2I$

  3. $O$

  4. $-I$

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Correct Option: C
Explanation:

${ A }^{ -1 }{ B }^{ r-1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A$

$=A^{-1}B^rB^{-1}A - A^{-1}B^{-1}A $
$={ A }^{ -1 }I{ B }^{ -1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A       (\because B^{r}=I)$
$={ A }^{ -1 }{ B }^{ -1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A$
$= O$

Hence, option C.

If $\begin{pmatrix}1 & -tan  \theta\ tan  \theta & 1\end{pmatrix} \begin{pmatrix} 1 & tan  \theta\ - tan  \theta & 1\end{pmatrix}^{-1} = \begin{bmatrix} a& -b\ b & a\end{bmatrix}$, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $a = cos 2 \theta$

  2. $a = 1$

  3. $b = sin 2 \theta$

  4. $b = -1$

Show answer
Correct Option: A,C
Explanation:
we have 

$ \begin{pmatrix} 1 & tan  \theta\\ - tan  \theta & 1\end{pmatrix}^{-1} = \dfrac{1}{1+tan^2\theta} \begin{pmatrix}1 & -tan  \theta\\ tan  \theta & 1\end{pmatrix} $

$\therefore \begin{bmatrix} a& -b\\ b & a\end{bmatrix}=cos^2\theta \begin{pmatrix}1 & -tan  \theta\\ tan  \theta & 1\end{pmatrix} \begin{pmatrix}1 & -tan  \theta\\ tan  \theta & 1\end{pmatrix}$

We get 

$\begin{pmatrix}cos2\theta & -sin2\theta\\ sin2\theta & cos2\theta \end{pmatrix}$

$\therefore a=cos2\theta, b=sin2\theta$

$A = \begin{bmatrix} 1& 0 & 0\0 &  1& 1\ 0 & -2 & 4\end{bmatrix}, I = \begin{bmatrix}1 & 0 & 0\ 0& 1 & 0\ 0 & 0 & 1\end{bmatrix}$ and $A^{-1} = \left [ \dfrac{1}{6} (A^2 + cA + dI) \right]$

The value of $(c,d)$ is

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $(-6, -11)$

  2. $(6, 11)$

  3. $(-6, 11)$

  4. $(6, -11)$

Show answer
Correct Option: C
Explanation:

Given $A = \begin{bmatrix} 1& 0 & 0\0 &  1& 1\ 0 & -2 & 4\end{bmatrix}$
The characteristic equation of $A$ is given by 
$|A-\lambda I|=0$
$\begin{vmatrix} 1-\lambda  & 0 & 0 \ 0 & 1-\lambda  & 1 \ 0 & -2 & 4-\lambda  \end{vmatrix}=0$

$\Rightarrow {\lambda}^{3}-6{\lambda}^{2}+11\lambda-6=0$
$\Rightarrow A^{3}-6A^{2}+11A-6=0$    ($\because$ Every square matrix satisfies its characteristic equation )
$\Rightarrow A^{2}-6A+11I-6A^{-1}=0$
$\Rightarrow A^{-1}=\displaystyle \frac{1}{6}(A^{2}-6A+11I)$

Comparing this with $A^{-1} = \left [ \frac{1}{6} (A^2 + cA + dI) \right]$, we get $c=-6, d=11$
$\therefore (c,d) = (-6,11)$

Hence, option C.

Two $n \times n$ square matrices $A$ and $B$ are said to be similar if there exists a non-singular matrix $P$ such that  $P^{-1}A: P=B$
If $A$ and $B$ are two non-singular matrices, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $A$ is similar to $B$

  2. $AB$ is similar to $BA$

  3. $AB$ is similar to $A^{-1}B$

  4. none of these

Show answer
Correct Option: B
Explanation:

$AB = (B^{-1}B)AB = B^{-1}(BA)B $

$\therefore$ $AB$ is similar to $BA$.

Hence, option B.

Two $n \times n$ square matrices $A$ and $B$ are said to be similar if there exists a non-singular matrix $P$ such that  $P^{-1}A: P=B$
If $A$ and $B$ are similar matrices such that $det :(A) =1$, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $det : (B) = 1$

  2. $det: (A)+det: (B)=0$

  3. $det : (B) = -1$

  4. none of these

Show answer
Correct Option: A
Explanation:

As $A$ and $B$ are similar matrices there exists a non-singular matrix $P$ such that

                                      $A=P^{-1}:BP$

$\Rightarrow det : (A) = det: (P^{-1}:BP)$

                $=det: (P^{-1}): det : (B) : det : (P)$

                 $\displaystyle =\frac{1}{det: (P)}det : (B) : (det P)$

                $= det : B$

Thus, $det : (A) = 0 \Leftrightarrow det : (B) = 0 : and : det : (A) =1 \Leftrightarrow  det : (B) = 1$

Hence, option A.

Two $n \times n$ square matrices $A$ and $B$ are said to be similar if there exists a non-singular matrix $P$ such that  $P^{-1}A: P=B$
If $A$ and $B$ are similar and $B$ and $C$ are similar, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $AB$ and $BC$ are similar

  2. $A$ and $C$ are similar

  3. $A + C$ and $B$ are similar

  4. none of these

Show answer
Correct Option: B
Explanation:

$A = P^{-1}:BP, B = Q^{-1} : C : Q$,

$\Rightarrow      A = P^{-1}(Q^{-1} : C : Q)P = (QP)^{-1}: C : QP$

Thus, $A$ is similar to $C$

Hence, option B.


The number of values of $\theta \in (0,\pi )$ for which the system of linear equations
x+3y+7z=0
x+4y+7z=0
$(\sin { 3\theta  } )x+(\cos { 2\theta  } )y+2z=0$
has a non trivial solution is :

maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. one

  2. three

  3. four

  4. two

Show answer
Correct Option: A

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. $0$

  2. $\displaystyle\frac { 1 }{ 3 } $

  3. $\displaystyle\frac { 1 }{ 12 } $

  4. $\displaystyle\frac { 1 }{ 36 } $

Show answer
Correct Option: D
Explanation:

When two dice are rolled, the number of outcomes is $36$.
The only even prime number is $2$.
Let $E$ be the event of getting an even prime number on each die.
$\therefore E=\left{ \left( 2,2 \right)  \right} $
$\Rightarrow P\left( E \right) =\displaystyle\frac { 1 }{ 36 } $
Therefore, the correct answer is (D).

If the letters of the word  $"ATTEMPT"$  are written down at random. The probability that all the  $T's$  come together is

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. $1/21$

  2. $6/7$

  3. $1/7$

  4. $1/42$

Show answer
Correct Option: C
Explanation:

Number of ways of arranging the words keeping all $T's$ together is  $5!$

Number of ways of arranging the words  is  $\dfrac{7!}{3!}$
Probability that all the $T's$  together is $\dfrac{5! }{\dfrac{7!}{3!}}=\dfrac{1}{7}$

The probability of getting number 10 in a throw of a dice is ____.

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. 0

  2. 1

  3. 0.5

  4. 0.75

Show answer
Correct Option: A
Explanation:

Since the outcome of a throw of dice can never be 10, the probability is 0.