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

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

  1. $I$
  2. $2I$
  3. $O$
  4. $-I$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $a = cos 2 \theta$
  2. $a = 1$
  3. $b = sin 2 \theta$
  4. $b = -1$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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 

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

$\therefore a=cos2\theta, b=sin2\theta$
Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

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

The value of $(c,d)$ is

  1. $(-6, -11)$
  2. $(6, 11)$
  3. $(-6, 11)$
  4. $(6, -11)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

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

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

Hence, option B.

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

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

  1. $det \: (B) = 1$
  2. $det\: (A)+det\: (B)=0$
  3. $det \: (B) = -1$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. $AB$ and $BC$ are similar
  2. $A$ and $C$ are similar
  3. $A + C$ and $B$ are similar

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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.


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

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

  1. $0$
  2. $\displaystyle\frac { 1 }{ 3 } $
  3. $\displaystyle\frac { 1 }{ 12 } $
  4. $\displaystyle\frac { 1 }{ 36 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
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).

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

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

  1. $1/21$
  2. $6/7$
  3. $1/7$
  4. $1/42$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\dfrac{5! }{\dfrac{7!}{3!}}=\dfrac{1}{7}$