Tag: properties of inverses of matrices

Questions Related to properties of inverses of matrices

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. $b = -1$

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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

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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.