Tag: inverse of a matrix and linear equations

Questions Related to inverse of a matrix and linear equations

Let A and B be two non-singular matrices which commute. The $A^{-1}$, $B^{-1}$

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

  1. do not commute

  2. commute

  3. $AB = A^{-1}B^{-1}$

  4. $(AB)^{-1}=AB$

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Correct Option: B
Explanation:

$A^{-1}B^{-1}=(BA)^{-1}=(AB)^{-1}$ since $A,B$ commute
$\Rightarrow A^{-1}B^{-1}=B^{-1}A^{-1}$
Hence $A^{-1}, B^{-1}$ also commute

$\mathrm{A}\mathrm{B}\mathrm{A^{-1}}$ $=\mathrm{X}$ then $\mathrm{B}^{2}=$

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

  1. $\mathrm{x}^{2}$

  2. $\mathrm{A}\mathrm{x}\mathrm{A}^{-1}$

  3. $\mathrm{A}\mathrm{x}^{2}\mathrm{A}^{-1}$

  4. $\mathrm{A}^{-1}\mathrm{x}^{2}\mathrm{A}$

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Correct Option: D
Explanation:

$ABA^{-1}=X$
$\Rightarrow (ABA^{-1})(ABA^{-1})=XX=X^2$
$\Rightarrow ABAA^{-1}BA^{-1} =X^2$
$\Rightarrow ABBA^{-1} =X^2[\because AA^{-1}=I]$
$\Rightarrow AB^2A^{-1} =X^2$
Pre and post multiplying both sides with $A^{-1}$ and $A$ respectively we get,
$\Rightarrow B^2 =A^{-1}X^2A$

If $A = \begin{bmatrix} 2 & 3\ 5 & 1 \end{bmatrix},$ then find $A^{-1}$

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

  1. $\begin{bmatrix} - \frac {1}{13} & \frac {3}{13}\ - \frac {5}{13} & - \frac {2}{13} \end{bmatrix}$

  2. $\begin{bmatrix} - \frac {1}{13} & \frac {3}{13}\ \frac {5}{13} & \frac {2}{13} \end{bmatrix}$

  3. $\begin{bmatrix} - \frac {1}{13} & \frac {3}{13}\ \frac {5}{13} & - \frac {2}{13} \end{bmatrix}$

  4. $\begin{bmatrix} \frac {1}{13} & \frac {3}{13}\ \frac {5}{13} & - \frac {2}{13} \end{bmatrix}$

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Correct Option: C
Explanation:
$A=\begin{bmatrix} 2 & 3\\ 5 & 1\end{bmatrix}$

Cofactor matrix of $A=\begin{bmatrix} 1 & -5\\ -3 & 2\end{bmatrix}$

adj. $A=\begin{bmatrix} 1 & -5\\ -3 & 2\end{bmatrix}'=\begin{bmatrix} 1 & -3\\ -5 & 2\end{bmatrix}$

$|A|=2\times 1-(3\times 5)$

$=2-15=-13$

$A^{-1}=\dfrac{adj. A}{|A|}=\dfrac{-1}{13}\begin{bmatrix} 1 & -3\\ -5 & 2\end{bmatrix}$

$=\begin{bmatrix} \dfrac{-1}{13} & \dfrac{3}{13}\\ \dfrac{5}{13} & \dfrac{-2}{13}\end{bmatrix}$.

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. $b = -1$

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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. $(6, -11)$

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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

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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

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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.