Tag: inverse of a matrix and linear equations

Questions Related to inverse of a matrix and linear equations

If $A =\begin{bmatrix}a &b \c &d \end{bmatrix}$ such that $A$ satisfies the relation $A^2- (a + d)A = 0$, then inverse of $A$ is

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. $A$

  3. $(a + d)A$

  4. none of these

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Correct Option: D
Explanation:

 $A =\begin{bmatrix}a &b \c &d \end{bmatrix}$ such that $A$ satisfies the relation $A^2- (a + d)A = 0$
$\Rightarrow A^2-(a+d)A=0$
$\Rightarrow \begin{bmatrix}a &b \c &d \end{bmatrix}\begin{bmatrix}a &b \c &d \end{bmatrix}-(a+d)\begin{bmatrix}a &b \c &d \end{bmatrix}=\begin{bmatrix}0 &0 \0 &0 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}a^2+bc &ab+bd \ac+cd &bc+d^2 \end{bmatrix}-(a+d)\begin{bmatrix}a &b \c &d \end{bmatrix}=\begin{bmatrix}0 &0 \0 &0 \end{bmatrix}$
$\Rightarrow a^2+bc-a^2-ad=0$
$\Rightarrow ad-bc=\begin{vmatrix}a &b \c &d \end{vmatrix}=0$
$\therefore$ Inverse of A doesnot exist.
Hence, option D.

Let the matrix A and B be defined as $A =\begin{bmatrix}3 &2 \ 2 &1 \end{bmatrix}$ and $B= \begin{bmatrix}3 &1 \ 7 &3 \end{bmatrix}$ then the value of Det.$(2A^9B^{-1})$, is 

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

  1. $2$

  2. $1$

  3. $-1$

  4. $-2$

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Correct Option: D
Explanation:

$A =\begin{bmatrix}3 &2 \ 2 &1 \end{bmatrix}$ and $B= \begin{bmatrix}3 &1 \ 7 &3 \end{bmatrix}$

$|A| = \begin{vmatrix} 3 & 2 \ 2 & 1 \end{vmatrix} = -1$

$|B| = \begin{vmatrix} 3 & 1 \ 7 & 3 \end{vmatrix} = 2$

$\displaystyle |2A^9 B^{-1}| = 2^2|A|^9\frac{1}{|B|}$

                     $\displaystyle= 4\times (-1)\times \frac{1}{2}$

$\therefore |2A^9 B^{-1}|=-2$

Hence, option D.

If $P$ is a two-rowed matrix satisfying $P^T = P^{-1}$, then $P$ can be

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

  1. $\begin{bmatrix}cos\, \theta & -sin\, \theta \ -sin\,\theta & cos\, \theta \end{bmatrix}$

  2. $\begin{bmatrix}cos\, \theta & sin\, \theta \ -sin\,\theta & cos\, \theta \end{bmatrix}$

  3. $\begin{bmatrix}-cos\, \theta & sin\, \theta \ sin\,\theta & -cos\, \theta \end{bmatrix}$

  4. none of these

Show answer
Correct Option: B
Explanation:

$A=\begin{bmatrix} cos\theta  & -sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},{ A }^{ T }=\begin{bmatrix} cos\theta  & -sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},$

${ A }^{ -1 }=\cfrac { 1 }{ { cos }^{ 2 }\theta -{ sin }^{ 2 }\theta  } \begin{bmatrix} cos\theta  & sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}\ B=\begin{bmatrix} cos\theta  & sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},{ B }^{ T }=\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix},$
${ B }^{ -1 }=\cfrac { 1 }{ { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta  } \begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}=\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}\ C=\begin{bmatrix} -cos\theta  & sin\theta  \ sin\theta  & -cos\theta  \end{bmatrix},{ C }^{ T }=\begin{bmatrix} -cos\theta  & sin\theta  \ sin\theta  & -cos\theta  \end{bmatrix},$
${ C }^{ -1 }=\cfrac { 1 }{ { cos }^{ 2 }\theta -{ sin }^{ 2 }\theta  } \begin{bmatrix} -cos\theta  & sin\theta  \ sin\theta  & -cos\theta  \end{bmatrix}$
hence $P=B=\begin{bmatrix} cos\theta  & sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix}$

Let A be an invertible matrix then which of the following is/are true

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

  1. $|A^{-1}| = |A|^{-1}$

  2. $(A^2)^{-1} = (A^{-1})^2$

  3. $(A^T)^{-1} = (A^{-1})^T$

  4. none of these

Show answer
Correct Option: A,B,C
Explanation:
Option A
$\left| { A }^{ -1 } \right| ={ \left| A \right|  }^{ -1 }$
$det\left( A \right) (det\left( B \right) )$
$d\left( A{ A }^{ -1 } \right) =detAdet\left( { A }^{ -1 } \right) $
$det\left( I \right) =1$
$\Rightarrow det\left( A \right) \ast det\left( { A }^{ -1 } \right) =I$
$det\left( { A }^{ -1 } \right) ={ \left( detA \right)  }^{ -1 }$

Option B:
A is invertible $A{ A }^{ -1 }={ A }^{ -1 }A=I$
$\Rightarrow { A }^{ 2 }$ is also invertible
${ \left( A{ A }^{ -1 } \right)  }^{ 2 }={ I }^{ 2 }$
${ A }^{ 2 }{ \left( { A }^{ -1 } \right)  }^{ 2 }=I$
${ \left( { A }^{ -1 } \right)  }^{ 2 }={ ({ A }^{ 2 }) }^{ -1 }$

Option C:
${ \left( { A }^{ T } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ T }$
$\left( { A }^{ T } \right) { \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 }A \right)  }^{ T }={ I }^{ T }=I$
Also,
${ \left( { A }^{ -1 } \right)  }^{ T }\left( { A }^{ T } \right) ={ \left( A{ A }^{ -1 } \right)  }^{ T }={ I }^{ T }=I$
${ A }^{ 1 }{ \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 } \right)  }^{ T }\left( { A }^{ T } \right) =I$
$\Rightarrow { \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ T } \right)  }^{ -1 }$

Option A,B,C are correct

If A and B are invertible matrices, which one of the following statement is/are correct 

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

  1. $Adj(A) = |A|A^{-1}$

  2. $det(A^{-1}) =|det(A)|^{-1}$

  3. $(A + B)^{-1}= B^{-1 }+ A^{-1}$

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

Show answer
Correct Option: A,B,D
Explanation:
Option A
${ A }^{ -1 }=\cfrac { AdjA }{ \left| A \right|  } $
$\Rightarrow AdjA=\left| A \right| { A }^{ -1 }$
Option A is true

Option B
$det\left( AB \right) =\left( detA \right) \left( detB \right) $
$\Rightarrow A{ A }^{ -1 }=I$
$det\left( A{ A }^{ -1 } \right) =detI$
$\Rightarrow detA\left( det{ A }^{ -1 } \right) =1$
$\Rightarrow det{ A }^{ -1 }={ \left( detA \right)  }^{ -1 }$
Option B is true

Option C
${ \left( A+B \right)  }^{ -1 }={ A }^{ -1 }+{ B }^{ -1 }$
Option C is true

Option D
${ \left( AB \right)  }^{ -1 }=?$
$AB\left( { B }^{ -1 }{ A }^{ -1 } \right) =A\left( B{ B }^{ -1 } \right) { A }^{ -1 }$
$=AI{ A }^{ -1 }=\left( A{ A }^{ -1 } \right) =I$
$\Rightarrow { B }^{ -1 }{ A }^{ -1 }={ \left( AB \right)  }^{ -1 }$
Option D is true

If $A=\begin{bmatrix} 1 & -2 \ 3 & 0 \end{bmatrix}$, $B=\begin{bmatrix} -1 & 4 \ 2 & 3 \end{bmatrix}$, and $ABC=\begin{bmatrix} 4 & 8 \ 3 & 7 \end{bmatrix}$, then $C$ equals

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

  1. $\cfrac { 1 }{ 66 } \begin{bmatrix} 54 & 110 \ 3 & 11 \end{bmatrix}$

  2. $\cfrac { 1 }{ 66 } \begin{bmatrix} -54 & -110 \ 3 & 11 \end{bmatrix}$

  3. $\cfrac { 1 }{ 66 } \begin{bmatrix} -54 & 110 \ 3 & -11 \end{bmatrix}$

  4. None of these

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Correct Option: B

If $A _{3X3}$ and $ det A= 2$ then $det A^{-1}=$ 

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

  1. $\dfrac {1}{2}$

  2. $-2$

  3. $\dfrac {1}{4}$

  4. $-4$

Show answer
Correct Option: A
Explanation:

$AA^{-1}=I$
So, $|AA^{-1}=|I|=1$
$|A||A^{-1}=1|$
So, $det A^{-1}=\dfrac{1}{|A|}=\dfrac{1}{2}$

The value of $(\mathrm{A}$dj $\mathrm{A})^{-1}$ is equal to 

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

  1. $\mathrm{A}$dj $(\mathrm{A}^{-1})$

  2. $\mathrm{A}$dj $[-\mathrm{A}]$

  3. $(\mathrm{A}$dj$\mathrm{A})^{\mathrm{T}}$

  4. $\mathrm{A}$dj $(\mathrm{A}^{\mathrm{T}})$

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Correct Option: A

lf the value of a third order determinant is 11, then the value of the determinant of $A^{-1}=$ 

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

  1. 11

  2. 121

  3. $1/11$

  4. $1/121$

Show answer
Correct Option: C
Explanation:

$det(A _{3 \times 3})=11$
$AA^{-1}=I$
$\therefore det(A)=\frac{1}{det(A^{-1})}$
$\therefore det(A^{-1})=\frac{1}{11}$

. $\mathrm{If}$ $\mathrm{A}$ is non-singular matrix such that $A^{2}=A^{-1}$ then $adjA=$ 

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

  1. $\mathrm{A}$

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

  3. $\mathrm{A}^{3}$

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

Show answer
Correct Option: B
Explanation:

$A^{2}=A^{-1}$

$A.A^{2}=A.A^{-1}$
$A^{3}=I$
$detA.A^{3}=detA.I$
$detA.A^{3}=A.adjA$
$detA.A^{2}=adjA$
Therefore 
$adjA=A^{-1}$.