Tag: inverse of a matrix and linear equations

Quizzes (5) Questions (140)

Questions Related to inverse of a matrix and linear equations

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

  1. $I$
  2. $A$
  3. $(a + d)A$

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

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

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 

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $P=B=\begin{bmatrix} cos\theta  & sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix}$

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

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

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

  4. none of these
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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
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 invertible matrices, which one of the following statement is/are correct 

  1. $Adj(A) = |A|A^{-1}$
  2. $det(A^{-1}) =|det(A)|^{-1}$
  3. $(A + B)^{-1}= B^{-1 }+ A^{-1}$
  4. $(AB)^{-1} = B^{-1}A^{-1}$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
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
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 $A=\begin{bmatrix} 1 & -2 \ 3 & 0 \end{bmatrix}$, $B=\begin{bmatrix} -1 & 4 \ 2 & 3 \end{bmatrix}$, and $ABC=\begin{bmatrix} 4 & 8 \ 3 & 7 \end{bmatrix}$, then $C$ equals

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given ABC = D, then C = (AB)^-1 * D. Calculate AB first, then find its inverse, and multiply by the matrix D.

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

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

  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 $(\mathrm{A}^{\mathrm{T}})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The property (Adj A)^-1 = Adj(A^-1) is a standard identity in matrix algebra.

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

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

  1. $\mathrm{A}$
  2. $\mathrm{A}^{-1}$
  3. $\mathrm{A}^{3}$
  4. $(\mathrm{A}^{-1})^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 
$adjA=A^{-1}$.