Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

Matrices A and B satisfy $AB = B^{-1}$, where $ B\quad =\quad \begin{bmatrix} 2 & -1 \ 2 & 0 \end{bmatrix}$, then find without finding $A^{-1}$, the matrix X satisfying $A^{-1}XA = ?$

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

  1. $B$

  2. $B^2$

  3. $A$

  4. None of these

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Correct Option: A
Explanation:

Given, $A^{-1} XA = B$


$AA^{-1} XA = AB$

$IXA =AB$

$XAB =AB^2$ 

$XAB =I$ since $\left[AB=B^{-1}\Rightarrow AB^2=I\right]$

$XAB^2 = B$

$XI = B$

$\therefore X = B$

Hence option $'A'$ is the answer.

If $A$ satisfies the equation $x^3-5x^2+4x+kI=0,$ then $A^{-1}$ exists if

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

  1. $k\neq -1$

  2. $k\neq 0$

  3. $k\neq 1$

  4. none of these

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Correct Option: B
Explanation:

Since A satisfies the given equation, therefore ${ A }^{ 3 }-5{ A }^{ 2 }+4A+kI=0$

${ A }^{ -1 }$ exits if $k\neq 0$ since if $k=0$ then the above equation gives $A=0$ and in that case ${ A }^{ -1 }$ wont exist.

If $A^3 = O$, then $I + A + A^2$ equals

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

  1. $I - A$

  2. $(I + A^1)^{-1}$

  3. $(I - A)^{-1}$

  4. none of these

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Correct Option: C
Explanation:

Given, $A^{3} = 0$
$\Rightarrow I-A^{3} = I$

Using the identity, we get
$\Rightarrow (I-A)(I+A+A^{2}) = I$
$\therefore I+A+A^{2} = (I-A)^{-1}$

If $A$ and $B$ are symmetric matrices and $AB=BA$, then ${ A }^{ -1 }B$ is a

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

  1. Symmetric matrix

  2. Skew-symmetric matrix

  3. Identity matrix

  4. None of these

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Correct Option: A
Explanation:

We have $AB=BA=B'A'=\left( AB \right) '$

$\Rightarrow AB$ is symmetric.
Also, $AB{ A }^{ -1 }=BA{ A }^{ -1 }=B\quad \quad \left( \because AB=BA \right) $
$\Rightarrow { A }^{ -1 }AB{ A }^{ -1 }={ A }^{ -1 }B\Rightarrow B{ A }^{ -1 }={ A }^{ -1 }B$
Therefore, $\left( { A }^{ -1 }B \right) '=\left( B{ A }^{ -1 } \right) =\left( { A }^{ -1 } \right) 'B'=A'B$   $[\because { A }^{ -1 }$ and $B$ are symmetric $]$
Thus, the matrix ${ A }^{ -1 }B$ is symmetric.

If $A^2 + A - I = 0$, then $A^{-1}$ =

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

  1. $I + A$

  2. $I - A$

  3. $-I + A$

  4. $-I - A$

Show answer
Correct Option: A
Explanation:

$A^2 + A - I = 0$


multiplying both sides with $A^{-1}$ gives

$A^{-1}A^2+A^{-1}A-A^{-1}I=0$

$\Rightarrow A+I-A^{-1}=0$

$\Rightarrow A^{-1}=I+A$

Hence, option A.

IF $A,B,C$ are non-singular $n\times n$ matrices, then $(ABC)^{-1}$ = ____________.

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

  1. $A^{-1}C^{-1}B^{-1}$

  2. $C^{-1}B^{-1}A^{-1}$

  3. $C^{-1}A^{-1}B^{-1}$

  4. $B^{-1}C^{-1}A^{-1}$

Show answer
Correct Option: B
Explanation:

using the property $(AB)^{-1}=B^{-1}A^{-1}$


$(ABC)^{-1}=(BC)^{-1}A^{-1}=C^{-1}B^{-1}A^{-1}$

Hence, option B.

If $A^{-1}=\begin{bmatrix} 1 & -2 \ -2 & 2 \end{bmatrix}$, then what is $det(A)$ 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. $2$

  2. $-2$

  3. $1/2$

  4. $-1/2$

Show answer
Correct Option: D
Explanation:

$det{\left( {A}^{-1} \right)} = \left( 1 \times 2 \right) - \left( -2 \right) \times \left( -2 \right) = 2 - 4 = -2$

As we know that,
$det{\left( {A}^{-1} \right)} = \cfrac{1}{det{\left( A \right)}}$
$\Rightarrow det{\left( A \right)} = \cfrac{1}{det{\left( {A}^{-1} \right)}} = \cfrac{1}{-2} = - \cfrac{1}{2}$

A square, non-singular matrix $A$ satifies $A^2 - A + 2I = 0$, then $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. $I - A$

  2. $\dfrac {(I - A) }{2}$

  3. $I + A$

  4. $\dfrac {(I + A)}{2}$

Show answer
Correct Option: B
Explanation:

Given, $A^{2}-A+2I = 0$

$\Rightarrow 2I = A-A^{2}$

$\Rightarrow 2A^{-1}I = A^{-1}A-A^{-1}A^{2}$

$\therefore A^{-1} = \displaystyle\frac{I-A}{2}$

If matrix $A=\left| \begin{matrix} sin\theta  & cosec\theta  & 1 \ cosec\theta  & 1 & sin\theta  \ 1 & sin\theta  & cosec\theta  \end{matrix} \right| $ a non invertible matrix. then possible value of $\theta$ is-

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

  1. $n\pi+(-1)^n\dfrac{\pi}{4}$

  2. $n\pi+(-1)^n\dfrac{\pi}{3}$

  3. $n\pi+(-1)^n\dfrac{\pi}{6}$

  4. $2n\pi+\dfrac{\pi}{2}$

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Correct Option: A

If $A$ be a $3\times 3$ matrix and $I$ be the unit matrix of that order such that $\displaystyle A=A^{2}+I$ then $A^{-1}$ 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. $A$

  2. $A+I$

  3. $I-A$

  4. $A-I$

Show answer
Correct Option: C
Explanation:

Given : $\displaystyle A=A^{2}+I$

$\Rightarrow A^{2}-A+I=O$

$\Rightarrow A^{2}A^{-1}-AA^{-1}+IA^{-1}=O$

$\Rightarrow A-I+A^{-1}=O$

$\Rightarrow A^{-1}=I-A$