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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 p be a nonsingular matrix, and $I + p + p^2 + ..... + p^n = 0$, then find $p^{-1}$.

  1. $I$
  2. $p^{n+1}$
  3. $p^n$
  4. $\left( p^{n+1} - I\right) \left( p-I\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have $I + p + p^2 + ..... + p^n = O$  ----------$(1)$
Since p is a nonsingular matrix, p is invertible.
Multiplying both sides of (1) by $p^{-1}$, we get
$p^{-1} + I + Ip + ..... + p^{n - 1} I = O. p^{-1}$
or $ p^{-1} + I (1 + p + ...... + p^{n-1}) = O$
or $ p^{-1} = - I (I + p + p^2 + ..... + p^{n - 1}) = - I(-p^n) = p^n$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices 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 $A^{-1}XA = ?$

  1. $B$
  2. $B^2$
  3. $A$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

  1. $k\neq -1$
  2. $k\neq 0$
  3. $k\neq 1$

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

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

  1. Symmetric matrix

  2. Skew-symmetric matrix

  3. Identity matrix

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

  1. $A^{-1}C^{-1}B^{-1}$
  2. $C^{-1}B^{-1}A^{-1}$
  3. $C^{-1}A^{-1}B^{-1}$
  4. $B^{-1}C^{-1}A^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

  1. $2$
  2. $-2$
  3. $1/2$
  4. $-1/2$
Reveal answer Fill a bubble to check yourself
D Correct answer
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,
$det{\left( {A}^{-1} \right)} = \cfrac{1}{det{\left( A \right)}}$
$\Rightarrow det{\left( A \right)} = \cfrac{1}{det{\left( {A}^{-1} \right)}} = \cfrac{1}{-2} = - \cfrac{1}{2}$

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

A square, non-singular matrix $A$ satifies $A^2 - A + 2I = 0$, then $A^{-1} = $

  1. $I - A$
  2. $\dfrac {(I - A) }{2}$
  3. $I + A$
  4. $\dfrac {(I + A)}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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

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 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-

  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. $2n\pi+\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The matrix is non-invertible if its determinant is zero. Calculating the determinant of the given matrix and setting it to zero leads to a trigonometric equation that yields the specified value for theta.