Mathematics

Inverses in Mathematics

90 Questions

Inverses in mathematics cover additive, multiplicative, and matrix operations. These concepts test the ability to reverse mathematical functions and transformations. Such questions frequently appear in various competitive examinations to evaluate fundamental algebraic skills.

Additive inverseMultiplicative inverseMatrix inversionInverse functionsLinear transformations

Inverses in Mathematics Questions

Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

With $1,\omega, \omega^2$ as cube roots of unity, inverse of which of the following matrices exists

  1. $\begin{bmatrix}1 & \omega \\ \omega & \omega^2\end{bmatrix}$
  2. $\begin{bmatrix}\omega^2 & 1 \\ 1 & \omega\end{bmatrix}$
  3. $\begin{bmatrix} \omega & \omega^2 \\ \omega^2 & 1\end{bmatrix}$

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

The inverse of a matrix exists if its determinant is not equal $0$.
For option A, 
Let $A=\begin{bmatrix}1 & {\omega} \ {\omega} & {\omega}^2\end{bmatrix}$
Here, $|A|=0$
Hence, inverse does not exists.

For option B, 
Let $A=\begin{bmatrix}{\omega}^{2} & 1 \ 1 & {\omega}\end{bmatrix}$
Here, $|A|=0$
Hence, inverse does not exists.

For option C, 
Let $A=\begin{bmatrix}{\omega} & {\omega}^{2} \ {\omega}^{2} & 1\end{bmatrix}$
Here, $|A|=0$
Hence, inverse does not exists.

Hence, the inverse does not exist for any of the given matrices

Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

The inverse of a skew-symmetric matrix of odd order is

  1. a symmetric matrix

  2. a skew-symmetric matrix

  3. diagoinal matrix

  4. does not exists
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let A be a skew-symmetric matrix of order n.

By definition ${ A }^{ ' }=-A$
$\Rightarrow \left| { A }^{ ' } \right| =\left| -A \right| \Rightarrow \left| A \right| ={ \left( -1 \right)  }^{ n }\left| A \right| $
$\Rightarrow \left| { A } \right| =\left| -A \right| $ as n is odd
$\Rightarrow 2\left| { A } \right| =0\Rightarrow \left| { A } \right| =0$
Thus ${ A }^{ -1 }$ does not exist

Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

If $\left |\begin{matrix}1 & -1 &x \ 1 & x & 1\ x & -1 & 1\end{matrix} \right|$ has no inverse, then the real value of $x$ can be is

  1. 2

  2. 3

  3. 0

  4. 1
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}$
No inverse
$\Rightarrow \left| A \right| =0$
$x+1+1(1-x)+x(-1-{ x }^{ 2 })=0$
$x+1+1-x-x-{ x }^{ 3 }=0$
${ x }^{ 3 }-x-2=0$
$\Rightarrow { x }^{ 3 }+2x-x-2=0$
$\left( { x }^{ 2 }-1 \right) \left( x+2 \right) =0$
$\Rightarrow x=\pm 1,x=-2$
Real value of $x=1$
Option D
Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

If the matrix $\begin{bmatrix} -1& 3 &2 \1&k&-3\1&4&5\end{bmatrix}$ has an inverse then the values of $k$.

  1. $k$ is any real number
  2. $k = -4$
  3. $k \neq -4$
  4. $k \neq 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For a matrix to be invertible, its determinant must not be zero.
$\therefore -1(5k + 12) - 3(5 - (-3)) + 2(4 - k) \neq 0$
$\therefore -5k - 12 - 24 + 8 - 2k \neq 0$
$\therefore -7k \neq 28$
$\therefore k \neq -4$
Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

The matrix $A=\begin{bmatrix}1&3&2\1&x-1&1\2&7&x-3\end{bmatrix}$ will have inverse for every real number x except for

  1. $x=\dfrac{11\pm \sqrt{5}}{2}$
  2. $x=\dfrac{9\pm \sqrt{5}}{2}$
  3. $x=\dfrac{11\pm \sqrt{3}}{2}$
  4. $x=\dfrac{9\pm \sqrt{3}}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solution:

For $A$ have a inverse, $|A|\neq 0.$
$|A|={(x-1)(x-3)-7}+3{2-(x-3)}+2{7-2(x-1)}$
$=x^2-4x+3-7+6-3x+9+14-4x+4$
$=x^2-11x+29$
$=\left(x-\cfrac{11+\sqrt5}{2}\right)\left(x-\cfrac{11-\sqrt5}{2}\right)$
So, $|A|\neq0$
$x\neq\left(\cfrac{11\pm\sqrt5}{2}\right)$
i.e. for $x=\cfrac{11\pm\sqrt5}{2}$ matrix $A$ will not have its inverse.
Hence, A is the correct option.

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 $B$ will be inverse of each other only if

  1. $AB=BA$
  2. $AB=0,BA=I$
  3. $AB=BA=0$
  4. $AB=BA=I$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that if $A$ is a square of order $m$, and if there exists another square matrix $B$ of the same order $m$, such that $AB=I$, then $B$ is said to be the inverse of $A$. 

In this case, it is clear that $A$ is the inverse of $B$.
Thus , matrices $A$ and $B$ will be inverses of each other only if $AB=BA=I.$

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 $3\times 3$ matrix $A$ has its inverse equal to $A$, then ${A}^{2}$ is equal to

  1. $\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
  2. $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}$
  3. $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
  4. $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $A^{-1}=A$

$AA^{-1}=I$
$\implies A.A=I$
$\implies A^{2}=I=\begin{bmatrix}1&0&0\0&1&0\0&0&1\end{bmatrix}$

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

Which of the following matrix is inverse of itself

  1. $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
  2. $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
  3. $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}$
  4. $\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Inverse of unit matrix also unit matrix.

Ans: 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 $\begin{bmatrix} 1 & -1 & x \ 1 & x & 1 \ x & -1 & 1 \end{bmatrix}$ has no inverse, then the real value of $x$ is 

  1. $2$
  2. $3$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A matrix has no inverse if its determinant is zero. Setting the determinant of the given matrix to zero results in an equation that solves for x = 0.

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 inverse of a symmetric matrix (if it exists) is

  1. a symmetric matrix

  2. a skew symmetric matrix

  3. a diagonal matrix

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

 Let $A$ be an invertible symmetric matrix.
 $\therefore AA^{-1}=A^{-1}:A=I _n$
$\Rightarrow  ( AA^{-1})'=(A^{-1}:A)'=(I _n)'$
$\Rightarrow   ( A^{-1})'A'=A'(A^{-1})'=(I _n)$
$\Rightarrow   ( A^{-1})'A=A(A^{-1})'=(I _n)$
$\Rightarrow   ( A^{-1})'=A^{-1}$    [inverse of a matrix is unique]
i.e $A^{-1}$ is symmetric.
Hence, option A.

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 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 rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

The multiplicative inverse of $-1 + \sqrt{2}$ is

  1. $-1-\sqrt{2}$
  2. $1-\sqrt{2}$
  3. $1+\sqrt{2}$
  4. $\sqrt{2}$
  5. $2-\sqrt{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiplicative inverse of $\sqrt2-1$ is $\dfrac{1}{\sqrt2-1}$
Now multiply and divide with $\sqrt2+1$
We get $\dfrac{\sqrt2+1}{(\sqrt2+1)(\sqrt2-1)} $

$=\dfrac {\sqrt 2+1}{(\sqrt 2)^2-1^2}$
$= \sqrt2+1$

Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Find the inverse f the following matrices by using transformation method.

  1. $\begin{bmatrix}
    1 &2 \\
    2 &-1
    \end{bmatrix}$
  2. $\begin{bmatrix}
    2 &-3 \\
    -1 &2
    \end{bmatrix}$
  3. $\begin{bmatrix}
    0& 1 &2 \\
    1& 2 &3 \\
    3& 1 &1
    \end{bmatrix}$
  4. $\begin{bmatrix}
    2& 0 &-1 \\
    5& 1 &0 \\
    0& 1 &3
    \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer