Mathematics

Inverses in Mathematics

106 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 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
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A$ is a $2\times 2$ matrix such that $A^{2}-4A+3I=0$, then the inverse of $A+3I$ is equal to

  1. $\dfrac{1}{24}S-\dfrac{7}{24}I$
  2. $\dfrac{1}{21} A-\dfrac{7}{21}I$
  3. $\dfrac{7}{24}I+\dfrac{1}{24}A$
  4. $A-3I$`
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From A^2 - 4A + 3I = 0, we can write A^2 - 4A = -3I. Factoring gives A(A - 4I) = -3I, so A(4I - A) = 3I. The inverse of (A + 3I) is found by manipulating the characteristic equation; the result is 1/24 A - 7/24 I.

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

The inverse of the matrix  $\left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \ { 3 } & { 3 } & { 0 } \ { 5 } & { 2 } & { - 1 } \end{array} \right]$  is

  1. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \\ { 3 } & { 1 } & { 0 } \\ { 9 } & { 2 } & { - 3 } \end{array} \right]$
  2. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { 0 } \\ { - 9 } & { - 2 } & { 3 } \end{array} \right]$
  3. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { 3 } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { 0 } \\ { - 9 } & { - 2 } & { 3 } \end{array} \right]$
  4. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \\ { - 3 } & { - 1 } & { 0 } \\ { - 9 } & { - 2 } & { 3 } \end{array} \right]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The inverse of a matrix M is (1/det(M)) * adj(M). The determinant of the given matrix is 1(3*-1 - 0) = -3. Calculating the adjugate matrix and multiplying by -1/3 yields the correct option.

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

Inverse of $\begin{bmatrix} -1 & 5 \ -3 & 2 \end{bmatrix}$ is

  1. $\begin{bmatrix} 2/13 & -5/13 \\ 3/13 & -1/13 \end{bmatrix}$
  2. $\begin{bmatrix} -2/13 & 5/13 \\ -3/13 & 1/13 \end{bmatrix}$
  3. $\begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$
  4. $Cannot\ be\ determined$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a 2x2 matrix [[a, b], [c, d]], the inverse is (1/(ad-bc)) * [[d, -b], [-c, a]]. Here, det = (-1)(2) - (5)(-3) = -2 + 15 = 13. The inverse is (1/13) * [[2, -5], [3, -1]].

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

Use the method of elementary row transformation to compute the inverse of 
$\quad \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}$

  1. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$
  2. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{1}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{11}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{2}{21}\end{bmatrix}$
  3. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{16}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{4}{21}\end{bmatrix}$
  4. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{2}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{4}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $\quad A = \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}$

$\Rightarrow \quad Write \space A A^{-1}= I$

$\quad \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix} A^{-1}= \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}$

$\quad \begin{matrix}R _{21}(-2)\ \mbox{~}\ R _{31}(1)\end{matrix}\begin{bmatrix}1 & 0 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \ -2 & 1 & 0 \ 1 & 0 & 1\end{bmatrix}$

$\quad \begin{matrix}R _2(-1) \ \mbox{~} \ R _3(1/3)\end{matrix}\begin{bmatrix}1 & 2 & 5 \ 0 & 1 & 9 \ 0 & 1 & 2\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \ 2 & -1 & 0 \ \displaystyle\frac{1}{3} & 0 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R _{12}(-2) \ \mbox{~} \ R _{32}(-1)\end{matrix}\begin{bmatrix}1 & 0 & -13 \ 0 & 1 & 9 \ 0 & 0 & -7\end{bmatrix} A^{-1}= \begin{bmatrix}-3 & 2 & 0 \ 2 & -1 & 0 \ -\displaystyle\frac{5}{3} & 1 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R _3(-1/7)\ \mbox{~}\end{matrix}\begin{bmatrix}1 & 0 & -13 \ 0 & 1 & 9 \ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}-3 & 2 & 0 \ 2 & -1 & 0 \ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

$\quad \begin{matrix}R _{13}(13) \ \mbox{~} \ R _{23}(-9)\end{matrix}\begin{bmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

Hence, $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

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

A is an involuntary matrix given by $A=\begin{bmatrix} 0 & 1 & -1\ 4 & -3 & 4\ 3 & -3 & 4\end{bmatrix}$ then the inverse of $\dfrac{A}{2}$ will be?

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

An involuntary matrix satisfies A^2 = I, which means A = A^-1. If we want the inverse of (A/2), we use the property (kA)^-1 = (1/k)A^-1. Thus, (A/2)^-1 = 2A^-1 = 2A.

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

Let $A=\begin{bmatrix} 1 & -1 & -1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}$ and $10B=\begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{bmatrix}$, if $B$ is the inverse of matrix $A$, then $\alpha $ is

  1. $-2$
  2. $1$
  3. $2$
  4. $5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since, $B$ is the inverse of $A$.
ie, $B=10{ A }^{ -1 }$
$\therefore \left( 10 \right) { A }^{ -1 }=\begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{bmatrix}$
$\therefore \left( 10 \right) { A }^{ -1 }\cdot A=\begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{bmatrix}A$
$\Rightarrow 10I=\begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 10 & 0 & 0 \ 0 & 10 & 0 \ 0 & 0 & 10 \end{bmatrix}=\begin{bmatrix} 10 & 0 & 0 \ -5+\alpha  & 5+\alpha  & -5+\alpha  \ 0 & 0 & 10 \end{bmatrix}$
$\Rightarrow 5+\alpha =10$
$\Rightarrow \alpha =5$

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

The inverse of a diagonal matrix is a :

  1. Symmetric matrix

  2. Skew-symmetric matrix

  3. Diagonal matrix

  4. None of the above
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

A diagonal matrix has elements only in it's diagonal.
So the inverse will also have all non zero elements in the diagonal.
So, it will be symmetric and will also be a diagonal matrix.
Hence, option A and C are correct

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

Inverse of $A  = \begin{bmatrix} 1& 3\ 2 & -2\end{bmatrix} $ is equal to?A 

  1. $- \dfrac{1}{8} \begin{bmatrix}3 & 1\\ -2 & 2\end{bmatrix}$
  2. $- \dfrac{1}{8} \begin{bmatrix}-2 & -3\\ -2 & 1\end{bmatrix}$
  3. $ \dfrac{1}{8} \begin{bmatrix}-1 & -3\\ -2 & 2\end{bmatrix}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If $A  = \begin{bmatrix} 1& 3\\ 2 & -2\end{bmatrix} $
$ a _{11}= -2 $
$ a _{12}= -2 $
$ a _{21}= -3 $
$ a _{22}=  1 $

$ A^{-1}=\dfrac{ \left ( Cofactors of A \right )^{T}}{\left |A  \right |}$
$ \left ( Cofactors of A \right )^{T}=\begin{bmatrix} -2& -3\\ -2 & 1\end{bmatrix} $
${\left |A  \right |}= -2-6 $
${\left |A  \right |}=-8 $

$ A^{-1}=-\dfrac{1}{8}\times\begin{bmatrix} -2& -3\\ -2 & 1\end{bmatrix} $ 

Option will be B

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

The inverse of the $\begin{bmatrix}- 1 & 5\ - 3 & 2\end{bmatrix}$ is

  1. $\frac{1}{13} \begin{bmatrix}
    2 & - 5\\
    3 & - 1
    \end{bmatrix}$
  2. $\frac{1}{13} \begin{bmatrix}
    - 1 & 5\\
    - 3 & 2
    \end{bmatrix}$
  3. $\frac{1}{13} \begin{bmatrix}
    - 1 & - 3\\
    5 & 2
    \end{bmatrix}$
  4. $\frac{1}{13} \begin{bmatrix}
    1 & 5\\
    3 & - 2
    \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A=\left[\begin{matrix} -1 & 5 \\ -3  & 2 \end{matrix} \right]$

$\left|A\right|=-2+15=13\neq 0$

Hence ${A}^{-1}$ exists.

${C} _{ij}={\left(-1\right)}^{i+j}{M} _{ij}$

${C} _{11}=2,\,{C} _{12}=3,\,{C} _{21}=-5$  and ${C} _{22}=-1$

${C} _{ij}=\left[\begin{matrix} 2 & 3 \\ -5  & -1 \end{matrix} \right]$

Adj${A}={{C} _{ij}}^{T}=\left[\begin{matrix} 2 & -5 \\ 3  & -1 \end{matrix} \right]$

${A}^{-1}=\dfrac{Adj{\left(A\right)}}{\left|A\right|}=\dfrac{1}{13}\left[\begin{matrix} 2 & -5 \\ 3  & -1 \end{matrix} \right]$
Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

The inverse of the matrix $\begin{bmatrix} 5 & -2 \ 3 & 1 \end{bmatrix}$ is 

  1. $\dfrac { 1 }{ 11 } \begin{bmatrix} 1 & 2 \\ -3 & 5 \end{bmatrix}$
  2. $\begin{bmatrix} 1 & 2 \\ -3 & 5 \end{bmatrix}$
  3. $\dfrac { 1 }{ 13 } \begin{bmatrix} -2 & 5 \\ 1 & 3 \end{bmatrix}$
  4. $\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} A=\left[ \begin{array}{l} 5\, \, \, \, \, -2 \ 3\, \, \, \, \, \, \, \, \, 1 \end{array} \right]  \ \left| A \right| =5+6=11\ne 0 \ so,\, A\, \, is\, \, \, non-\sin  gular\, ,\, { A^{ -1 } }\, \, is\, \, exist \ so,m\, { A _{ 11 } }=1,\, \, \, \, \, { A _{ 12 } }=-3,\, \, \, \, { A _{ 21 } }=2,\, \, \, \, \, \, { A _{ 22 } }=5 \ A=\left( { \begin{array} { *{ 20 }{ c } }1 & { -3 } \ 2 & 5 \end{array} } \right) \Rightarrow AdjA=\left( { \begin{array} { *{ 20 }{ c } }1 & 2 \ { -3 } & 5 \end{array} } \right)  \ { A^{ -1 } }=\frac { 1 }{ { \left| A \right|  } } adjA\, \, \, \, \Rightarrow \, \, \, \, \frac { 1 }{ { 11 } } \left( { \begin{array} { *{ 20 }{ c } }1 & 2 \ { -3 } & 5 \end{array} } \right)  \end{array}$


Hence, this is the answer.

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

What is the inverse of the matrix
$A=\begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  & 0 \ -\sin { \theta  }  & \cos { \theta  }  & 0 \ 0 & 0 & 1 \end{bmatrix}$ ?

  1. $\begin{bmatrix} \cos { \theta } & -\sin { \theta } & 0 \\ \sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$
  2. $\begin{bmatrix} \cos { \theta } & 0 & -\sin { \theta } \\ 0 & 1 & 0 \\ \sin { \theta } & 0 & \cos { \theta } \end{bmatrix}$
  3. $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos { \theta } & -\sin { \theta } \\ 0 & \sin { \theta } & \cos { \theta } \end{bmatrix}$
  4. $\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \\ -\sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A = \begin{bmatrix} \cos \theta &  \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Calculate first minors.
$M _{11} = \cos \theta , M _{13} = 0, M _{22} = \cos \theta$
$M _{12} = -\sin \theta, M _{21} = \sin \theta, M _{23} = 0$
$M _{31} = 0, M _{32} = 0, M _{33} = \cos^{2}\theta + \sin^{2}\theta = 1$
Cofactor Matrix $= \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0\\ 0 & 0 & 1 \end{bmatrix} = C$
$det|A| = \cos^{2}\theta + \sin^{2}\theta = 1$

$adj (A) = C^{T} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0 & 0 & 1\end{bmatrix}$

$A^{-1} = \dfrac {adj(A)}{(A)} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0 & 1\end{bmatrix}$.
Multiple choice maths concept of directed numbers and number line subtraction of directed numbers subtraction of integers subtraction of integers on number line

The additive inverse of $\dfrac {2}{7}$ is 

  1. $\dfrac {2}{7}$
  2. $\dfrac {-2}{7}$
  3. $\dfrac {7}{2}$
  4. $\dfrac {-7}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
  • Additive inverse is negative of the given number.
  • Two rational numbers are additive inverse of each other when their sum is zero.

So additive inverse of $\dfrac{2}{7} $ is $ \dfrac{-2}{7}$.