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 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 $\displaystyle\frac{-a}{b}$ is __________.

  1. $\displaystyle\frac{a}{b}$
  2. $\displaystyle\frac{b}{a}$
  3. $\displaystyle\frac{-b}{a}$

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

The additive inverse of any number $x$ is defined as

$x +y =0$
Then $y$ is the additive inverse of $x$.
Say, the additive inverse of $ -\dfrac{a}{b}$ is $z$.
$\Rightarrow  -\dfrac{a}{b} +z =0$
$\Rightarrow z= \dfrac{a}{b}$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

The inverse of the matrix $\begin{bmatrix}1 & 0 & 1\ 0 & 2 & 3\ 1 & 2& 1\end{bmatrix}$ is

  1. $\dfrac {-1}{6} \begin{bmatrix}-4 & 2 & -2\\ 3 & 0 & -3\\ -2 & -2& 2\end{bmatrix}$
  2. $\dfrac {1}{6} \begin{bmatrix}-4 & 2 & -2\\ 3 & 0 & -3\\ -2 & -2& 2\end{bmatrix}$
  3. $\begin{bmatrix}-2 & 1 & -1\\ 1 & 0 & -1\\ -2 & -2& 2\end{bmatrix}$
  4. $\begin{bmatrix}2 & -1 & 1\\ -1 & 0 & 1\\ 2 & 2& -2\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The inverse of a matrix A is (1/det(A)) * adj(A). For the given matrix, the determinant is 1(2-6) - 0 + 1(0-2) = -4 - 2 = -6. Calculating the adjugate matrix and dividing by -6 yields the correct inverse.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If inverse of $A=\left[ \begin{matrix} 1 & 1 & 1 \ 2 & -1 & -1 \ 1 & -1 & 1 \end{matrix} \right] $ is $\cfrac { -1 }{ 6 } \left[ \begin{matrix} -2 & -2 & 0 \ -3 & 0 & \alpha  \ -1 & 2 & -3 \end{matrix} \right] $ then $\alpha=$

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

$A=\begin{bmatrix}1&1&1\\ 2&-1&-1\\ 1&-1&1\end{bmatrix}$

$A^{-1}=\begin{bmatrix}1&1&1\\ 2&-1&-1\\ 1&-1&1\end{bmatrix}^{-1}$

$=\begin{bmatrix}1&1&1&\mid \:&1&0&0\\ 2&-1&-1&\mid \:&0&1&0\\ 1&-1&1&\mid \:&0&0&1\end{bmatrix}$

$\:R _1\:\leftrightarrow \:R _2$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 1&1&1&\mid \:&1&0&0\\ 1&-1&1&\mid \:&0&0&1\end{bmatrix}$

$R _2\:\leftarrow \:R _2-\frac{1}{2}\cdot \:R _1$

$R _3\:\leftarrow \:R _3-\frac{1}{2}\cdot \:R _1$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 0&\frac{3}{2}&\frac{3}{2}&\mid \:&1&-\frac{1}{2}&0\\ 0&-\frac{1}{2}&\frac{3}{2}&\mid \:&0&-\frac{1}{2}&1\end{bmatrix}$

$R _3\:\leftarrow \:R _3+\frac{1}{3}\cdot \:R _2$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 0&\frac{3}{2}&\frac{3}{2}&\mid \:&1&-\frac{1}{2}&0\\ 0&0&2&\mid \:&\frac{1}{3}&-\frac{2}{3}&1\end{bmatrix}$

$R _3\:\leftarrow \frac{1}{2}\cdot \:R _3$

$R _2\:\leftarrow \:R _2-\frac{3}{2}\cdot \:R _3$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 0&\frac{3}{2}&0&\mid \:&\frac{3}{4}&0&-\frac{3}{4}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$R _1\:\leftarrow \:R _1+1\cdot \:R _3$

$R _2\:\leftarrow \frac{2}{3}\cdot \:R _2$

$=\begin{bmatrix}2&-1&0&\mid \:&\frac{1}{6}&\frac{2}{3}&\frac{1}{2}\\ 0&1&0&\mid \:&\frac{1}{2}&0&-\frac{1}{2}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$R _1\:\leftarrow \:R _1+1\cdot \:R _2$

$=\begin{bmatrix}2&0&0&\mid \:&\frac{2}{3}&\frac{2}{3}&0\\ 0&1&0&\mid \:&\frac{1}{2}&0&-\frac{1}{2}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$R _1\:\leftarrow \frac{1}{2}\cdot \:R _1$

$=\begin{bmatrix}1&0&0&\mid \:&\frac{1}{3}&\frac{1}{3}&0\\ 0&1&0&\mid \:&\frac{1}{2}&0&-\frac{1}{2}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$=\begin{bmatrix}\frac{1}{3}&\tfrac{1}{3}&0\\ \tfrac{1}{2}&0&-\tfrac{1}{2}\\ \tfrac{1}{6}&-\tfrac{1}{3}&\tfrac{1}{2}\end{bmatrix}$

$=-\dfrac{1}{6}\begin{bmatrix}-2 &-2  &0 \\  -3& 0 &3 \\  -1& 2 &-3 \end{bmatrix}$

$\therefore \alpha =3$
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  1. $\begin{bmatrix}3&-1 \\-5 &-3\end{bmatrix}$
  2. $\begin{bmatrix}2&-1 \\-5 &3\end{bmatrix}$
  3. $\begin{bmatrix}-3&5 \\1 &-2\end{bmatrix}$
  4. $\begin{bmatrix}-2&5 \\1 &-3\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $A=\begin{bmatrix}3& 1\\5&2\end{bmatrix}$
$\left|A\right|=6-5=1\neq 0$
$\therefore {A}^{-1}$ exists.
${C} _{11}={\left(-1\right)}^{1+1}{M} _{11}={\left(-1\right)}^{2}2=2$
${C} _{12}={\left(-1\right)}^{1+2}{M} _{12}={\left(-1\right)}^{3}5=-5$
${C} _{13}={\left(-1\right)}^{1+3}{M} _{13}={\left(-1\right)}^{4}1=1$
${C} _{14}={\left(-1\right)}^{1+4}{M} _{14}={\left(-1\right)}^{5}3=-3$
${C} _{ij}=\begin{bmatrix}2& -5\\-1 & 3\end{bmatrix}$
Adj$\left(A\right)={\begin{bmatrix}2& -5\\-1 & 3\end{bmatrix}}^{T}$
$=\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$
${A}^{-1}=\dfrac{adj\left(A\right)}{\left|A\right|}=\dfrac{1}{1}\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$
$\therefore {A}^{-1}=\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  1. a symmetric matrix

  2. a skew-symmetric matrix

  3. diagonal matrix

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

Let A be a skew-symmeteic matric of order $n.$

By definition $\displaystyle { A }^{ T }=-A$ 
$\displaystyle\Rightarrow \left| { A }^{ T } \right| =\left| -A \right| \Rightarrow \left| A \right| ={ \left( -1 \right)  }^{ n }\left| A \right| \$
$\displaystyle \Rightarrow \left| A \right| =-\left| A \right|\quad\quad[\because $ n is odd $]$
$\displaystyle \Rightarrow 2\left| A \right| =0\Rightarrow\left| A \right| =0$
$\therefore{ A }^{ -1 }$ does not exist.