Tag: multiplicative inverse of a matrix

Questions Related to multiplicative inverse of a matrix

$A=\begin{bmatrix} 1&-2&3\7&-8&9\4&-5&6\end{bmatrix}$ the new matrix formed by adding $\ 2^{nd}\ row \ to \ 1^{st} $ row  will be

  1. $\begin{bmatrix}8&-10&12\7&-8&9\4&-5&6\end{bmatrix}$

  2. $\begin{bmatrix} 6&6&6\7&8&9\4&5&6\end{bmatrix}$

  3. $\begin{bmatrix} 1&2&3\7&8&9\11&-13&14\end{bmatrix}$

  4. $\begin{bmatrix} 1&-2&3\7&8&-29\4&-2&6\end{bmatrix}$


Correct Option: A
Explanation:
Given $A=\begin{bmatrix} 1 & -2 & 3 \\ 7 & -8 & 9 \\ 4 & -5 & 6 \end{bmatrix}$
Now$\Rightarrow { R } _{ 1 }\rightarrow { R } _{ 1 }+{ R } _{ 2 }$
Resultant matrix$=\begin{bmatrix} 1+7 & -2-8 & 3+9 \\ 7 & -8 & 9 \\ 4 & -5 & 6 \end{bmatrix}$
$=\begin{bmatrix} 8 & -12 & 12 \\ 7 & -8 & 9 \\ 4 & -5 & 6 \end{bmatrix}$
Option A is correct

 A=$\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$
The new matrix formed  after interchanging $2^{nd}$ and $3^{rd}$rows  will be 

  1. $-\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$

  2. $\begin{bmatrix} 4&5&6\1&2&3\7&8&9\end{bmatrix}$

  3. $-\begin{bmatrix} 1&2&3\7&8&9\4&5&6\end{bmatrix}$

  4. $\begin{bmatrix} 1&2&3\7&8&9\4&5&6\end{bmatrix}$


Correct Option: D
Explanation:
Given
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
Now second and third rows are interchanged $\Rightarrow $all elements are interchanged
Resultant matrix$=\begin{bmatrix} 1 & 2 & 3 \\ 7 & 8 & 9 \\ 4 & 5 & 6 \end{bmatrix}$
Option D is correct

For a matrix $A \begin{pmatrix} 1& 0 & 0\ 2 & 1 & 0\ 3 & 2 & 1\end{pmatrix}$, if $U _{1}, U _{2}$ and $U _{3}$ are $3\times 1$ column matrices satisfying $AU _{1} = \begin{pmatrix}1\ 0 \ 0
\end{pmatrix}, AU _{2} \begin{pmatrix}2\3 \ 0
\end{pmatrix}, AU _{3} = \begin{pmatrix}2\ 3\ 1
\end{pmatrix}$ and $U$ is $3\times 3$ matrix whose columns are $U _{1}, U _{2}$ and $U _{3}$
Then sum of the elements of $U^{-1}$ is

  1. $6$

  2. $0 (zero)$

  3. $1$

  4. $2/3$


Correct Option: B
Explanation:

Let $U _{i} = \begin{pmatrix}a _{i}\b _{i} \c _{i} \end{pmatrix} i = 1, 2, 3$
$AU _{1} = \begin{pmatrix}a _{1}\2a _{1} + b _{1} \ 3a _{1} + 2b _{1} + c _{1}
\end{pmatrix} = \begin{pmatrix}1\0 \ 0
\end{pmatrix}$, So $a _{1} = 1, b _{1} = -2, c _{1} = 1$
$AU _{2} = \begin{pmatrix}a _{2}\2a _{2} + b^{2} \ 3a _{2} + 2b _{2} + c _{2}
\end{pmatrix} = \begin{pmatrix}2\ 3\ 0
\end{pmatrix}$,
So, $a _{2} = 2, b _{2} = -1, c _{2} = -4$. Similarly, $a _{3} = 2, b _{3} = -1, c _{3} = -3$
So, $U = \begin{pmatrix} 1& 2 & 2\ -2 & -1 & -1\ 1 & -4 & -3\end{pmatrix}$. So, sum of elements of $U^{-1}$ is zero.

The inverse of a diagonal matrix is a :

  1. Symmetric matrix

  2. Skew-symmetric matrix

  3. Diagonal matrix

  4. None of the above


Correct Option: A,C
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

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

  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


Correct Option: B
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

If a matrix A is such that $3{A^3} + 2{A^2} + 5A + I = 0$ , then $A^{-1}$ is equal to

  1. $ - (3{A^2} + 2A + 5)$

  2. $3{A^2} + 2A + 5$

  3. $3{A^2} - 2A - 5$

  4. None of these


Correct Option: A
Explanation:

$3A^3+2A^2+5A+I=0$
$3A^3+2A^2+5A+AA^{-1}=0$
$A^{-1}=-3A^2-2A-5$

If $A$ is a non zero square matrix of order $n$ with $det\left( I+A \right) \neq 0$, and ${A}^{3}=0$, where $I,O$ are unit and null matrices of order $n\times n$ respectively, then ${ \left( I+A \right)  }^{ -1 }=$

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

  2. $I+A+{ A }^{ 2 }$

  3. $I+{ A }^{ 2 }$

  4. $I+A$


Correct Option: A
Explanation:
$det(I+A)\neq 0$
$A^3=0$   where $0$ is null matrix, $I$ is the identity matrix
$A^3+I=I$ [adding $I$ on both sides]
$(A+I)(A^2-IA+I^2)=I$ [by the formula of $a^3+b^3$]
$(A+I)(A^2-A+I)=I$
$(I+A)(I+A)^{-1}=I$ [by the rule of inverse matrix]
hence $(I+A)^{-1}=(A^2-A+I)$
Ans: $I-A+A^2$

If $A=\begin{bmatrix} 3 & -2 \ 5 & 8 \end{bmatrix}$, then $A^{-1}=$

  1. $\frac{1}{30}\begin{bmatrix} 8 & 2 \ -5 & 3 \end{bmatrix}$

  2. $\frac{1}{34}\begin{bmatrix} 8 & 2 \ -5 & 3 \end{bmatrix}$

  3. $-\frac{1}{34}\begin{bmatrix} -8 & -2 \ -5 & 3 \end{bmatrix}$

  4. None of these


Correct Option: B
Explanation:

$A=\left[{\begin{array}{cc}3&2\5&8\end{array}}\right]$

$A^{-1}=\cfrac{1}{ad-bc}\left[{\begin{array}{cc}d&-b\-c&a\end{array}}\right]$  (determinant)
$=\cfrac{1}{3\times 8-(-2\times 5)}\left[{\begin{array}{cc}8&2\-5&3\end{array}}\right]$  
$=\cfrac{1}{34}\left[{\begin{array}{cc}8&2\-5&3\end{array}}\right]$  

If the matrix $\begin{bmatrix} 0 & 2\beta & \Upsilon \ \alpha & \beta & -\Upsilon \ \alpha & -\beta & \Upsilon \end{bmatrix}$is orthogonal, then

  1. $\alpha = \pm\dfrac{1}{\sqrt{2}}$

  2. $\beta = \pm\dfrac{1}{\sqrt{6}}$

  3. $\gamma = \pm\dfrac{1}{\sqrt{3}}$

  4. all of these


Correct Option: D
Explanation:
$\begin{bmatrix} 0 & 2\beta  & \gamma  \\ \alpha  & \beta  & -\gamma  \\ \alpha  & -\beta  & \gamma  \end{bmatrix}$
for orthogonal matrix we have 
$A.A^{T}=I$
$\begin{bmatrix} 0 & 2\beta  & \gamma  \\ \alpha  & \beta  & -\gamma  \\ \alpha  & -\beta  & \gamma  \end{bmatrix}\begin{bmatrix} 0 & \alpha  & \alpha  \\ 2\beta  & \beta  & -\beta  \\ \gamma  & -\gamma  & \gamma  \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 0+4{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } & 0+2{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } & 0-2{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } \\ 0+2{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }-{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } \\ 0-2{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }-{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$4\beta^{2}+\gamma^{2}=1, 2\beta^{2}-\gamma^{2}=0$, 
$4\left(\dfrac{\gamma^{2}}{2}\right)+\gamma^{2}=1$        $\beta^{2}=\dfrac{r^{2}}{2}$
$r^{2}[3]=1$
$r=\pm \dfrac{1}{\sqrt{3}}$
$2\beta^{2}-\gamma^{2}=0, \alpha^{2}+\beta^{2}+\gamma^{2}+\gamma^{2}=1, \alpha^{2}-\beta^{2}-\gamma^{2}=0$
$\beta^{2}=\dfrac{\gamma^{2}}{2},  \alpha^{2}+\dfrac{\gamma^{2}}{2}+\dfrac{\gamma^{2}}{1}=1$
$\alpha^{2}+\dfrac{3\gamma^{2}}{2}=1$
$\beta^{2}=\dfrac{1}{6}\alpha^{2}+\dfrac{3}{2}\times \dfrac{1}{3}=1$
$\beta=\pm \dfrac{1}{\sqrt{6}}$            $\alpha=\pm \dfrac{1}{\sqrt{2}}$

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. $\frac{1}{13} \begin{bmatrix}
    1 & 5\
    3 & - 2
    \end{bmatrix}$


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

${A}^{-1}=\dfrac{Adj{\left(A\right)}}{\left|A\right|}=\dfrac{1}{13}\left[\begin{matrix} 2 & -5 \\ 3  & -1 \end{matrix} \right]$