Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

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?

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

  1. $2A$

  2. $\dfrac{A^{-1}}{2}$

  3. $\dfrac{A}{2}$

  4. $A^{-2}$

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Correct Option: A

If $A\begin{bmatrix} 1 & 1\ 2 & 0\end{bmatrix}=\begin{bmatrix} 3 & 2\ 1 & 1\end{bmatrix}$, then $A^{-1}$ is given by?

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

  1. $\begin{bmatrix} 0 & -1\ 2 & -4\end{bmatrix}$

  2. $\begin{bmatrix} 0 & -1\ -2 & -4\end{bmatrix}$

  3. $\begin{bmatrix} 0 & 1\ 2 & -4\end{bmatrix}$

  4. None of these

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Correct Option: A

If $A=\left[ \begin{matrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{matrix} \right] $, then value of $A^{-1}$ is equal to 

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

  1. $A$

  2. $A^{2}$

  3. $A^{3}$

  4. $A^{4}$

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Correct Option: A

If A and B are any $2\times2$ matrices, then det. (A+B) =0 implies

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

  1. None of these

  2. det A=0 and det B=0

  3. det A=0 or det B=0

  4. det A=0 + det B=0

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Correct Option: A

If $A^2-A+1=0$, then the inverse of A is?

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

  1. A

  2. $A+I$

  3. $I-A$

  4. $A-I$

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Correct Option: A

Let $\begin{bmatrix} 1 & 1\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 3\ 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & n-1\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\ 0 & 1\end{bmatrix}$
If $A=\begin{bmatrix} 1 & n\ 0 & 1\end{bmatrix}$ then $A^{-1}=?$

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

  1. $\begin{bmatrix} 1 & 12\ 0 & 1\end{bmatrix}$

  2. $\begin{bmatrix} 1 & -13\ 0 & 1\end{bmatrix}$

  3. $\begin{bmatrix} 1 & -12\ 0 & 1\end{bmatrix}$

  4. $\begin{bmatrix} 1 & 0\ -13 & 1\end{bmatrix}$

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Correct Option: B
Explanation:

$\begin{bmatrix} 1 & 1\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 3\ 0 & 1\end{bmatrix}..\begin{bmatrix} 1 & n-1\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\ 0 & 1\end{bmatrix}$
$\Rightarrow \dfrac{n(n-1)}{2}=78\Rightarrow n=13$
$A=\begin{bmatrix} 1 & 13\ 0 & 1\end{bmatrix}$
so $A^{-1}=\begin{bmatrix} 1 & -13\ 0 & 1\end{bmatrix}$.

If $\displaystyle A=\begin{bmatrix} 0 & 0 & 1\ 0 & 1&0 \ 1& 0 & 0\end{bmatrix}$, then $A^{-1}$ is.

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

  1. $-A$

  2. $A$

  3. $1$

  4. None of these

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Correct Option: B
Explanation:

We have, $A=\begin{bmatrix} 0 & 0&1\ 0 &1 &0\ 1&0 &0\end{bmatrix}$
$\Rightarrow |A|=0(0-0)-0(0-0)+1(0-1)$
$\Rightarrow |A|=-1$
and cofactors of A are
$A _{11}=0, A _{12}=0, A _{13}=-1,$
$A _{21}=0, A _{22}=-1, A _{23}=0,$
$A _{31}=-1, A _{32}=0, A _{33}=0$
$\therefore A^{-1}=\displaystyle\frac{adj(A)}{|A|}$
$=-\displaystyle\frac{1}{1}\begin{bmatrix} 0 & 0 & -1\0 & -1 &0\ -1 &0 &0\end{bmatrix}$

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

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

  1. $-2$

  2. $1$

  3. $2$

  4. $5$

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Correct Option: D
Explanation:

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

If $\begin{bmatrix} 1 & 2 \ 3 & -5 \end{bmatrix}$, then ${A}^{-1}$ is equal to

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

  1. $\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \ \cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$

  2. $\begin{bmatrix} -\cfrac { 5 }{ 11 } & -\cfrac { 2 }{ 11 } \ -\cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$

  3. $\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \ \cfrac { 3 }{ 11 } & \cfrac { 1 }{ 11 } \end{bmatrix}$

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

Show answer
Correct Option: A
Explanation:

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


$\therefore \left| A \right| =\begin{bmatrix} 1 & 2 \ 3 & -5 \end{bmatrix}=-5-6=-11$

and $adj(A)=\begin{bmatrix} -5 & -2 \ -3 & 1 \end{bmatrix}$

$\therefore { A }^{ -1 }=\cfrac { 1 }{ \left| A \right|  } adj(A)$

$=-\cfrac { 1 }{ 11 } \begin{bmatrix} -5 & -2 \ -3 & 1 \end{bmatrix}=\cfrac { 1 }{ 11 } \begin{bmatrix} 5 & 2 \ 3 & -1 \end{bmatrix}$

$\quad =\begin{bmatrix} \cfrac { 5 }{ 11 }  & \cfrac { 2 }{ 11 }  \ \cfrac { 3 }{ 11 }  & -\cfrac { 1 }{ 11 }  \end{bmatrix}$

If you switch the first row with the fourth row, what will the new first row be?
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$

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

  1. $3, 4, 2, 11$

  2. $9, 1, 0, 0$

  3. $0, 1, 0, 2$

  4. $0, 0, 6, 1$

  5. $0, 2, 0, 3$

Show answer
Correct Option: D
Explanation:
Given Matrix$=\begin{bmatrix} 3 & 4 & 2 & 11 \\ 9 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 6 & 1 \end{bmatrix}$
First and fourth row are interchanged
New matrix obtained $=\begin{bmatrix} 0 & 0 & 6 & 1 \\ 9 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 3 & 4 & 2 & 1 \end{bmatrix}$
New first row$=\begin{bmatrix} 0 & 0 & 6 & 1 \end{bmatrix}$ Option D