Tag: multiplicative inverse of a matrix

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Questions Related to multiplicative inverse of a matrix

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The determinant of a sum of matrices is not generally related to the sum of their determinants or the individual determinants being zero.

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

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

  1. A
  2. $A+I$
  3. $I-A$
  4. $A-I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given the matrix equation A^2 - A + I = 0 (where 1 represents the identity matrix I), we can rearrange it to isolate I: I = A - A^2 = A(I - A), or multiply through by A^-1. Multiplying A^2 - A + I = 0 by A^-1 gives A - I + A^-1 = 0, which rearranges to A^-1 = I - A.

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

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 $A^{-1}=?$

  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. $\begin{bmatrix} 1 & 0\\ -13 & 1\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $A^{-1}=\begin{bmatrix} 1 & -13\ 0 & 1\end{bmatrix}$.

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

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

  1. $-A$
  2. $A$
  3. $1$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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
$A _{11}=0, A _{12}=0, A _{13}=-1,$
$A _{21}=0, A _{22}=-1, A _{23}=0,$
$A _{31}=-1, A _{32}=0, A _{33}=0$
$\therefore A^{-1}=\displaystyle\frac{adj(A)}{|A|}$
$=-\displaystyle\frac{1}{1}\begin{bmatrix} 0 & 0 & -1\0 & -1 &0\ -1 &0 &0\end{bmatrix}$

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

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

  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. $\begin{bmatrix} 5 & 2 \\ 3 & -1 \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $adj(A)=\begin{bmatrix} -5 & -2 \ -3 & 1 \end{bmatrix}$

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

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

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

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

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

  1. $3, 4, 2, 11$
  2. $9, 1, 0, 0$
  3. $0, 1, 0, 2$
  4. $0, 0, 6, 1$
  5. $0, 2, 0, 3$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Which of the following is the new row that results when you add rows $1$ and $3$?
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$

  1. $6, 8, 4, 22$
  2. $3, 5, 2, 13$
  3. $3, 4, 2, 11$
  4. $3, 4, 8, 12$
  5. $4, 5, 3, 12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given Matrix$=\begin{bmatrix} 3 & 4 & 2 & 11 \\ 9 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 6 & 1 \end{bmatrix}$
Row $1=\begin{bmatrix} 3 & 4 & 2 & 11 \end{bmatrix}$
Row $3=\begin{bmatrix} 0 & 1 & 0 & 2 \end{bmatrix}$
Sum$=\begin{bmatrix} 3+0 & 4+1 & 2+0 & 11+2 \end{bmatrix}$
$\begin{bmatrix} 3 & 5 & 2 & 13 \end{bmatrix}$
$\therefore $Option $2$ is correct
Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Use a transformation matrix to find the image of $D(-7,6)$ after a rotation of $180^0$ counterclockwise around the origin.

  1. $(7,6)$
  2. $(-7,-6)$
  3. $(7,-6)$
  4. $(-7,6)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The transformation matrix for rotation  is $\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}$

For $\theta=180^{0}$ , the transformation matrix will be $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$
So the image of point $(-7,6)$ is $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}\begin{bmatrix} -7 \ 6 \end{bmatrix}=\begin{bmatrix} 7 \ -6 \end{bmatrix}$
Therefore the correct option is $C$