Tag: applications of matrices and determinants

Quizzes (3) Questions (176)

Questions Related to applications of matrices and determinants

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

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

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

Let X = [[1, 1], [2, 0]] and B = [[3, 2], [1, 1]]. We have AX = B, so A = BX^-1. Then A^-1 = (BX^-1)^-1 = XB^-1. Calculating B^-1 and multiplying by X gives the result.

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