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

Use the method of elementary row transformation to compute the inverse of 
$\quad \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}$

  1. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$
  2. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{1}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{11}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{2}{21}\end{bmatrix}$
  3. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{16}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{4}{21}\end{bmatrix}$
  4. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{2}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{4}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

$\Rightarrow \quad Write \space A A^{-1}= I$

$\quad \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix} A^{-1}= \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}$

$\quad \begin{matrix}R _{21}(-2)\ \mbox{~}\ R _{31}(1)\end{matrix}\begin{bmatrix}1 & 0 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \ -2 & 1 & 0 \ 1 & 0 & 1\end{bmatrix}$

$\quad \begin{matrix}R _2(-1) \ \mbox{~} \ R _3(1/3)\end{matrix}\begin{bmatrix}1 & 2 & 5 \ 0 & 1 & 9 \ 0 & 1 & 2\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \ 2 & -1 & 0 \ \displaystyle\frac{1}{3} & 0 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R _{12}(-2) \ \mbox{~} \ R _{32}(-1)\end{matrix}\begin{bmatrix}1 & 0 & -13 \ 0 & 1 & 9 \ 0 & 0 & -7\end{bmatrix} A^{-1}= \begin{bmatrix}-3 & 2 & 0 \ 2 & -1 & 0 \ -\displaystyle\frac{5}{3} & 1 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R _3(-1/7)\ \mbox{~}\end{matrix}\begin{bmatrix}1 & 0 & -13 \ 0 & 1 & 9 \ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}-3 & 2 & 0 \ 2 & -1 & 0 \ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

$\quad \begin{matrix}R _{13}(13) \ \mbox{~} \ R _{23}(-9)\end{matrix}\begin{bmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

Hence, $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

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

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 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 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