Tag: business maths

Quizzes (44) Questions (670)

Questions Related to business maths

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

$A=\begin{bmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos\theta\end{bmatrix}$ and $AB=BA=I$, then B is equal to

  1. $\begin{bmatrix} -\cos\theta & \sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$
  2. $\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$
  3. $\begin{bmatrix} -\sin\theta & \cos\theta \\ \cos\theta & \sin\theta\end{bmatrix}$
  4. $\begin{bmatrix} \sin\theta & -\cos\theta \\ -\cos\theta & \sin\theta\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $A=\begin{bmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos\theta \end{bmatrix}$ and $AB=BA=I$
$\Rightarrow B=A^{-1}I=A^{-1}$
$=\displaystyle\frac{1}{\cos^2\theta +\sin^2\theta}\begin{bmatrix} \cos\theta & \sin\theta \ -\sin\theta & \cos\theta\end{bmatrix}$
$\Rightarrow B=\begin{bmatrix}\cos\theta & \sin\theta \ -\sin\theta & \cos\theta\end{bmatrix}$

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

$A=\begin{bmatrix} 2&2&1\0&1&4\0&2&6\end{bmatrix}$, $B=\begin{bmatrix} 2&2&1\0&1&4\0&0&1\end{bmatrix}$

To obtain B from the matrix A, order of operations would be   

  1. $R _3 \rightarrow R _3-3R _1$, $R _3\rightarrow R _2-R _1$
  2. $R _3 \rightarrow R _1-2R _2$, $R _3 \rightarrow (R _3 \times {-2})$
  3. $R _2 \rightarrow R _2-2R _2$, $R _3 \rightarrow (R _3 \div {2})$
  4. $R _3 \rightarrow R _3-2R _2$, $R _3 \rightarrow (R _3 \div {-2})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$A=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 2 & 6 \end{bmatrix}B=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$
From A to B
${ R } _{ 1 }$ & ${ R } _{ 2 }$ are same
changing ${ R } _{ 3 }$ w.r.t. ${ R } _{ 2 }$
Clearly the term $'2'$ in ${ R } _{ 3 }{ C } _{ 2 }$ in A is change to $'0'$ in comparing them we find
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 2 }$
We get the matrix
$\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & -2 \end{bmatrix}$
Now if we divide ${ R } _{ 3 }$ w.r.t. $'-2'$
We get the derived term $1$ in ${ R } _{ 3 }{ C } _{ 3 }$ of B
$\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}=B$
$\therefore $Operations are
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 2 }$
 and ${ R } _{ 3 }\rightarrow { R } _{ 3 }-(-2)$
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

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

B is matrix obtained by subtracting $4 \ times \ 1^{st}\ row\ from \ 2^{nd} \ row$ of A. Find matrix B

  1. $\begin{bmatrix} 1&2&3\\0&3&6\\0&6&12\end{bmatrix}$
  2. $\begin{bmatrix} 1&2&3\\7&0&0\\4&5&6\end{bmatrix}$
  3. $\begin{bmatrix} 1&2&3\\0&1&2\\3&4&5\end{bmatrix}$
  4. $\begin{bmatrix} 1&2&3\\0&-3&-6\\7&8&9\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given $A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
$\Rightarrow { R } _{ 2 }\rightarrow { R } _{ 2 }-4{ R } _{ 1 }$ (for matrix B)
$\Rightarrow B-\begin{bmatrix} 1 & 2 & 3 \\ 4-4(1) & 5-4(2) & 6-4(3) \\ 7 & 8 & 9 \end{bmatrix}=\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 7 & 8 & 9 \end{bmatrix}$
Option D is correct
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

$A=\begin{bmatrix} 2&2&1\4&5&6\6&8&9\end{bmatrix}$, $B=\begin{bmatrix} 2&2&1\0&1&4\0&2&6\end{bmatrix}$

To convert matrix A into matrix B, the order of row operations are 

  1. $R _1\rightarrow R _2-2R _1$, $R _3 \rightarrow R _3-R _1$
  2. $R _2\rightarrow R _2-2R _1$, $R _3 \rightarrow R _3-3R _1$
  3. $R _1\rightarrow R _1-2R _2$, $R _3 \rightarrow R _1-R _3$
  4. $R _2\rightarrow R _3-2R _3$, $R _3 \rightarrow R _1-R _1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$A=\begin{bmatrix} 2 & 2 & 1 \\ 4 & 5 & 6 \\ 6 & 8 & 9 \end{bmatrix}\quad B=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 2 & 6 \end{bmatrix}$
Clearly in order to convert matrix A to B. Both ${ R } _{ 2 }$ and ${ R } _{ 3 }$ are changed w.r.t. ${ R } _{ 1 }$
${ R } _{ 2 }^{ 1 }\leftrightarrow X{ R } _{ 2 }-Y{ R } _{ 1 }$
For ${ R } _{ 2 }=4$ and ${ R } _{ 1 }=2$ then ${ R } _{ 2 }^{ 1 }=0$
$\Rightarrow 0=X4-2$Y$
$\Rightarrow 2X=Y$
For ${ R }_{ 2 }=5$ and ${ R }_{ 1 }=1$ then ${ R }_{ 2 }^{ 1 }=1$
$1=X(5)-Y(2)$
$\Rightarrow X=1\quad Y=2$
Relation is ${ R }_{ 2 }\leftrightarrow { R }_{ 2 }-2{ R }_{ 21}$
Parallely ${ R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }$
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

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

The new matrix obtained after  adding $2^{nd} \  row \ to\  3\ times\  3^{rd} \ row $ is

  1. $\begin{bmatrix} 1&2&3\\4&5&6\\25&29&33\end{bmatrix}$
  2. $\begin{bmatrix} 1&2&3\\25&-29&-33\\4&5&6\end{bmatrix}$
  3. $\begin{bmatrix} 1&2&3\\7&8&9\\4&5&6\end{bmatrix}$
  4. $\begin{bmatrix} 1&2&3\\-25&-29&-33\\4&5&6\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
$\Rightarrow { R } _{ 3 }\rightarrow 3{ R } _{ 3 }+{ R } _{ 2 }$ for new matrix
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3(7)+4 & 3(8)+5 & 3(9)+6 \end{bmatrix}$
New matrix $=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 25 & 29 & 33 \end{bmatrix}$
Option A is correct
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

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

B is matrix obtained by subtracting $4\ times 1^{st}\ row from \ 2^{nd} \ row$ of A.
C is matrix obtained by subtracting $7 \ times \ 1^{st}\ row\ from\ 3^{rd} row$, then $C$ is 

  1. $\begin{bmatrix} 1&2&3\\0&3&6\\0&6&12\end{bmatrix}$
  2. $\begin{bmatrix} 1&2&3\\7&0&0\\4&5&6\end{bmatrix}$
  3. $\begin{bmatrix} 1&2&3\\0&1&2\\3&4&5\end{bmatrix}$
  4. $\begin{bmatrix} 1&2&3\\0&-3&-6\\0&-6&-12\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given $A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
B is obtained by substracting $4{ R } _{ 1 }$ from ${ R } _{ 2 }$
$\Rightarrow { R } _{ 2 }\leftrightarrow { R } _{ 2 }-4{ R } _{ 1 }$$\rightarrow B\sim \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 7 & 8 & 9 \end{bmatrix}$
C is obtained by subtracting $7{ R } _{ 1 }$ from ${ R } _{ 3 }$
$\Rightarrow { R } _{ 3 }\leftrightarrow { R } _{ 3 }-7{ R } _{ 1 }$
$C\sim \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{bmatrix}$
$\therefore $option D is correct
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

The system $\begin{pmatrix} 1 & -1 & 2 \ 3 & 5 & -3 \ 2 & 6 & a \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} 3 \ b \ 2 \end{pmatrix}$ has no solution, if

  1. $a=-5,b\ne 5$
  2. $a=-5, b=5$
  3. $a\ne -5, b=5$
  4. $a\ne -5,b\ne 5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 
$\begin{pmatrix} 1 & -1 & 2 \ 3 & 5 & -3 \ 2 & 6 & a \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} 3 \ b \ 2 \end{pmatrix}$
The augment matrix of given system is
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 3 & 5 & -3 & | & b \ 2 & 6 & a & | & 2 \end{bmatrix}$
${ R } _{ 2 }\rightarrow { R } _{ 2 }-3{ R } _{ 1 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 2 & 6 & a & | & 2 \end{bmatrix}$
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 1 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 0 & 8 & a-4 & | & -4 \end{bmatrix}$
${ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 2 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 0 & 0 & a+5 & | & 5-b \end{bmatrix}$
We know that
For no solution
Rank A < Rank $\left[ A|B \right] $
$\therefore$ $a+5=0$ and $5-b\ne 0$
$a=-5, b\ne 5$

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

If $A = \left[ {\begin{array}{*{20}{c}}1&2\3&4\end{array}} \right]$, then $8A^{-4}$ is equal to

  1. $145A^{-1}+27I$
  2. $145A^{-1}-27I$
  3. $27I - 145A^{-1}$
  4. $29A^{-1} +9I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the Cayley-Hamilton theorem for matrix A = [[1, 2], [3, 4]], the characteristic equation is A^2 - 5A - 2I = 0. By manipulating this equation, one can express higher powers of A in terms of A and I, eventually leading to the expression 27I - 145A^-1.