Tag: business maths

Questions Related to business maths

If the matrix $\displaystyle \left[ \begin{matrix} a \ c \end{matrix}\begin{matrix} b \ d \end{matrix} \right] $ is commutative with the matrix $\displaystyle \left[ \begin{matrix} 1 \ 0 \end{matrix}\begin{matrix} 1 \ 1 \end{matrix} \right] $, then

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $a=0, b=c$

  2. $b=0, c=d$

  3. $c=0, d=a$

  4. $d=0, a=b$

Show answer
Correct Option: C
Explanation:

Since matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is commutative with the matrix $\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$


So,

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$\begin{bmatrix} a & a+b \\ c & c+d \end{bmatrix}=\begin{bmatrix} a+c & b+d \\ c & d \end{bmatrix}$

$a=a+c,\quad a+b=b+d$
$c=0,\quad a=d$

If $A$ is a square matrix of order $3$ and det $A = 5$, then what is det $[(2A)^{-1}]$ equal to?

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $\dfrac{1}{10}$

  2. $\dfrac{2}{5}$

  3. $\dfrac{8}{5}$

  4. $\dfrac{1}{40}$

Show answer
Correct Option: D
Explanation:

If $A$ is of order $3$ then,$A^-1$ is also of order $3$.

Now, $\text{det} (cA)=c^n(\text{ det} A)$ where $n$ is the order of the matrix.
And, $\text {det } A=\dfrac{1}{\text{det} A^{-1}}$
Thus $\text{det} [(2A)^{-1}]=2^3\text{det}[A]$
and $\text{det}[A^{-1}]=\dfrac{1}{8.5}$
$=\dfrac{1}{40}$

If A is a square matrix such that $A^2 = I $ where I is the identity matrix, then what is $A^{-1}$ equal to ?

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. A + 1

  2. Null matrix

  3. A

  4. Transpose of A

Show answer
Correct Option: C
Explanation:
$A\rightarrow $square matrix
${ A }^{ 2 }=I$
To find ${ A }^{ -1 }=?$
Given ${ A }^{ 2 }=I$
$\Rightarrow A\times A=I$
$\Rightarrow A=\cfrac { I }{ A } $
$\Rightarrow A={ A }^{ -1 }$
$\therefore { A }^{ -1 }=A$
Option C is correct

If A is an orthogonal matrix of order 3 and $B=\begin{bmatrix}1&2&3\-3&0&2\2&5&0\end{bmatrix}$, then which of the following is/are correct?
1. $|AB|= \pm 47$
2. $AB=BA$
Select the correct answer using the code given below :

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2

Show answer
Correct Option: A
Explanation:

Solution:

$A$ is an orthogonal matrix.
$A^2=1$
$|A|^2=1$
$|A|=\pm 1$
$|AB|=\pm|B|$
Now, $|B|=\begin{bmatrix}1&2&3\-3&0&2\2&5&0\end{bmatrix}=47$
$\therefore |AB|=\pm47$
And $AB=\begin{bmatrix}3&7&3\-3&0&2\2&5&0\end{bmatrix}$ and 
$BA=\begin{bmatrix}1&2&3\-3&0&2\2&5&0\end{bmatrix}$
Since, $AB\neq BA$
Hence, A is the correct option.

If A is a non singular matrix satisfying $A=AB-BA$, then which one of the following holds true

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $det. B=0$

  2. $B=0$

  3. $det. A=1$

  4. $det(B+I) =det(B-I)$

Show answer
Correct Option: D
Explanation:

$A=AB-BA$
$AI+BA=AB$
$A(I+B)A=AB$
$\left| I+B \right| \left| A \right| =\left| A \right| \left| B \right| $
$\left| B \right| =\left| I+B \right| \longrightarrow 1$
$A=AB-BA$
$BA=AB-A$
$BA=A(B-I)$
$\left| B \right| \left| A \right| =\left| A \right| \left| B-I \right| $
$\left| B \right| =\left| B-I \right| \longrightarrow 2$
From equation 1 and 2
$\left| B-I \right| =\left| B+I \right| $

If A is a square matrix of order 3,then $|Adj\left( Adj{ A }^{ 2 } \right) |=$

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. ${ |A| }^{ 2 }$

  2. ${ |A| }^{ 4 }$

  3. ${ |A| }^{ 8 }$

  4. ${ |A| }^{ 16 }$

Show answer
Correct Option: C
Explanation:

$|adj\left( adj{ A }^{ 2 } \right) |$
$Q={ { |A }^{ 2 }| }^{ { \left( 3-1 \right)  }^{ 2 } }={ { |A }^{ 2 } }|^{ 4 }={ |A| }^{ 8 }$


If $AB=0$ for the matrices
$A=\left[ \begin{matrix} \cos ^{ 2 }{ \theta  }  & \cos { \theta  } \sin { \theta  }  \ \cos { \theta  } \sin { \theta  }  & \sin ^{ 2 }{ \theta  }  \end{matrix} \right] $ and $B=\left[ \begin{matrix} \cos ^{ 2 }{ \phi  }  & \cos { \phi  } \sin { \phi  }  \ \cos { \phi  } \sin { \phi  }  & \sin ^{ 2 }{ \phi  }  \end{matrix} \right] $ then $\theta-\phi $ is

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. an odd multiple of $\dfrac{\pi}{2}$

  2. an odd multiple of ${\pi}$

  3. an odd even of $\dfrac{\pi}{2}$

  4. $0$

Show answer
Correct Option: A
Explanation:
$A=\begin{bmatrix} { \cos {  }  }^{ 2 }\theta  & \cos { \theta  } \sin { \theta  }  \\ \cos { \theta  } \sin { \theta  }  & { \sin {  }  }^{ 2 }\theta  \end{bmatrix}$      $B=\begin{bmatrix} { \cos {  }  }^{ 2 }  & \cos { \phi  } \sin { \phi  }  \\ \cos { \phi  } \sin { \phi  }  & { \sin {  }  }^{ 2 }\phi  \end{bmatrix}$

$AB = \begin{bmatrix} { \cos {  }  }^{ 2 }\theta  & \cos { \theta  } \sin { \theta  }  \\ \cos { \theta  } \sin { \theta  }  & { \sin {  }  }^{ 2 }\theta  \end{bmatrix}$  $ \begin{bmatrix} { \cos {  }  }^{ 2 }\phi  & \cos { \phi  } \sin { \phi  }  \\ \cos { \phi  } \sin { \phi  }  & { \sin {  }  }^{ 2 }\phi  \end{bmatrix}$

$=\begin{bmatrix} { \cos {  }  }^{ 2 }\theta \cos^{2}\phi + \cos { \theta  } \sin { \theta  } \cos\phi \sin\phi & {\cos^2\theta \cos\phi \sin\phi+\sin^2\phi \sin\theta \cos\theta} \\ \cos { \theta  } \sin { \theta  } \cos^{2}\phi  + { \sin {  }  }^{ 2 }\theta \cos\phi  \sin\phi  & \cos\theta \sin\theta \cos\phi  { \sin {  }  }^{ 2 }\phi  +\sin^{2}\theta\sin^{2}\phi   \end{bmatrix}$

$= \begin{bmatrix} 0  & 0  \\ 0  & 0  \end{bmatrix}$  

$\Rightarrow $ $\begin{bmatrix} { \cos {  }  }^{ 2 }\theta \cos \phi \cos (\theta -\phi)  & \cos { \theta  } \sin { \phi  } \cos (\theta-\phi) \\ \sin {\phi  } \cos {\phi   }  & { \sin {  }  }^{ 2 }\theta \sin \phi \cos (\theta -\phi)  \end{bmatrix}$ = $\begin{bmatrix} 0  & 0  \\ 0  & 0  \end{bmatrix}$

$\Rightarrow$ $\cos (\theta - \phi )$$\begin{bmatrix} { \cos {  }  }^{ 2 }\theta \cos \phi  & \cos { \theta  } \sin { \phi  }  \\ \sin {\phi  } \cos {\phi   }  & { \sin {  }  }^{ 2 }\theta \sin \phi   \end{bmatrix}$ = $\begin{bmatrix} 0  & 0  \\ 0  & 0  \end{bmatrix}$

$\Rightarrow \cos (\theta-\phi) (\cos \theta \cos \phi \sin \theta \sin \phi - \cos \theta \cos \phi \sin \theta \sin \phi ) = 0$

$\Rightarrow \cos (\theta - \phi) = 0$

$\theta - \phi = (2n+1)\dfrac{\pi}{2}$

i.e an odd multiple of $\dfrac{\pi}{2}$ 
Let $A$ be a matrix of order $2 \times 2$ such that $A^2 = 0$ then $A^2 - (a + d)A + (ad - bc) I$ is equal to
business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $I$

  2. $0 _{2\times 2}$

  3. $-I$

  4. none of these

Show answer
Correct Option: B
Explanation:
Given $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$

$A^2=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}=\begin{pmatrix} a^2+bc & ab+bd \\ ca+cd & bc+d^2 \end{pmatrix}............................(1)$

$-(a+d)A=\begin{pmatrix} -a^2-ad & -ab-bd \\ -ac-cd & -ad-d^2 \end{pmatrix}..............(2)$

$(ad-bc)I=\begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix}.............(3)$
Adding 1,2,3 we get,
$A^2-(a+d)A+(ad-bc)I=\begin{pmatrix} a^2+bc-a^2-ad+ad-bc & ab+bd-ab-bd \\ ac+cd-ac-cd & bc+d^2-ad-d^2+ad-bc \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

Hence, the value of $A^2-(a+d)A+(ad-bc)I=0 _{2\times 2}$

Let $A$ and $B$ are two matrices such that $AB =BA$, then for every $n\in N$,

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $A^nB=BA^n$

  2. $(AB)^n = A^nB^n$

  3. $(A+B)^n=$ $^nC _0A^n+$ $^nC _1A^{n-1}B^1+$ $^nC _2A^{n-2}B^2+ ... + ^nC _n:B^n$.

  4. $A^{2n}-B^{2n}=(A^n-B^n)(A^n+B^n)$

Show answer
Correct Option: A,B,C,D
Explanation:

$A^2B =A(AB) =A(BA) =(AB)A$

$= (BA)A =BA^2$

Similarly, $A^3B=BA^3$

In general $A^nB=BA^n: \forall : n\geq 1$

(b) and (c) hold as $AB =BA$.

Also, $(A^n -B^n) (A^n + B^n)$

$=A^nA^n-B^n:A^n+A^n:B^n-B^n:B^n$

$=A^{2n}-B^{2n}$

Hence, options A,B,C and D.

If $D _1$ and $D _2$ are two $3\times 3$ diagonal matrices, then

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $D _1:D _2$ is diagonal matrix

  2. $D _1:D _2=D _2:D _1$

  3. $D _1^2+D _2^2$ is a diagonal matrix

  4. none of these

Show answer
Correct Option: A,B,C
Explanation:

Let ${ D } _{ 1 }=\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix},{ D } _{ 2 }=\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}$

Then

${ D } _{ 1 }{ D } _{ 2 }=\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}=\begin{bmatrix} ax & 0 & 0 \ 0 & by & 0 \ 0 & 0 & cz \end{bmatrix}\$


$ { D } _{ 2 }{ D } _{ 1 }=\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}=\begin{bmatrix} xa & 0 & 0 \ 0 & yb & 0 \ 0 & 0 & zc \end{bmatrix}$

As $ax=xa,by=yb,cz=zc$

${ D } _{ 1 }{ D } _{ 2 }={ D } _{ 2 }{ D } _{ 1 }$

${ { D } _{ 1 } }^{ 2 }+{ { D } _{ 2 } }^{ 2 }=\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}+\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}\$

$ =\begin{bmatrix} { a }^{ 2 } & 0 & 0 \ 0 & { b }^{ 2 } & 0 \ 0 & 0 & { c }^{ 2 } \end{bmatrix}+\begin{bmatrix} { x }^{ 2 } & 0 & 0 \ 0 & { y }^{ 2 } & 0 \ 0 & 0 & { z }^{ 2 } \end{bmatrix}=\begin{bmatrix} { { a }^{ 2 }+x }^{ 2 } & 0 & 0 \ 0 & { { b }^{ 2 }+y }^{ 2 } & 0 \ 0 & 0 & { { c }^{ 2 }+z }^{ 2 } \end{bmatrix}$