Tag: business maths

Questions Related to business maths

Let $A$ and $B$ be two $2 \times 2$ matrices. Consider the statements
          $(i)$ $AB =0 \Rightarrow A = 0 :or :B = 0$
         $ (ii)$ $AB =I \Rightarrow A =B^{-1}$
          $(iii)$ $(A + B)^2 = A^2 + 2AB + B^2$

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

  1. $(i)$ is false, $(ii)$ and $(iii)$ are true

  2. $(i)$ and $(iii)$ are false, $(ii)$ is true

  3. $(i)$ and $(ii)$ are false, $(iii)$ is true

  4. $(ii)$ and $(iii)$ are false, $(i)$ is true

Show answer
Correct Option: B
Explanation:

(i) is false.


If $A=\begin{bmatrix}0& 1\0 & -1\end{bmatrix}$ and $B =\begin{bmatrix}1& 1\0 & 0\end{bmatrix}$, then

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

Thus,  $AB = 0$ but neither $A = 0$ nor $B = 0$

(iii) is false since matrix multiplication is not commutative.

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

(ii) is true as the product $AB$ is an identity matrix, if and only $B$ is inverse of the matrix $A$.

Hence, option B.

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

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

  1. $\begin{bmatrix}\dfrac {3}{7}
    &\dfrac {-1}{7} \
    \dfrac {1}{7} & \dfrac {2}{7}
    \end{bmatrix}$

  2. $\begin{bmatrix}\dfrac {3}{7}
    &\dfrac {1}{7} \
    \dfrac {-1}{7} & \dfrac {2}{7}
    \end{bmatrix}$

  3. $\begin{bmatrix}\dfrac {3}{7}
    &\dfrac {1}{7} \
    \dfrac {1}{7} & \dfrac {2}{7}
    \end{bmatrix}$

  4. None of these

Show answer
Correct Option: B
Explanation:

$|A| = \begin{vmatrix}2 & -1\ 1 & 3\end{vmatrix} = (6 + 1) = 7\neq 0$
$M _{11} = 3, M _{12} = 1, M _{21} = -1$ and $M _{22} = 2$


$\therefore C _{11} = 3, C _{12} = -1, C _{21} = 1$ and $C _{22} = 2$

$\Rightarrow Adj\ A = \begin{bmatrix} 3& -1\ 1 & 2\end{bmatrix}^T = \begin{bmatrix} 3& 1\ -1 & 2\end{bmatrix}$

$\Rightarrow A^{-1} = \dfrac {1}{|A|} (adj\ A) = \dfrac {1}{7} \begin{bmatrix}3 & 1\ -1 & 2\end{bmatrix} = \begin{bmatrix}\dfrac {3}{7} & \dfrac {1}{7}\ -\dfrac {1}{7} & \dfrac {2}{7}\end{bmatrix}$.

If $A$ and $B$ are invertible square matrices of the same order then $(AB)^{-1} = ?$

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

  1. $AB^{-1}$

  2. $A^{-1}B$

  3. $A^{-1}B^{-1}$

  4. $B^{-1}A^{-1}$

Show answer
Correct Option: D

If $A$ and $B$ are two square matrices of the same order and $m$ is a positive integer, then
$(A + B)^m =$ $^mC _0A^m +$ $^mC _1 A^{m -1} B + ^mC _2A^{m-2} B^2 + ... +$ $^mC _{m- 1} AB^{m-1}+$ $^mC _m B^m$ if

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

  1. $AB =BA$

  2. $AB + BA =0$

  3. $A^m = 0, :B^m = 0$

  4. none of these.

Show answer
Correct Option: A
Explanation:

Binomial theorem is applicable if and only if $AB=BA$.

Let $A, : B : and : C$ be $2\times 2$ matrices with entries from the set of real numbers. Define $\ast $ as follows: $\displaystyle A\ast B=\frac{1}{2}(AB + BA)$, then

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

  1. $A\ast B=B\ast A$

  2. $A\ast A=A^2$

  3. $A\ast (B+C)=A\ast B+A\ast C$

  4. $A\ast I=A$

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

$\displaystyle A*1=\frac { 1 }{ 2 } \left( AI+IA \right) =\frac { 1 }{ 2 } \left( A+A \right) =A$

$\therefore A*I=A,\therefore$ (a)holds.
$\displaystyle A*A=\frac { 1 }{ 2 } \left( AA+AA \right) =\frac { 1 }{ 2 } \left( { A }^{ 2 }+{ A }^{ 2 } \right) ={ A }^{ 2 }$
$\therefore$ (b) holds.
$\displaystyle \left( c \right) A*B=\frac { 1 }{ 2 } \left( AB+BA \right) B*A=\frac { 1 }{ 2 } \left( BA+AB \right) =\frac { 1 }{ 2 } \left( AB+BA \right) $
$[\because $ addition is commutative$]$
$\therefore A*B=B*A,\therefore$ (c)holds
$\displaystyle A*\left( B+C \right) =\frac { 1 }{ 2 } \left( A\left( B+C \right) +\left( B+C \right) A \right) =\frac { 1 }{ 2 } \left( AB+CA+BA+CA \right)$
$\displaystyle  =\frac { 1 }{ 2 } \left( AB+BA \right) +\frac { 1 }{ 2 } \left( AC+CA \right) =A*B+A*C$
$\therefore$ (d) holds.

If $A$ and $B$ are square matrices of the same order such that $A^2=A,:B^2=B, :AB = BA = 0$, then

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

  1. $AB^2=0$

  2. $(A + B)^2 = A + B$

  3. $(A - B)^2 = A - B$

  4. none of these.

Show answer
Correct Option: A,B
Explanation:

$AB^2 =(AB)B = 0B =0$  

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

$=A+0+0+B =A+B$

Also, $(A +B)^2 = A + B\neq A-B$

If $A^k=0$ for some value of $k$ and $B=1+A+A^2+...+A^{k-1},$ then $B^{-1}$ equal

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

  1. $I-A$

  2. $I+A$

  3. $I-A^{k-1}$

  4. None of these

Show answer
Correct Option: A
Explanation:

Let $\displaystyle B=I+A+{ A }^{ 2 }+{ A }^{ 3 }+...+{ A }^{ k-1 }$

$\displaystyle \therefore B\left( I-A \right) =\left( I+A+{ A }^{ 2 }+...+{ A }^{ k-1 } \right) \left( I-A \right) $
$\displaystyle =I-A+A-{ A }^{ 2 }+{ A }^{ 2 }+...+{ A }^{ k-1 }-{ A }^{ k }=I={ A }^{ k }=I\quad \quad \quad \left( \because { A }^{ k }=O \right) $
Hence, $\displaystyle { \left( I-A \right)  }^{ -1 }=I+A+{ A }^{ 2 }+...+{ A }^{ k }-1=B$
$\displaystyle \Rightarrow { B }^{ -1 }=I-A$

Let $A, : B : and : C$ be $2\times 2$ matrices with entries from the set of real numbers. Define $\ast $ as follows:
  $\displaystyle A \ast B=\frac{1}{2}(AB\,'+A'B)$. Which of the given is true?

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

  1. $A\ast B= B \ast A$

  2. $A\ast A=A^2$

  3. $A\ast (B+C)=A\ast B+A \ast C$

  4. $A\ast I =A+A'$

Show answer
Correct Option: A,C
Explanation:

$\displaystyle A \ast B=\frac{1}{2}(AB'+A'B)$

1) $\displaystyle B \ast A=\frac{1}{2}(BA'+B'A)=\frac{1}{2}(AB'+A'B)=A \ast B$

2)$\displaystyle A \ast A=\frac{1}{2}(AA'+A'A)$

3)$\displaystyle A \ast (B+C)=\frac{1}{2}(A(B+C)'+A'(B+C))$

                             $=\displaystyle\frac{1}{2}(AB'+A'B)+\frac{1}{2}(AC'+A'C)$

                             $=A\ast B+A \ast C$

4)$\displaystyle A \ast I=\frac{1}{2}(AI'+A'I)=\frac{1}{2}(A+A')$

Hence, options A and C.

Say true or false:

Let A, B be two matrices such that they commute, then $(AB)^n = A^nB^n$.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$A$ and $B$ commute each other then $AB=BA$
${ \left( AB \right)  }^{ n }={ \left( BA \right)  }^{ n }\ \Rightarrow { \left( AB \right)  }^{ n }={ B }^{ n }{ A }^{ n }={ A }^{ n }{ B }^{ n }={ \left( BA \right)  }^{ n }$

If $A$ is a non-singular matrix, then 

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

  1. ${ A }^{ -1 }$ is symmetric if $A$ is symmetric

  2. ${ A }^{ -1 }$ is skew-symmetric if $A$ is symmetric

  3. $\left| { A }^{ -1 } \right| =\left| A \right| $

  4. $\left| { A }^{ -1 } \right| ={ \left| A \right|  }^{ -1 }$

Show answer
Correct Option: A,D
Explanation:

Since $\left| A \right| \neq 0$, therefore ${ A }^{ -1 }$ exists.

Now, $A{ A }^{ -1 }=I={ A }^{ -1 }A$
$\Rightarrow \left( A{ A }^{ -1 } \right) '=I'=\left( { A }^{ -1 }A \right) '\Rightarrow \left( { A }^{ -1 } \right) 'A'=I=A'\left( { A }^{ -1 } \right) '\quad \quad \quad \left( \because A'=A \right) $
$\Rightarrow \left( { A }^{ -1 } \right) 'A=I=A\left( { A }^{ -1 } \right) '\Rightarrow { A }^{ -1 }=\left( { A }^{ -1 } \right) '\Rightarrow { A }^{ -1 }$ is symmetric
Also, since $\left| A \right| \neq 0,\therefore { A }^{ -1 }$ exists such that
$A{ A }^{ -1 }=I={ A }^{ -1 }A\Rightarrow \left| A{ A }^{ -1 } \right| =\left| I \right| $
$\Rightarrow \left| A \right| \left| { A }^{ -1 } \right| =1\quad \quad \left( \because \left| AB \right| =\left| A \right| \left| B \right|  \right) $
$\displaystyle \Rightarrow \left| { A }^{ -1 } \right| =\frac { 1 }{ \left| A \right|  } $