Tag: business maths

Questions Related to business maths

The rod of fixed length $k$ slides along the coordinate axes. If it meets the axes at $A(a,0)$ and $B(0,b)$, then the minimum value of $\left(a+\dfrac 1a\right) ^2+\left(b+\dfrac 1b\right) ^2$ is

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $0$

  2. $8$

  3. $k^2-4+\dfrac 4{k^2}$

  4. $k^2+4+\dfrac 4{k^2}$

Show answer
Correct Option: B
Explanation:

since, $A.M. \geq G.M. \implies \dfrac{a+b}{2}\geq \sqrt{ab}$


let $a=1 $ and $b=a^2$

Therefore, $\dfrac{1+a^2}{2}\geq \sqrt{a^2}$

$\implies \dfrac{1+a^2}{2}\geq \sqrt{a^2}$

$\implies \dfrac{1+a^2}{2}\geq a$

$\implies \dfrac{1+a^2}{a}\geq 2$

$\implies \dfrac{1}{a}+\dfrac{a^2}{a}\geq 2$

$\implies a+\dfrac{1}{a}\geq 2$
 squaring on both sides
$\implies (a+\dfrac{1}{a})^2\geq 4$ ------- (1)

similarly, $ (b+\dfrac{1}{b})^2\geq 4$ --------(2)

adding (1) and (2) we get

$(a+\dfrac{1}{a})^2+ (b+\dfrac{1}{b})^2\geq 4+4$

$(a+\dfrac{1}{a})^2+ (b+\dfrac{1}{b})^2\geq 8$

Therefore the minimum value is $8$

If $a>0$, then least value of $(a^3+a^2+a+1) ^2$ is

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $64a^2$

  2. $16a^4$

  3. $16a^3$

  4. None of the above.

Show answer
Correct Option: C
Explanation:

we know that $A.M.\geq G.M.$


therefore, $\dfrac{a^3+a^2+a+1}{4}\geq \sqrt[4]{a^3*a^2*a*1}$

$\implies \dfrac{a^3+a^2+a+1}{4}\geq \sqrt[4]{a^6}$

squaring on both sides 

$\implies (\dfrac{a^3+a^2+a+1}{4})^2\geq ({a^{\dfrac{6}{4}}})^2$

$\implies ({a^3+a^2+a+1})^2\geq 16a^3$

The number of value of $x$ in the closed interval $[-4,-1]$, the matrix $\begin{bmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{bmatrix}$ is singular is 

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: C
Explanation:
For a singular matrix the value of the determinant = 0 
$ [3(-1)(2)+(-1+x)(x+2)(x+3)+(2)(3)(-1)] $
$\times  [-(2)(-1)(x+3)-(-1+x)(3)(2)-(3)(x+2)(x+3)] = 0 $
$-6+x^3+4x^2+x-6-6+2x+6+6-6x-3x^2-15x-18 = 0 $
$ x^3+x^2-18x-24 = 0 $
$ x = -4 $
$x =(3-\sqrt{33})/2=-1.372 $
$x =(3+\sqrt{33})/2= 4.372 $

Value of x in the closed interval $[-4,-1]$ are $-4,   -1.372$

If $\left[ {\begin{array}{*{20}{c}}1&{ - 1}&x\1&x&1\x&{ - 1}&1\end{array}} \right]$ has no inverse, then the real value of $x$ is 

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $2$

  2. $3$

  3. $0$

  4. $1$

Show answer
Correct Option: D
Explanation:

$A=\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}has\quad no\quad inverse\quad x$

$if\,\left| A \right| = 0$

$\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}=0$

$1\left( {x + 1} \right) - 1\left( { - 1 + x} \right) + x\left( { - 1 - {x^2}} \right) = 0$

$x + 1 + 1 - x - x - {x^3} = 0$

$ - {x^3} - x + 2 = 0$

${x^3} + x - 2 = 0$

$\left( {x - 1} \right)\left( {{x^2} + x + 2} \right) = 0$

$x = 1\,is\,real\,value\,$

The matrix $\begin{bmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \ 3 & 1 & 1 \end{bmatrix}$ is:

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. nonsingular

  2. singular

  3. skew symmetric

  4. symmetric

Show answer
Correct Option: B
Explanation:

$\begin{vmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \end{vmatrix}=1\begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix}-0\begin{vmatrix} 2 & 0 \ 3 & 1 \end{vmatrix}+1\begin{vmatrix} 2 & 1 \ 3 & 1 \end{vmatrix}$

$=1(1-0)-0+1(2-3)$
$=1-1$
$=0$
Therefore, it is a singular matrix.

If $\begin{bmatrix} 1 & 2 & x \  4 & -1 & 7  \  2 & 4 & 6  \end{bmatrix}$ is a singular matrix, then $x=$

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $0$

  2. $1$

  3. $-3$

  4. $3$

Show answer
Correct Option: C
Explanation:

$\begin{vmatrix} 1 & 2 & x \ 4 & -1 & 7 \ 2 & 4 & -6  \end{vmatrix}=0$

$\Rightarrow 1\begin{vmatrix} -1 & 7 \ 4 & -6 \end{vmatrix}-2\begin{vmatrix} 4 & 7 \ 2 & -6 \end{vmatrix}+x\begin{vmatrix} 4 & -1 \ 2 & 4 \end{vmatrix}=0$

$\Rightarrow (6-28)-2(-24-14)+x(16+2)=0$
$\Rightarrow -22+76+18x=0$
$\Rightarrow 18x=-54$

$\Rightarrow x=-3$

$A$ and $B$ are two non-zero square matrices such that $AB = 0$. Then

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. Both $A$ and $B$ are singular

  2. Either of them is singular

  3. Neither matrix is singular

  4. None of these

Show answer
Correct Option: B
Explanation:

Given $AB =0$

$\Rightarrow |AB|=0$

$\Rightarrow |A||B|=0$

$\Rightarrow |A|=0 \,or\, |B|=0$

If the matrix $\begin{bmatrix} \alpha  & 2 & 2 \ -3 & 0 & 4 \ 1 & -1 & 1 \end{bmatrix}$ is not invertible, then:

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $\alpha =-5$

  2. $\alpha =5$

  3. $\alpha =-0$

  4. $\alpha =-1$

Show answer
Correct Option: A
Explanation:

For a matrix to be not invertible, determinant of the matrix should be $0$
So, $\left| \begin{matrix} \alpha  & 2 & 2 \ -3 & 0 & 4 \ 1 & -1 & 1 \end{matrix} \right| =0$

$\Rightarrow -(-3)(2-(-2))-4(-\alpha -2)=0\ \Rightarrow 12+4\alpha +8=0\ \Rightarrow 4\alpha =-20\ \Rightarrow \alpha =-5$
Hence, option A is correct

Consider the following statements:
1. The matrix
               $\begin{pmatrix} 1 & 2 & 1 \ a & 2a & 1 \ b & 2b & 1 \end{pmatrix}$ is singular.
2. The matrix
              $\begin{pmatrix} c & 2c & 1 \ a & 2a & 1 \ b & 2b & 1 \end{pmatrix}$ is non-singular.
Which of the above statements is/are correct?

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2

Show answer
Correct Option: A
Explanation:

For matrix

  $\begin{vmatrix} 1 & 2 & 1 \ a & 2a & 1 \ b & 2b & 1 \end{vmatrix}$


$C _2\rightarrow C _2-2C _1 $

$\begin{vmatrix} 1 & 0 & 1 \\ a & 0 & 1 \\ b & 0 & 1 \end{vmatrix}$ The determinant is $0$. So the matrix is singular.

For matrix,
$ \begin{vmatrix} c & 2c & 1 \\ a & 2a & 1 \\ b & 2b & 1 \end{vmatrix}$

$C _2\rightarrow C _2-2C _1 $

$\begin{vmatrix} c & 0 & 1 \\ a & 0 & 1 \\ b & 0 & 1 \end{vmatrix}$ The determinant is $0$. So the matrix is singular.

Thus, only $1$ is true.
Hence, option A is correct.

Let $A$ be a square matrix all of whose entries are integers. Then which one of the following is true?

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. If $det(A)=\pm 1$, then ${A}^{-1}$ exists but all its entries are not necessarily integers.

  2. If $det(A)=\pm 1$, then ${A}^{-1}$ exists and all its entries are non integers

  3. If $det(A)=\pm 1$, then ${A}^{-1}$ exists and all its entries are integers

  4. If $det(A)=\pm 1$, then ${A}^{-1}$ need not exist

Show answer
Correct Option: C