Tag: business maths

Questions Related to business maths

If $A$ and $B$ are two non-zero square matrices of the same order such that the product $AB=0$, then

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

  1. both $A$ and $B$ must be singular

  2. exactly one of them must be singular

  3. both of them are non singular

  4. none of these

Show answer
Correct Option: D
Explanation:

$AB=O$
taking determinant on both sides
$|AB|=|A||B|=0$
$\Rightarrow$ Either $A$ or $B$ should be a singular matrix.
Hence, option D.

Let $A=\begin{bmatrix} a & b\ c & d\end{bmatrix}$ be a $2\times 2$ matrix, where a, b, c and d take the values $0$ or $1$ only. The number of such matrices which have inverses is?

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

  1. $8$

  2. $7$

  3. $6$

  4. $5$

Show answer
Correct Option: A
Explanation:

${ A }^{ -1 }=\frac { adj\quad A }{ \left| A \right|  } $

so $\left| A \right| \neq 0\ ad-bc\neq 0$
for  $ ad-bc\neq 0$ 
either $ad=1 ,bc=0$ or $ad=0 ,bc=1$
$ad=1 ,bc=0$ can be chosen in 4 ways 
similarly $ad=0 ,bc=1$ can be chosen in 4 ways
so number of matrices  formed$=4+4=8$

If $A$ is a nonsingular matrix satisfying $AB=BA+A$ then

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

  1. $\left|B\right|=\left|I+B\right|$

  2. $\left|B\right|=\left|2I+B\right|$

  3. $\left|B\right|=\left|B-I\right|$

  4. $\left|B\right|=\left|B-2I\right|$

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Correct Option: A

If $A$ and $B$ and square matrix of the same order such that $AB=A$ and $BA=B$, then $A$ and $B$ are both:

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

  1. Singular

  2. Non-singular

  3. Idempotent

  4. Involutory

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Correct Option: A

The number of $3\times 3$ non-singular matrices, with four entries as $1$ and all other entries as $0$ is 

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

  1. Less than $4$

  2. $5$

  3. $6$

  4. At least $7$

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Correct Option: D

If A and B are two non-singular square matrices and AB=I, then which of the following is true ?

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

  1. $BA = I$

  2. ${ A }^{ -1 }=B$

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

  4. ${ A }^{ 2 }=B$

Show answer
Correct Option: A

If $A$ and $B$ are non-singular matrices, then _____

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

  1. $(AB)^{-1} = A^{-1}B^{-1}$

  2. $AB = BA$

  3. $(AB)^T = A^T. B^T$

  4. $(AB)^{-1} = B^{-1} A^{-1}$

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Correct Option: A

The matrix $\left[ \begin{matrix} \lambda  & 7 & -2 \ 4 & 1 & 3 \ 2 & -1 & 2 \end{matrix} \right]$ is a singular matrix if $\lambda$ is

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

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

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

  3. $-5$

  4. $none\ of\ these$

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Correct Option: A

If 3, -2 are the Exigent values of non-singular matrix A and |A|=4. Then Exigent values of Adj(A) are

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

  1. 3/4, -1/2

  2. 4/3, -2

  3. 12, -8

  4. -12, 8

Show answer
Correct Option: B
Explanation:

$\begin{array}{l} { \lambda _{ 1 } }=3,\, \, { \lambda _{ 2 } }=-2 \ \left| A \right| =4 \ adj\left( A \right) =\left( A \right) \cdot { A^{ -1 } } \ Ax=\lambda x \ \frac { 1 }{ \lambda  } x={ A^{ -1 } }x \ \left( { \lambda I\cdot { A^{ -1 } } } \right) =\frac { { \left( A \right) \cdot \left( { \lambda I-{ A^{ -1 } } } \right)  } }{ { \left( A \right)  } }  \ exigent\, value\, of\, adj\left( A \right) \, is\, \frac { { \left( A \right)  } }{ { exigent\, value\, of\, A } }  \ =\frac { 4 }{ 3 } ,\frac { 4 }{ { -8 } }  \ =\left( { \frac { 4 }{ 3 } ,-2 } \right)  \ Hence,\, option\, B\, is\, correct\, answer. \end{array}$

The values of K for which matrix $A = \begin{bmatrix} 1& 0 & - K\ 2 & 1 & 3\ K & 0 & 1\end{bmatrix}$ is invertible are

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

  1. $\displaystyle {-1,1 }$

  2. $\displaystyle R$

  3. $\displaystyle R\backslash {-1,1}$

  4. $\displaystyle no\space real\space values$

Show answer
Correct Option: B
Explanation:

Matrix A is invertible if $|A| \neq 0$, i.e.,
$\begin{bmatrix}1 & 0 & -K\ 2 & 1 & 3\ K & 0 & 1\end{bmatrix} \neq 0$
or $1(1) - K (-K) \neq 0$
Expanding along second column
$|A| =-0+1(1-(-K)(K))=1+K^2 \neq 0$ which is true for all real K.
Hence, A is invertible for all real values of K.