Tag: singular & non-singular matrix

Questions Related to singular & non-singular matrix

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

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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

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

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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.

With $1,\omega, \omega^2$ as cube roots of unity, inverse of which of the following matrices exists

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

  1. $\begin{bmatrix}1 & \omega \ \omega & \omega^2\end{bmatrix}$

  2. $\begin{bmatrix}\omega^2 & 1 \ 1 & \omega\end{bmatrix}$

  3. $\begin{bmatrix} \omega & \omega^2 \ \omega^2 & 1\end{bmatrix}$

  4. None of these

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Correct Option: D
Explanation:

The inverse of a matrix exists if its determinant is not equal $0$.
For option A, 
Let $A=\begin{bmatrix}1 & {\omega} \ {\omega} & {\omega}^2\end{bmatrix}$
Here, $|A|=0$
Hence, inverse does not exists.

For option B, 
Let $A=\begin{bmatrix}{\omega}^{2} & 1 \ 1 & {\omega}\end{bmatrix}$
Here, $|A|=0$
Hence, inverse does not exists.

For option C, 
Let $A=\begin{bmatrix}{\omega} & {\omega}^{2} \ {\omega}^{2} & 1\end{bmatrix}$
Here, $|A|=0$
Hence, inverse does not exists.

Hence, the inverse does not exist for any of the given matrices

$\displaystyle \begin{bmatrix} 1 & -2 & 3 \ 2 & -1 & 4 \ 3 & 4 & 1 \end{bmatrix}$ is a

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

  1. rectangular matrix

  2. singular matrix

  3. square matrix

  4. nonsingular matrix

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Correct Option: C,D
Explanation:

It is a $3 \times 3$ so it is a square matrix,


$\displaystyle \begin{bmatrix} 1 & -2 & 3 \ 2 & -1 & 4 \ 3 & 4 & 1 \end{bmatrix}$


$=1(-17)+2(-10)+3(11)$

$=-17-20+33$

$=33-37=-4$

so, it is not singular