Tag: singular & non-singular matrix

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Questions Related to singular & non-singular matrix

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

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

  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. $\left|B\right|=\left|B-2I\right|$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given AB = BA + A, we can rewrite this as AB - BA = A. This is a commutation relation. Using properties of determinants and the given condition, it can be shown that |B| = |I + B|.

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

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

  1. Singular

  2. Non-singular

  3. Idempotent

  4. Involutory
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If AB = A and BA = B, then A^2 = A(BA) = (AB)A = AA = A^2. Also, A = AB implies |A| = |A||B|. If |A| is not 0, then |B| = 1. However, if A and B are singular, |A| = 0 and |B| = 0, which satisfies the equations.

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

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

  1. Less than $4$
  2. $5$
  3. $6$
  4. At least $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A 3x3 matrix with four 1s and five 0s has a determinant that can be calculated by expanding along rows or columns. By testing various configurations, one finds that many such matrices are singular, but there are indeed more than 7 non-singular configurations possible.

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

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

  1. $BA = I$
  2. ${ A }^{ -1 }=B$
  3. ${ B }^{ -1 }=A$
  4. ${ A }^{ 2 }=B$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For square matrices A and B, if AB = I, then A and B are inverses of each other. By the properties of inverse matrices, if AB = I, then BA = I must also hold.

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

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

  1. $(AB)^{-1} = A^{-1}B^{-1}$
  2. $AB = BA$
  3. $(AB)^T = A^T. B^T$
  4. $(AB)^{-1} = B^{-1} A^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For any two invertible matrices A and B of the same order, the inverse of their product is the product of their inverses in reverse order, which is given by the formula (AB)^-1 = B^-1 A^-1.

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

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

  1. 3/4, -1/2

  2. 4/3, -2

  3. 12, -8

  4. -12, 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

  1. $\displaystyle \{-1,1 \}$
  2. $\displaystyle R$
  3. $\displaystyle R\backslash \{-1,1\}$
  4. $\displaystyle no\space real\space values$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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

  1. rectangular matrix

  2. singular matrix

  3. square matrix

  4. nonsingular matrix
Reveal answer Fill a bubble to check yourself
C,D Correct answer
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