Tag: non singular matrix

Questions Related to non singular matrix

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: A

If the matrix $A = \begin{bmatrix}8 & -6 & 2 \ -6 & 7 & -4 \ 2 & -4 & \lambda\end{bmatrix}$ is singular, then $\lambda = $

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

  1. $3$

  2. $4$

  3. $2$

  4. $5$

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Correct Option: A
Explanation:

Given, the matrix $A$ is singular.

$\Rightarrow |A|=0$

$\begin{vmatrix} 8 & -6 & 2 \ -6 & 7 & -4 \ 2 & -4 & \lambda  \end{vmatrix}=0$

$8(7\lambda -16)+6(-6\lambda +8)+2(10)=0$

$\Rightarrow 20\lambda -60=0$

$\Rightarrow \lambda=3$

The inverse of a skew-symmetric matrix of odd order is

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

  1. a symmetric matrix

  2. a skew-symmetric matrix

  3. diagoinal matrix

  4. does not exists

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Correct Option: D
Explanation:

Let A be a skew-symmetric matrix of order n.

By definition ${ A }^{ ' }=-A$
$\Rightarrow \left| { A }^{ ' } \right| =\left| -A \right| \Rightarrow \left| A \right| ={ \left( -1 \right)  }^{ n }\left| A \right| $
$\Rightarrow \left| { A } \right| =\left| -A \right| $ as n is odd
$\Rightarrow 2\left| { A } \right| =0\Rightarrow \left| { A } \right| =0$
Thus ${ A }^{ -1 }$ does not exist

State whether the following statement is true or false.

The matrix $\begin{bmatrix}1&3&0\ 4&0&-2\ 2&6&0\end{bmatrix}$ is singular matrix.

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

  1. True

  2. False

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Correct Option: A
Explanation:

Let $A=$ $\begin{bmatrix}1&3&0\ 4&0&-2\ 2&6&0\end{bmatrix}$


Now, 

$|A|$$=\begin{vmatrix}1&3&0\ 4&0&-2\ 2&6&0\end{vmatrix}$

$=-2(1\times6-2\times3)$      [ Expanding about the second row]

$=0$.

So the given matrix $A$ is singular.

Suppose $  A  $ is any $  3 \times 3  $ non-singular matrix and $  (A-3 I)(A-5 I)=0,  $ where $  {I}={I} _{3}  $ and $  {O}={O} _{3} .  $ If $  \alpha {A}+\beta {A}^{-1}=4 {I},  $ then $  \alpha+\beta  $ is equal to :

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

  1. 8

  2. 7

  3. 13

  4. 12

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Correct Option: D

Suppose $A$ is any $3\times3$ non-singular matrix and $(A-3I)(A-5I)=O$,where $I=I _{3}$ and $O=O _{3}$.If $\alpha A+\beta A^{-1}=8I$ ,then $\alpha+\beta$ is equal to:

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

  1. $8$

  2. $7$

  3. $16$

  4. $12$

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Correct Option: C
Explanation:

Given,  $(A-3I)(A-5I)=O$ for a $3\times 3$ non-singular matrix.

or, $A^2-8A+15I=O$
or, $A-8I+15A^{-1}=O$ [ Since $A$ is non-singular ($A^{-1}$ exists) then multiplying both sides with $A^{-1}$]
or, $A+15A^{-1}=8I$.
Comparing this with the given equation we get, $\alpha=1, \beta=15$.
So $\alpha+\beta=1+15=16$.


Let $A$ and $B$ are two matrices of same order $\displaystyle 3\times 3$ given by $\displaystyle A=\begin{bmatrix}1 &3  &\lambda+2 \2  &4  &6 \3  &5  &8 \end{bmatrix}$ $\displaystyle B= \begin{bmatrix}3 &2  &4 \3  &2  &5 \2
 &1  &4 \end{bmatrix}$If $2A + B$ is singular, then $\displaystyle 2\lambda$ equals

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

  1. $3$

  2. $5$

  3. $7$

  4. $9$

Show answer
Correct Option: A
Explanation:

Here, $2A+B=\begin{bmatrix} 2 & 6 & 2\lambda +4 \ 4 & 8 & 12 \ 6 & 10 & 16 \end{bmatrix}+\begin{bmatrix} 3 & 2 & 4 \ 3 & 2 & 5 \ 2 & 1 & 4 \end{bmatrix}$

$\Rightarrow 2A+B=\begin{bmatrix} 5 & 8 & 2\lambda +8 \ 7 & 10 & 17 \ 8 & 11 & 20 \end{bmatrix}$

Given , $|2A+B|=0$


$\begin{vmatrix} 5 & 8 & 2\lambda +8 \ 7 & 10 & 17 \ 8 & 11 & 20 \end{vmatrix}=0$

$\Rightarrow 9-6\lambda=0$

$\Rightarrow 2\lambda=3$

Let $A$ be a square matrix all of whose entries are integers, 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. If $\displaystyle \left | A\right | \neq \pm 1 $, then $\displaystyle A^{-1} $ exist & all its entries are non-integer

  2. If $\displaystyle \left | A\right | = \pm 1 $, then $\displaystyle A^{-1} $ exist & all its entries are integer

  3. If $\displaystyle \left | A\right | = \pm 1 $,then $\displaystyle A^{-1} $ need not exist

  4. If $\displaystyle \left | A\right | = \pm 1 $, then $\displaystyle A^{-1} $ exist but all its entries are not necessarily integers.

Show answer
Correct Option: B
Explanation:

Given A is a square matrix with all entries as integer
(a) Now $\displaystyle \left|A\right|\neq 1,-1$ mean it may be zero also $\displaystyle \left|A\right| =0,$then $\displaystyle A^{-1}$ does'not exist $\displaystyle \therefore $ choice (a) is false.

(b) If $\displaystyle \left|A\right| =1,-1$ then $\displaystyle A^{-1}$ certainly exist but A is a square matrix with all integral entries so all cofactors are integers .So adj. A matrix has all integral entries .
$\displaystyle A^{-1}=\frac{1}{\left|A\right| }(adj A)=\pm(adj A) $
$\therefore $ choice (b) is correct.

(c) $\displaystyle \left|A\right| =1,-1 $ $\displaystyle \therefore $ $\displaystyle A^{-1}$ must exist but given $\displaystyle A^{-1}$ does not exist which is false result $\displaystyle \therefore $  choice (c) is incorrect

(d) $\displaystyle \left|A\right| =1,-1$ it is true that $\displaystyle A^{-1}$ exist but we have to follow that A has all integral entries but choice (d) says that it is not necessarily that entries are integer which mean adj. A, may or may not be with integral entries. Hence choice (d) is false.

Let $A$ be a square matrix of order $n \times  n$. A constant $\lambda $ is said to be characteristic root of $A$ if there exists a $n \times  1$ matrix $X$ such that  $AX=\lambda X$

If $0$ is a characteristic root of $A$, then :

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

  1. $A$ is non-singular

  2. $A$ is singular

  3. $A = 0$

  4. $A = I _n$

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Correct Option: B
Explanation:

Since $X\neq 0$ is such that $(A-\lambda I)X=0,:|A-\lambda I|=0\Leftrightarrow A-\lambda I$ is singular. If $A-\lambda I$ is non-singular the then equation $(A-\lambda I)X=0\Rightarrow X=0$
If $\lambda= 0$, we get $|A|=0\Rightarrow A$ is singular.

If $A = \begin{bmatrix}1 & k & 3\ 3 & k & -2 \ 2 & 3 & -4\end{bmatrix}$ is singular then $k = ?$

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

  1. $\dfrac {16}{3}$

  2. $\dfrac {34}{5}$

  3. $\dfrac {33}{2}$

  4. None of these

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Correct Option: C