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

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

  1. $3$
  2. $4$
  3. $2$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, the matrix $A$ is singular.

$\Rightarrow |A|=0$

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

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

$\Rightarrow 20\lambda -60=0$

$\Rightarrow \lambda=3$

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

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

  1. a symmetric matrix

  2. a skew-symmetric matrix

  3. diagoinal matrix

  4. does not exists
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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:

  1. $8$
  2. $7$
  3. $16$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\alpha+\beta=1+15=16$.

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


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

  1. $3$
  2. $5$
  3. $7$
  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 , $|2A+B|=0$


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

$\Rightarrow 9-6\lambda=0$

$\Rightarrow 2\lambda=3$

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

Let $A$ be a square matrix all of whose entries are integers, then which of the following is true?

  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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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.