Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

Which of the following is the new row that results when you add rows $1$ and $3$?
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$

  1. $6, 8, 4, 22$

  2. $3, 5, 2, 13$

  3. $3, 4, 2, 11$

  4. $3, 4, 8, 12$

  5. $4, 5, 3, 12$


Correct Option: B
Explanation:
Given Matrix$=\begin{bmatrix} 3 & 4 & 2 & 11 \\ 9 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 6 & 1 \end{bmatrix}$
Row $1=\begin{bmatrix} 3 & 4 & 2 & 11 \end{bmatrix}$
Row $3=\begin{bmatrix} 0 & 1 & 0 & 2 \end{bmatrix}$
Sum$=\begin{bmatrix} 3+0 & 4+1 & 2+0 & 11+2 \end{bmatrix}$
$\begin{bmatrix} 3 & 5 & 2 & 13 \end{bmatrix}$
$\therefore $Option $2$ is correct

Use a transformation matrix to find the image of $D(-7,6)$ after a rotation of $180^0$ counterclockwise around the origin.

  1. $(7,6)$

  2. $(-7,-6)$

  3. $(7,-6)$

  4. $(-7,6)$


Correct Option: C
Explanation:

The transformation matrix for rotation  is $\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}$

For $\theta=180^{0}$ , the transformation matrix will be $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$
So the image of point $(-7,6)$ is $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}\begin{bmatrix} -7 \ 6 \end{bmatrix}=\begin{bmatrix} 7 \ -6 \end{bmatrix}$
Therefore the correct option is $C$

$A=\begin{bmatrix} 1&-2&3\7&-8&9\4&-5&6\end{bmatrix}$ the new matrix formed by adding $\ 2^{nd}\ row \ to \ 1^{st} $ row  will be

  1. $\begin{bmatrix}8&-10&12\7&-8&9\4&-5&6\end{bmatrix}$

  2. $\begin{bmatrix} 6&6&6\7&8&9\4&5&6\end{bmatrix}$

  3. $\begin{bmatrix} 1&2&3\7&8&9\11&-13&14\end{bmatrix}$

  4. $\begin{bmatrix} 1&-2&3\7&8&-29\4&-2&6\end{bmatrix}$


Correct Option: A
Explanation:
Given $A=\begin{bmatrix} 1 & -2 & 3 \\ 7 & -8 & 9 \\ 4 & -5 & 6 \end{bmatrix}$
Now$\Rightarrow { R } _{ 1 }\rightarrow { R } _{ 1 }+{ R } _{ 2 }$
Resultant matrix$=\begin{bmatrix} 1+7 & -2-8 & 3+9 \\ 7 & -8 & 9 \\ 4 & -5 & 6 \end{bmatrix}$
$=\begin{bmatrix} 8 & -12 & 12 \\ 7 & -8 & 9 \\ 4 & -5 & 6 \end{bmatrix}$
Option A is correct

 A=$\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$
The new matrix formed  after interchanging $2^{nd}$ and $3^{rd}$rows  will be 

  1. $-\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$

  2. $\begin{bmatrix} 4&5&6\1&2&3\7&8&9\end{bmatrix}$

  3. $-\begin{bmatrix} 1&2&3\7&8&9\4&5&6\end{bmatrix}$

  4. $\begin{bmatrix} 1&2&3\7&8&9\4&5&6\end{bmatrix}$


Correct Option: D
Explanation:
Given
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
Now second and third rows are interchanged $\Rightarrow $all elements are interchanged
Resultant matrix$=\begin{bmatrix} 1 & 2 & 3 \\ 7 & 8 & 9 \\ 4 & 5 & 6 \end{bmatrix}$
Option D is correct

For a matrix $A \begin{pmatrix} 1& 0 & 0\ 2 & 1 & 0\ 3 & 2 & 1\end{pmatrix}$, if $U _{1}, U _{2}$ and $U _{3}$ are $3\times 1$ column matrices satisfying $AU _{1} = \begin{pmatrix}1\ 0 \ 0
\end{pmatrix}, AU _{2} \begin{pmatrix}2\3 \ 0
\end{pmatrix}, AU _{3} = \begin{pmatrix}2\ 3\ 1
\end{pmatrix}$ and $U$ is $3\times 3$ matrix whose columns are $U _{1}, U _{2}$ and $U _{3}$
Then sum of the elements of $U^{-1}$ is

  1. $6$

  2. $0 (zero)$

  3. $1$

  4. $2/3$


Correct Option: B
Explanation:

Let $U _{i} = \begin{pmatrix}a _{i}\b _{i} \c _{i} \end{pmatrix} i = 1, 2, 3$
$AU _{1} = \begin{pmatrix}a _{1}\2a _{1} + b _{1} \ 3a _{1} + 2b _{1} + c _{1}
\end{pmatrix} = \begin{pmatrix}1\0 \ 0
\end{pmatrix}$, So $a _{1} = 1, b _{1} = -2, c _{1} = 1$
$AU _{2} = \begin{pmatrix}a _{2}\2a _{2} + b^{2} \ 3a _{2} + 2b _{2} + c _{2}
\end{pmatrix} = \begin{pmatrix}2\ 3\ 0
\end{pmatrix}$,
So, $a _{2} = 2, b _{2} = -1, c _{2} = -4$. Similarly, $a _{3} = 2, b _{3} = -1, c _{3} = -3$
So, $U = \begin{pmatrix} 1& 2 & 2\ -2 & -1 & -1\ 1 & -4 & -3\end{pmatrix}$. So, sum of elements of $U^{-1}$ is zero.

The inverse of a diagonal matrix is a :

  1. Symmetric matrix

  2. Skew-symmetric matrix

  3. Diagonal matrix

  4. None of the above


Correct Option: A,C
Explanation:

A diagonal matrix has elements only in it's diagonal.
So the inverse will also have all non zero elements in the diagonal.
So, it will be symmetric and will also be a diagonal matrix.
Hence, option A and C are correct

Inverse of $A  = \begin{bmatrix} 1& 3\ 2 & -2\end{bmatrix} $ is equal to?

  1. $- \dfrac{1}{8} \begin{bmatrix}3 & 1\ -2 & 2\end{bmatrix}$

  2. $- \dfrac{1}{8} \begin{bmatrix}-2 & -3\ -2 & 1\end{bmatrix}$

  3. $ \dfrac{1}{8} \begin{bmatrix}-1 & -3\ -2 & 2\end{bmatrix}$

  4. None of these


Correct Option: B
Explanation:
If $A  = \begin{bmatrix} 1& 3\\ 2 & -2\end{bmatrix} $
$ a _{11}= -2 $
$ a _{12}= -2 $
$ a _{21}= -3 $
$ a _{22}=  1 $

$ A^{-1}=\dfrac{ \left ( Cofactors of A \right )^{T}}{\left |A  \right |}$
$ \left ( Cofactors of A \right )^{T}=\begin{bmatrix} -2& -3\\ -2 & 1\end{bmatrix} $
${\left |A  \right |}= -2-6 $
${\left |A  \right |}=-8 $

$ A^{-1}=-\dfrac{1}{8}\times\begin{bmatrix} -2& -3\\ -2 & 1\end{bmatrix} $ 

Option will be B

If a matrix A is such that $3{A^3} + 2{A^2} + 5A + I = 0$ , then $A^{-1}$ is equal to

  1. $ - (3{A^2} + 2A + 5)$

  2. $3{A^2} + 2A + 5$

  3. $3{A^2} - 2A - 5$

  4. None of these


Correct Option: A
Explanation:

$3A^3+2A^2+5A+I=0$
$3A^3+2A^2+5A+AA^{-1}=0$
$A^{-1}=-3A^2-2A-5$

If $A$ is a non zero square matrix of order $n$ with $det\left( I+A \right) \neq 0$, and ${A}^{3}=0$, where $I,O$ are unit and null matrices of order $n\times n$ respectively, then ${ \left( I+A \right)  }^{ -1 }=$

  1. $I-A+{ A }^{ 2 }$

  2. $I+A+{ A }^{ 2 }$

  3. $I+{ A }^{ 2 }$

  4. $I+A$


Correct Option: A
Explanation:
$det(I+A)\neq 0$
$A^3=0$   where $0$ is null matrix, $I$ is the identity matrix
$A^3+I=I$ [adding $I$ on both sides]
$(A+I)(A^2-IA+I^2)=I$ [by the formula of $a^3+b^3$]
$(A+I)(A^2-A+I)=I$
$(I+A)(I+A)^{-1}=I$ [by the rule of inverse matrix]
hence $(I+A)^{-1}=(A^2-A+I)$
Ans: $I-A+A^2$

If $A=\begin{bmatrix} 3 & -2 \ 5 & 8 \end{bmatrix}$, then $A^{-1}=$

  1. $\frac{1}{30}\begin{bmatrix} 8 & 2 \ -5 & 3 \end{bmatrix}$

  2. $\frac{1}{34}\begin{bmatrix} 8 & 2 \ -5 & 3 \end{bmatrix}$

  3. $-\frac{1}{34}\begin{bmatrix} -8 & -2 \ -5 & 3 \end{bmatrix}$

  4. None of these


Correct Option: B
Explanation:

$A=\left[{\begin{array}{cc}3&2\5&8\end{array}}\right]$

$A^{-1}=\cfrac{1}{ad-bc}\left[{\begin{array}{cc}d&-b\-c&a\end{array}}\right]$  (determinant)
$=\cfrac{1}{3\times 8-(-2\times 5)}\left[{\begin{array}{cc}8&2\-5&3\end{array}}\right]$  
$=\cfrac{1}{34}\left[{\begin{array}{cc}8&2\-5&3\end{array}}\right]$