Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

$A=\begin{bmatrix} 2&2&1\4&5&6\6&8&9\end{bmatrix}$, $B=\begin{bmatrix} 2&2&1\0&1&4\0&2&6\end{bmatrix}$

To convert matrix A into matrix B, the order of row operations are 

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $R _1\rightarrow R _2-2R _1$, $R _3 \rightarrow R _3-R _1$

  2. $R _2\rightarrow R _2-2R _1$, $R _3 \rightarrow R _3-3R _1$

  3. $R _1\rightarrow R _1-2R _2$, $R _3 \rightarrow R _1-R _3$

  4. $R _2\rightarrow R _3-2R _3$, $R _3 \rightarrow R _1-R _1$

Show answer
Correct Option: B
Explanation:
$A=\begin{bmatrix} 2 & 2 & 1 \\ 4 & 5 & 6 \\ 6 & 8 & 9 \end{bmatrix}\quad B=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 2 & 6 \end{bmatrix}$
Clearly in order to convert matrix A to B. Both ${ R } _{ 2 }$ and ${ R } _{ 3 }$ are changed w.r.t. ${ R } _{ 1 }$
${ R } _{ 2 }^{ 1 }\leftrightarrow X{ R } _{ 2 }-Y{ R } _{ 1 }$
For ${ R } _{ 2 }=4$ and ${ R } _{ 1 }=2$ then ${ R } _{ 2 }^{ 1 }=0$
$\Rightarrow 0=X4-2$Y$
$\Rightarrow 2X=Y$
For ${ R }_{ 2 }=5$ and ${ R }_{ 1 }=1$ then ${ R }_{ 2 }^{ 1 }=1$
$1=X(5)-Y(2)$
$\Rightarrow X=1\quad Y=2$
Relation is ${ R }_{ 2 }\leftrightarrow { R }_{ 2 }-2{ R }_{ 21}$
Parallely ${ R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }$

Multiply the fourth row by $3$.
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $0, 0, 18, 3$

  2. $0, 3, 0, 6$

  3. $0, 0, 24, 4$

  4. $9, 12, 6, 33$

  5. $0, 0, 12, 2$

Show answer
Correct Option: A
Explanation:

Fourth row = $0,0,6,1$
multiplying by $3$
$(0\times 3 , 0\times 3 , 6\times 3 , 1\times 3)$
$(0,0,18,3)$

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

The new matrix obtained after  adding $2^{nd} \  row \ to\  3\ times\  3^{rd} \ row $ is

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\begin{bmatrix} 1&2&3\4&5&6\25&29&33\end{bmatrix}$

  2. $\begin{bmatrix} 1&2&3\25&-29&-33\4&5&6\end{bmatrix}$

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

  4. $\begin{bmatrix} 1&2&3\-25&-29&-33\4&5&6\end{bmatrix}$

Show answer
Correct Option: A
Explanation:
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
$\Rightarrow { R } _{ 3 }\rightarrow 3{ R } _{ 3 }+{ R } _{ 2 }$ for new matrix
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3(7)+4 & 3(8)+5 & 3(9)+6 \end{bmatrix}$
New matrix $=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 25 & 29 & 33 \end{bmatrix}$
Option A is correct

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

B is matrix obtained by subtracting $4\ times 1^{st}\ row from \ 2^{nd} \ row$ of A.
C is matrix obtained by subtracting $7 \ times \ 1^{st}\ row\ from\ 3^{rd} row$, then $C$ is 

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\begin{bmatrix} 1&2&3\0&3&6\0&6&12\end{bmatrix}$

  2. $\begin{bmatrix} 1&2&3\7&0&0\4&5&6\end{bmatrix}$

  3. $\begin{bmatrix} 1&2&3\0&1&2\3&4&5\end{bmatrix}$

  4. $\begin{bmatrix} 1&2&3\0&-3&-6\0&-6&-12\end{bmatrix}$

Show answer
Correct Option: D
Explanation:
Given $A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
B is obtained by substracting $4{ R } _{ 1 }$ from ${ R } _{ 2 }$
$\Rightarrow { R } _{ 2 }\leftrightarrow { R } _{ 2 }-4{ R } _{ 1 }$$\rightarrow B\sim \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 7 & 8 & 9 \end{bmatrix}$
C is obtained by subtracting $7{ R } _{ 1 }$ from ${ R } _{ 3 }$
$\Rightarrow { R } _{ 3 }\leftrightarrow { R } _{ 3 }-7{ R } _{ 1 }$
$C\sim \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{bmatrix}$
$\therefore $option D is correct

In echelon form, which of the following is incorrect?

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. Every row of $A$ which has all its entries $0$ occurs below every row which has a non-zero entry

  2. The first non-zero entry in each non-zero row is $1$

  3. The number of zeros before the first non-zero element is a row is less than than the number of such zeros in the next row

  4. Two rows can have same number of zeros before the first non-zero entry

Show answer
Correct Option: D
Explanation:

Echelon form of matrix has the following characteristics:

1. The first non-zero entry of every row is 1.

2. The first non-zero entry in every row is one position right to the previous row.

3. The row with all zero elements will be below the rows having a non-zero element.

Therefore, according to the question statement in option D is incorrect.

Hence, the answer is option D.

The system $\begin{pmatrix} 1 & -1 & 2 \ 3 & 5 & -3 \ 2 & 6 & a \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} 3 \ b \ 2 \end{pmatrix}$ has no solution, if

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $a=-5,b\ne 5$

  2. $a=-5, b=5$

  3. $a\ne -5, b=5$

  4. $a\ne -5,b\ne 5$

Show answer
Correct Option: A
Explanation:

Given 
$\begin{pmatrix} 1 & -1 & 2 \ 3 & 5 & -3 \ 2 & 6 & a \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} 3 \ b \ 2 \end{pmatrix}$
The augment matrix of given system is
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 3 & 5 & -3 & | & b \ 2 & 6 & a & | & 2 \end{bmatrix}$
${ R } _{ 2 }\rightarrow { R } _{ 2 }-3{ R } _{ 1 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 2 & 6 & a & | & 2 \end{bmatrix}$
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 1 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 0 & 8 & a-4 & | & -4 \end{bmatrix}$
${ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 2 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 0 & 0 & a+5 & | & 5-b \end{bmatrix}$
We know that
For no solution
Rank A < Rank $\left[ A|B \right] $
$\therefore$ $a+5=0$ and $5-b\ne 0$
$a=-5, b\ne 5$

Let $A$ be a matrix of order $3\times 3$ such that $\left| \vec { A }  \right| =1$. Let $B=2{ A }^{ -1 }$ and $C=\dfrac { adj.A }{ 2 }$. Then the value of  $\left| { AB }^{ 2 }{ C }^{ 3 } \right|$, is ( where $\left| A \right|$ represent det. $A$)

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $1$

  2. $\dfrac { 1}{ 2 }$

  3. $8$

  4. $64$

Show answer
Correct Option: B
Explanation:
$adj(A)=|\vec{A}|^{n-1}$

$n=3,\therefore adj(A)=|\vec{A}|^{3-1}=|\vec{A}|^2$

$|\vec{A}|=1,\therefore adj(A)=1$

$adj(A)=|\vec{A}|^{n-1}$

$n=3,\therefore adj(A)=|\vec{A}|^{3-1}=|\vec{A}|^2$

$|\vec{A}|=1,\therefore adj(A)=1$

$|\vec{A}|=1$

$B=2|\vec{A}|=1\Rightarrow B=2\times 1=2$

$C=\dfrac{adjA}{2}=\dfrac{1}{2}$

$|AB^2C^3|=\left | 1\times 2^2\times \left ( \dfrac{1}{2} \right )^3 \right |$

$=\dfrac{1}{2}$

 $\begin{bmatrix}
              \cos\theta & -\sin\theta \[0.3em]
              \sin\theta & \cos\theta
              \end{bmatrix} = \begin{bmatrix}
              1 & -\tan\theta/2 \[0.3em]
             \tan\theta/2 & 1
              \end{bmatrix} \begin{bmatrix}
              1  & \tan\theta/2 \[0.3em]
              -\tan\theta/2 & 1
              \end{bmatrix}$

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. True

  2. False

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Correct Option: A

If $A = \begin{bmatrix} a & b\ c  & d \end{bmatrix} $ satisfies the equation $x^2 - (a+d)x+k=0$ then

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $k = bc$

  2. $ k =ad$

  3. $k = a^2+b^2+c^2+d^2$

  4. $k=ad-bc$

Show answer
Correct Option: D
Explanation:

$A=\begin{bmatrix}a&b\c&d\end{bmatrix}$


$\implies \begin{vmatrix}\lambda-a&b\c&\lambda-d\end{vmatrix}=0$

$(\lambda-a)(\lambda-d)-bc=0$

$\implies {\lambda}^{2}-(a+d)\lambda+(a{d}-b{c})=0$

Comparing with $x^2 - (a+d)x+k=0$

$\implies k=a{d}-b{c}$

The number of $2\times 2$ matrices $A=\left[ \begin{matrix} a & b \ c & d \end{matrix} \right] $ for which ${ \left[ \begin{matrix} a & b \ c & d \end{matrix} \right]  }^{ -1 }$ $=\left[ \begin{matrix} \frac { 1 }{ a }  & \frac { 1 }{ b }  \ \frac { 1 }{ c }  & \frac { 1 }{ d }  \end{matrix} \right] $, $(a,b,c,d\ \epsilon \ R)$ is

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $0$

  2. $1$

  3. $2$

  4. $Infinite$

Show answer
Correct Option: A
Explanation:

If $A=\begin{bmatrix} a & b \ c & d \end{bmatrix}\quad { \begin{bmatrix} a & b \ c & d \end{bmatrix} }^{ -1 }=\begin{bmatrix} 1/a & 1/b \ 1/c & 1/d \end{bmatrix}$                          Not possible for any values of $a, b, c$

$A=\begin{bmatrix} a & b \ c & d \end{bmatrix}\quad =Adj{ \begin{bmatrix} d & -c \ -b & a \end{bmatrix} }\ \Rightarrow { A }^{ -1 }=\begin{bmatrix} d & -c \ -b & a \end{bmatrix}$