Tag: algebra

Questions Related to algebra

Let $A+2B=\begin{bmatrix} 1 & 2 & 0 \ 6 & -3 & 3 \ -5 & 3 & 1 \end{bmatrix}$ and $2A-B=\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$, then $tr(A)-tr(B)$ has the value equal to

  1. 0

  2. 1

  3. 2

  4. none of these


Correct Option: C
Explanation:

$A+2B=\begin{bmatrix} 1 & 2 & 0 \ 6 & -3 & 3 \ -5 & 3 & 1 \end{bmatrix}$         .....(i)

$2A-B=\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$         .....(ii)

$\Rightarrow 4A-2B=\begin{bmatrix} 4 & -2 & 10 \ 4 & -2 & 12 \ 0 & 2 & 4 \end{bmatrix}$      ....(iii)

Adding (i) and (iii), we get
$5A=\begin{bmatrix} 5 & 0 & 10 \ 10 & -5 & 15 \ -5 & 5 & 5 \end{bmatrix}$

$\Rightarrow A=\begin{bmatrix} 1 & 0 & 2 \ 2 & -1 & 3 \ -1 & 1 & 1 \end{bmatrix}$
So, $tr(A)=1$

Now, by eq(ii),
$B=\begin{bmatrix} 2 & 0 & 4 \ 4 & -2 & 6 \ -2 & 2 & 2 \end{bmatrix}-\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$

$\Rightarrow B=\begin{bmatrix} 0 & 1 & -1 \ 2 & -1 & 0 \ -2 & 1 & 0 \end{bmatrix}$
So, $tr(B)=-1$
Now, $tr(A)-tr(B)=1+1=2$

If $A = \begin{bmatrix}2 & 3 & 4\ 5 & -3 & 8\ 9 & 2 & 16\end{bmatrix}$, then trace of A is,

  1. 17

  2. 25

  3. 8

  4. 15


Correct Option: D
Explanation:

Given $A = \begin{bmatrix}2 & 3 & 4\ 5 & -3 & 8\ 9 & 2 & 16\end{bmatrix}$

$tr(A)=2-3+16=15$

If $A=[a _{ij}] _{n\times n}$ and $a _{ij}=i(i+j)$ then trace of $A=$

  1. $\dfrac{n(n+1)(2n+1)}{6}$

  2. $\dfrac{n(n+1)(2n+1)}{3}$

  3. $\dfrac{n(n+1)}{2}$

  4. $\dfrac{n^{2}(n+1)^{2}}{4}$


Correct Option: A

Let $A$ be the $2\times2$ matrices given by $A=\left[a _{ij}\right]$ where $a _{ij} = \left{0,1,2,3,4\right}$ such that $a _{11} + a _{12} + a _{21} + a _{22} = 4$
Find the number of matrices $A$ such that the trace of $A$ is equal to 4

  1. 3

  2. 4

  3. 5

  4. 6


Correct Option: C
Explanation:

Given $tr(A)=4$
$\Rightarrow a _{11}+a _{22}=4$

$a _{ ij }={ { 0,1,2,3,4}  }$

So, diagonal entries of A can be 0 and 4 , 4 and 0, 1 and 3, 3 and 1, 2 and 2,
Hence, 5 matrices are possible

If $A=[a _{ij}]$ is a scalar matrix then the trace of $A$ is

  1. $\displaystyle \sum _{i}a _{ij}$

  2. $\displaystyle \sum _{f}a _{ij}$

  3. $\displaystyle \sum _{i}\sum _{i}a _{ij}$

  4. $\displaystyle \sum _{i}a _{ij}$


Correct Option: A

If $A=\begin{bmatrix} 2 & 1 \ 4 & 1 \end{bmatrix}; B=\begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix}$ and $C=\begin{bmatrix} 3 & -4 \ -2 & 3 \end{bmatrix}$ then $tr(A)+tr\left( \dfrac { ABC }{ 2 }  \right) +tr\left( \dfrac { A{ \left( BC \right)  }^{ 2 } }{ 4 }  \right) +tr\left( \dfrac { A{ \left( BC \right)  }^{ 3 } }{ 8 }  \right) +......\infty $ =

  1. $6$

  2. $9$

  3. $12$

  4. $15$


Correct Option: A

Consider three matrices A= $ \begin{bmatrix} 2 & 1 \ 4 & 1 \end{bmatrix} $, $ B = \begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix} $ and $ C = \begin{bmatrix} 3 & -4 \ -2 & 3 \end{bmatrix} $ Then the value of the sum 
$ tr(A)+tr \cfrac {(ABC) } {2} +tr \cfrac {A( {BC})^2} {4}+ \cfrac {A( {BC})^3} {2}  +...+ \infty               $is

  1. 6

  2. 9

  3. 12

  4. None of these


Correct Option: C

 $P=\left[ \begin{matrix} { 5a }^{ 2 }+2bc & 6 & 8 \ 13 & { 8b }^{ 2 }-10ac & -9 \ -7 & 5 & { 25c }^{ 2 } \end{matrix} \right]$ and $Q=\left[ \begin{matrix} { a }^{ 2 }+6bc & 3 & 5 \ 12 & { -b }^{ 2 } & 6 \ 1 & 4 & { 17bc }^{ 2 } \end{matrix} \right] a,b$ & $c \epsilon N$, if trace $\left(P\right)=trac\left(Q\right)$, and $a,b$ & $C$ are sides of $\Delta ABC$ with $BC=a,CA=b$ & $AB=C$ then $\cos A$ is:

  1. $\dfrac{-79}{120}$

  2. $\dfrac{-89}{120}$

  3. $\dfrac{-33}{40}$

  4. $\dfrac{-31}{40}$


Correct Option: A

If $\left( \begin{array} { l l } { 3 } & { 2 } \ { 7 } & { 5 } \end{array} \right) A \left( \begin{array} { c c } { - 1 } & { 1 } \ { - 2 } & { 1 } \end{array} \right) = \left( \begin{array} { c c } { 2 } & { - 1 } \ { 0 } & { 4 } \end{array} \right)$  then trace of  $A$  is equal to

  1. $-25$

  2. $-21$

  3. $-15$

  4. $-11$


Correct Option: A

Let  $A=\left[ \begin{matrix} p & q \ q & p \end{matrix} \right] $ such that det(A)=r where p,q,r all prime numbers, then trace of A is equal to 

  1. 6

  2. 5

  3. 2

  4. 3


Correct Option: A