Tag: matrices and determinants

Quizzes (6) Questions (70)

Questions Related to matrices and determinants

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

Let 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 $t _r(A)+t _r\left(\dfrac{ABC}{2}\right)+t _r\left(\dfrac{A(BC)^2}{4}\right)+t _r\left(\dfrac{A(BC)^3}{8}\right)+.....+\infty =?$

  1. $6$
  2. $9$
  3. $12$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is identical to the previous series summation problems involving matrix traces.

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

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 $\displaystyle tr(A)+tr\left(\frac{ABC}{2}  \right)+tr\left(\frac{A{(BC)}^{2}}{4}  \right)+tr\left(\frac{A{(BC)}^{2}}{8}  \right)+...+\infty=  $

  1. $6$
  2. $9$
  3. $12$
  4. $15$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This has a typo in the series - the 4th term should be A(BC)³/8, but it repeats A(BC)²/8. Assuming the intended pattern (powers of BC increase): BC = I, tr(A) = 3. Series = 3[1 + 1/2 + 1/4 + 1/8 + ...] = 3 x 2 = 6. Answer A is correct despite the typographical error.

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

Let $\displaystyle A=\begin{bmatrix}-1\2\3\end{bmatrix}$ and $\displaystyle B=\begin{bmatrix} -2 & -1 & -4 \end{bmatrix}$

If trace of matrix $AB$ is $-12$, then the value of $k$ 

  1. $7$
  2. $1$
  3. $2$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle A=\begin{bmatrix}-k\2\3\end{bmatrix}$ 


$\displaystyle B=\begin{bmatrix}-2-1-4\end{bmatrix}$

$ \therefore AB=\begin{bmatrix}2k &k  &4k \-4  &-2  &-8 \-6  &-3  &-12 \end{bmatrix}$ 

The trace (often abbreviated to tr) of a square matrix A is defined to be the sum of elements on the main diagonal (from the upper left to the lower right) of A.

$\displaystyle \therefore Tr(AB)=$Summation of diagonal elements$=12$

                    $\Rightarrow +2k-2-12=-12$...............(According to question)

                    $\Rightarrow 2k=14-12=2$

                    $\Rightarrow k=1$

                    $\therefore k=1$ 

                    Hence, option B.

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

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 $\displaystyle \lambda =4$,then $\displaystyle \frac {1}{6}\left \{tr(AB)+tr(BA)  \right \} $ is equal to
  1. $42$
  2. $37$
  3. $35$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$A=\begin{bmatrix} 1 & 3 & 6 \ 2 & 4 & 6 \ 3 & 5 & 8 \end{bmatrix},B=\begin{bmatrix} 3 & 2 & 4 \ 3 & 2 & 5 \ 2 & 1 & 4 \end{bmatrix}$

$ AB=\begin{bmatrix} 1 & 3 & 6 \ 2 & 4 & 6 \ 3 & 5 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 & 4 \ 3 & 2 & 5 \ 2 & 1 & 4 \end{bmatrix}=\begin{bmatrix} 24 & 14 & 43 \ 30 & 18 & 52 \ 40 & 24 & 69 \end{bmatrix}$


$\Rightarrow  tr(AB)=24+18+69=111$

$tr(BA)=tr(AB)=111$

 $\displaystyle \frac {1}{6}\left {tr(AB)+tr(BA)  \right }=\frac{1}{6}222 =37$

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

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 $A$ is a singular matrix, then $tr(A + B)$ is equal to

  1. $24$
  2. $11$
  3. $22$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

$A+B=\begin{bmatrix} 4 & 5 & \lambda +6 \ 5 & 6 & 11 \ 5 & 6 & 12 \end{bmatrix}$

$tr(A+B)=4+6+12=22$

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

The trace of the matrix $A = \begin{bmatrix}1 & -5 & 7\ 0 & 7 & 9\ 11 & 8 & 9\end{bmatrix}$ is

  1. $17$
  2. $25$
  3. $3$
  4. $12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,  $A = \begin{bmatrix}1 & -5 & 7\ 0 & 7 & 9\ 11 & 8 & 9\end{bmatrix}$ .


Now trace of $A=$ sum of the diagonal elements of $A$.

So trace of $A=1+7+9=17$.

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $A = [a _{ij}]$ is a scalar matrix of order $n\times n$ such that $a _{ii} = k$ for all $i$, then trace of $A$ is equal to

  1. $nk$
  2. $n + k$
  3. $\dfrac {n}{k}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $A = [a _{ij}]$ is a scalar matrix of order $n\times n$ such that $a _{ii} = k$ for all $i$.


The trace of a square matrix is defined to be the sum of the diagonal elements.

Now, trace $(A)=\displaystyle\sum\limits _{i=1}^n a _{ii}=$$\displaystyle\sum\limits _{i=1}^n k=nk.$

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $A$ is a $3\times 3$ skew-symmetric matrix, then trace of $A$ is equal to

  1. $1$
  2. $|A|$
  3. $-1$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As $A$ is a skew symmetric matrix
                   $A' = -A$
$\Rightarrow  a _{ii}=0 : \forall : i \Rightarrow trace : (A) =0$
Also             $|A|=|A'|=|-A|=(-1)^3|A|$
$\Rightarrow  2|A|=0\Rightarrow |A|=0$
Hence, option B.

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $A$ is $2\times 2$ matrix such that $A^2 = 0$, then $tr :(A)$ is

  1. 1

  2. 0

  3. -1

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $A = 0,tr(A)=0$.

Suppose $A\neq 0$ and $A =\begin{bmatrix}a

&b \c &d \end{bmatrix}$, then $|A|=0$ and


$A^2-(a+d)A+ad-bc=0$

$\Rightarrow a+d=0$

$\therefore tr(A)=0$

Hence, option B.

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $A =\begin{bmatrix} 1&9  & -7\ i & \omega^n & 8\ 1 & 6 &\omega^{2n} \end{bmatrix}$ where $i= \sqrt{-1} $ and $\omega$ is complex cube root of unity, then tr(A) will be 

  1. $1, \,if \,n = 3k,\, k \in\, N$
  2. $3, \,if \,n = 3k,\, k \in\, N$
  3. $0,\, if \,n\neq \,3k,\, k \epsilon \in N$
  4. $-1,\, if \,n\neq \,3k, \,k \epsilon \in N$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$tr(A)=1+\omega^n+\omega^{2n}$
if $n=3k$ i.e Mutiple of $3$.
$\Rightarrow tr(A)=1+\omega^{3k}+\omega^{6k}=1+1+1=3$
if $n\neq 3k$ i.e not a multiple of $3$.
then $n=3k+1$ or $n=3k+2$
$\Rightarrow tr(A)=1+\omega^{3k+1}+\omega^{2(3k+1)}=1+\omega+\omega^2=0$
Hence, options B and C.