Mathematics

Linear Algebra

510 Questions

Linear algebra involves the study of matrices, vectors, and linear transformations. Common topics include finding the rank of a matrix, calculating eigenvalues and eigenvectors, and performing LU decomposition. These advanced mathematical concepts are frequently tested in engineering, statistics, and civil service examinations.

Matrix rank calculationLU decompositionEigenvalues and eigenvectorsMatrix invertibilityDeterminant properties

Linear Algebra Questions

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

For $\alpha, \beta, \gamma \in R$, let $A=\begin{bmatrix} { \alpha  }^{ 2 } & 6 & 8 \ 3 & { \beta  }^{ 2 } & 9 \ 4 & 5 & { \gamma  }^{ 2 } \end{bmatrix}$ and $B=\begin{bmatrix} 2\alpha  & 3 & 5 \ 2 & 2\beta  & 6 \ 1 & 4 & 2\gamma -3 \end{bmatrix}$. If ${ T } _{ r }(A)={ T } _{ r }(B)$ then the value of $\left( \cfrac { 1 }{ \alpha  } +\cfrac { 1 }{ \beta  } +\cfrac { 1 }{ \gamma  }  \right) $ is-

${ T } _{ r }(A)$ is a Trace(A) of a matrix

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

${ T } _{ r }(A)={ \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 }$


${ T } _{ r }(B)=2\alpha +2\beta +2\gamma -3$

$\Rightarrow { \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 }=2\alpha +2\beta +2\gamma -3$

$\Rightarrow

{ \left( \alpha -1 \right)  }^{ 2 }+{ \left( \beta -1 \right)  }^{ 2

}+{ \left( \gamma -1 \right)  }^{ 2 }=0\quad $
$\Rightarrow \quad

\alpha =\beta =\gamma =1\Rightarrow \quad \cfrac { 1 }{ \alpha  }

+\cfrac { 1 }{ \beta  } +\cfrac { 1 }{ \gamma  } =3$

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

i. Trace of the matrix is called sum of the elements in a principle diagonal of the square matrix. 
ii. The trace of the matrix $\begin{bmatrix}
8 & 7 &5\
5 &8 & 2\
7 & 2 & 8
\end{bmatrix}$ is 24 Which of the following statement is correct. 

  1. Only i

  2. Only ii

  3. Both i and ii

  4. Neither i nor ii
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(i)\rightarrow$definition of matrix
$(ii)$ Sum of elements in principle diagonal
$=8+8+8=24$
Option is correct
Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $A=\begin{bmatrix}
1 &4  &7 \
2 &6  &5 \
3 &-1  &2
\end{bmatrix}$ and B $=$ diag (1 2 5), then
trace of matrix $AB^{2}$ is

  1. 74

  2. 75

  3. 529

  4. 23
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $B=diag(1    2    5)$
$\Rightarrow B^{2}=diag(1     4    25)$

Now, $AB^{2}=\begin{bmatrix}1 &0  &0 \ 0 &24  &0 \ 0 &0  &50 \end{bmatrix}$

$\Rightarrow tr(AB^{2})=1+24+50=75$

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 find
${ tr }\left( A \right) +{ 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 $, where $tr(A)$ represents trace of matrix $A$.

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

$A=\left| \begin{matrix} 2 & 1 \ 4 & 1 \end{matrix} \right| ,\quad B=\left| \begin{matrix} 3 & 4 \ 2 & 3 \end{matrix} \right| ,\quad C=\left| \begin{matrix} 3 & -4 \ -2 & 3 \end{matrix} \right| \ AB\quad =\quad \left| \begin{matrix} 2 & 1 \ 4 & 1 \end{matrix} \right| \left| \begin{matrix} 3 & 4 \ 2 & 3 \end{matrix} \right| \ \qquad =\quad \left| \begin{matrix} 6+2 & 8+3 \ 12+2 & 16+3 \end{matrix} \right| \ \quad \quad \quad =\quad \left| \begin{matrix} 8 & 11 \ 14 & 19 \end{matrix} \right| \ ABC\quad =\quad \left| \begin{matrix} 8 & 11 \ 14 & 19 \end{matrix} \right| \left| \begin{matrix} 3 & -4 \ -2 & 3 \end{matrix} \right| \ \quad \quad \quad \quad \quad =\quad \left| \begin{matrix} 24-22 & -32+33 \ 42-38 & -56+57 \end{matrix} \right| \ \qquad \quad \quad =\quad \left| \begin{matrix} 2 & 1 \ 4 & 1 \end{matrix} \right| \ { (BC) }^{ 2 }\quad =\quad \left| \begin{matrix} 3 & 4 \ 2 & 3 \end{matrix} \right| \left| \begin{matrix} 3 & -4 \ -2 & 3 \end{matrix} \right| \ \qquad \quad \quad =\quad { \left| \begin{matrix} 1 & 0 \ 0 & 1 \end{matrix} \right|  }^{ 2 }$

$tr(A)\quad +\quad tr(\frac { ABC }{ 2 } )\quad +\quad tr(\frac { A{ (BC) }^{ 2 } }{ 4 } )\quad +\quad tr(\frac { A{ (BC) }^{ 3 } }{ 8 } )\quad +\quad .\quad .\quad .\quad +\quad \infty $
$\quad =\quad 3\quad +\quad 2\quad +\quad 1\quad +\quad 0\quad +\quad 0\quad +\quad 0\quad +\quad .\quad .\quad .\quad +\quad \infty \ \quad =\quad 6$

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

Elements of a matrix $A$ of order $10\times10$ are defined as ${ a } _{ ij }={ w }^{ i+j }$(where $w$ is cube root of unity), then trace ($A$) of the matrix is

  1. $0$
  2. $1$
  3. $3$

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

Here, $A=\begin{bmatrix} { \omega  }^{ 2 } & { \omega  }^{ 3 } & ... & { \omega  }^{ 11 } \ { \omega  }^{ 3 } & { \omega  }^{ 4 } & ... & { \omega  }^{ 12 } \ ... & ... & ... & ... \ { \omega  }^{ 11 } & { \omega  }^{ 12 } &  & { \omega  }^{ 20 } \end{bmatrix}$

$tr(A)={ \omega  }^{ 2 }+{ \omega  }^{ 4 }+{ \omega  }^{ 6 }+.....+{ \omega  }^{ 20 }$

$=\displaystyle \frac { { \omega  }^{ 2 }(1-{ \omega  }^{ 20 }) }{ 1-{ \omega  }^{ 2 } } $

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

Let $A=\left[\begin{matrix}2&0&7\0&1&0\1&-2&1\end{matrix}\right]$ and $B=\left[\begin{matrix}-x&14x&7x\0&1&0\x&-4x&-2x\end{matrix}\right]$ are two matrices such that $AB = (AB)^{-1}$ and $AB\ne I$ (where $I$ is an identity matrix of order $3\times3$).
Find the value of $Tr.\left(AB+(AB)^2+(AB)^3+...+(AB)^{100}\right)$ where $Tr.(A)$ denotes the trace of matrix $A$.

  1. 98

  2. 99

  3. 100

  4. 101
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 $A=\left[\begin{matrix}2&0&7\0&1&0\1&-2&1\end{matrix}\right]$ and $B=\left[\begin{matrix}-x&14x&7x\0&1&0\x&-4x&-2x\end{matrix}\right]$
$AB=\left[\begin{matrix}2&0&7\0&1&0\1&-2&1\end{matrix}\right]\left[\begin{matrix}-x&14x&7x\0&1&0\x&-4x&-2x\end{matrix}\right]=\begin{bmatrix} 5x & 14x & 0 \ 0 & 1 & 0 \ 0 & 10x-2 & 5x \end{bmatrix}$
but,$AB=(AB)^{-1}\Rightarrow (AB)^2=I$
$\Rightarrow (AB)^2=\begin{bmatrix} 5x & 14x & 0 \ 0 & 1 & 0 \ 0 & 10x-2 & 5x \end{bmatrix}\begin{bmatrix} 5x & 14x & 0 \ 0 & 1 & 0 \ 0 & 10x-2 & 5x \end{bmatrix}=\begin{bmatrix} 25x^{ 2 } & 70x^{ 2 }+14x & 0 \ 0 & 1 & 0 \ 0 & (5x+1)(10x-2) & 25x^{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow x=\displaystyle\frac{-1}{5}$
$Tr.\left(AB+(AB)^2+(AB)^3+...+(AB)^{100}\right)=Tr.\left(AB+I+(AB)+...+I\right)$
$tr(AB)=10x+1=-1$ and $tr(I)=3$
$\therefore Tr.\left(AB+I+(AB)+...+I\right)=(-1+3-1+3......+3)=50(3-1)=100$
Hence, option C.

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

Let $A=\left[\begin{matrix}1 & \displaystyle\frac{3}{2}\1 & 2\end{matrix}\right], B = \left[\begin{matrix}4 & -3\-2 & 2\end{matrix}\right] \mbox{ and } C _r = \left[\begin{matrix}r.3^r & 2^r\0 & (r-1)3^r\end{matrix}\right]$ be 3 given matrices. Compute the value of $\sum _{r=1}^{50}{tr.\left((AB)^r C _r\right)}.($ where $tr.(A)$ denotes trace of matrix A $)$

  1. $3(49.3^{50}+1)$
  2. $3(49.3^{49}+1)$
  3. $3(49.3^{48}+1)$

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

Given, $A=\left[\begin{matrix}1 & \displaystyle\frac{3}{2}\1 & 2\end{matrix}\right], B = \left[\begin{matrix}4 & -3\-2 & 2\end{matrix}\right] $

So, $AB=\left[ \begin{matrix} 1 & 0 \ 0 & 1 \end{matrix} \right] $
$\Rightarrow AB=I$

Now, $(AB)^{r} C _r =I \left[\begin{matrix}r.3^r & 2^r\0 & (r-1)3^r\end{matrix}\right]$

$\Rightarrow (AB)^{r} C _r=\left[\begin{matrix}r.3^r & 2^r\0 & (r-1)3^r\end{matrix}\right]$

$\sum _{ r=1 }^{ 50 }{ tr.\left( (AB)^{ r }C _{ r } \right)  } =\sum _{ r=1 }^{ 50 }{ tr\left( C _{ r } \right)  } =\sum _{ r=1 }^{ 50 }{ (2r } -1)3^{ r }$

It is an arithmetico-geometric series
$S _{50}=1.3+3.3^{2}+5.3^{3}+......+99.3^{50}$
$3S _{50}=        3^{2}+3.3^{3}+......+97.3^{50}+99.3^{51}$

$\Rightarrow -2S=3+2.3^{2}+2.{3}^{3}+.....+2.3^{50}-99.3^{51}$

$\Rightarrow -2S=3-99.3^{ 51 }+2(3^{ 2 }+{ 3 }^{ 3 }+.....+3^{ 50 })$

$\Rightarrow -2S=3-99.3^{ 51 }+3^{ 51 }-3^{ 2 }$

$\Rightarrow S=3(1+49.3^{ 50 })$

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

Let $A=\left[\begin{matrix}3x^2\1\6x\end{matrix}\right], B=[a,b,c]$ and $C=\left[\begin{matrix}(x+2)^2&5x^2&2x\5x^2&2x&(x+2)^2\2x&(x+2)^2&5x^2\end{matrix}\right]$ be three given matrices, where $a,b,c$ and $x\in R$, Given that $tr.(AB) = tr.(C) \vee x\in R$, where $tr.(A)$ denotes trace of $A$. Find the value of $(a+b+c)$

  1. 6

  2. 7

  3. 8

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $A=\left[\begin{matrix}3x^2\1\6x\end{matrix}\right], B=[a,b,c]$ 

Here, $AB=\left[ \begin{matrix} 3x^{ 2 } \ 1 \ 6x \end{matrix} \right] [a,b,c]$

$\Rightarrow AB=\left[ \begin{matrix} 3ax^{ 2 } & 3bx^{ 2 } & 3cx^{ 2 } \ a & b & c \ 6ax & 6bx & 6cx \end{matrix} \right] $

$tr(AB)=3ax^{2}+b+6cx$

Now, given $C=\left[\begin{matrix}(x+2)^2&5x^2&2x\5x^2&2x&(x+2)^2\2x&(x+2)^2&5x^2\end{matrix}\right]$

$tr(C)=(x+2)^2+2x+5x^{2}$
$\Rightarrow tr(C)=6x^{2}+6x+4$

Since, we have $tr(AB)=tr(C)$
$\Rightarrow 3ax^{2}+b+6cx=6x^{2}+6x+4$

On comparing, we get $a=2,b=4,c=1$
So, $a+b+c=7$

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

$BC=\begin{bmatrix} 3& 4\\ 2 & 3\end{bmatrix} \begin{bmatrix} 3 & -4 \\-2 & -3 \end{bmatrix} = \begin{bmatrix}1& 0\\ 0 & 1\end{bmatrix}$ 

$\therefore \displaystyle S=t _r(A)+t _r\left ( \frac{A}{2} \right )+t _r\left ( \frac{A}{4} \right ) +....\infty $ 

$t _r(A)=2+1=3$

$\displaystyle S=3+\frac{3}{2}+\frac{3}{4}+....\infty$

The sum of infinite terms of a GP series is $S _\infty= \dfrac{a}{(1-r)}$

$ S=3 \left(\dfrac{1}{1-\dfrac{1}{2}}\right)=6$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

Find the value of $\begin{vmatrix} 2+i & 2-i \ 1+i & 1-i \end{vmatrix}$ if $i^2=-1$.

  1. A complex quantity

  2. real quantity
  3. $0$

  4. cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\left| \begin{matrix} 2+i & 2-i \ 1+i & 1-i \end{matrix} \right| \ =(2+i)(1-i)-(2+i)(1+i)\ =2-2i+i-{ i }^{ 2 }-2-2i-i-{ i }^{ 2 }\ =-4i+2$

So it is complex quantity

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If $\begin{vmatrix}6i & -3i & 1\4 & 3i & -1\20 & 3 & i\end{vmatrix} = x+ iy$, then 

  1. $x =3, y = 0$
  2. $x =1, y = 3$
  3. $x =0, y = 3$
  4. $x =0, y = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given:-

       $ \begin{vmatrix} 6i & -3i & 1 \ 4 & 3i & -1 \ 20 & 3 & i \end{vmatrix}=x+iy$
To find value of $x$ and $ y$.
By solving the given diterment.
$ \begin{vmatrix} 6i & -3i & 1 \ 4 & 3i & -1 \ 20 & 3 & i \end{vmatrix}=6i\left[ 3i\left( i \right) -\left( 3 \right) \left( -1 \right)  \right] -\left( -3i \right) \left[ 4\left( i \right) -\left( 20 \right) \left( -1 \right)  \right] +1\left[ (4)\left( 3 \right) -\left( 20 \right) \left( 3i \right)  \right] $
$ 6i\left[ 3{ i }^{ 2 }+3 \right] +3i\left[ 4i+20 \right] +1\left[ 12-60i \right] $
 We know that $i=\sqrt { -1 }$ hence,$ { i }^{ 2 }=-1$
By substituting the value of ${ i }^{ 2 }$ we get
$ \begin{vmatrix} 6i & -3i & 1 \ 4 & 3i & -1 \ 20 & 3 & i \end{vmatrix}=6i\left[ -3+3 \right] +12{ i }^{ 2 }+60i+12-60i$
$=0+12(-1)+60i+12-60i$
$ =0$
 By comparing with given we get.
$ x+iy=0$
 If $x& y$ are real no. then the only possible solution
 for$ x+iy=0$ is
$ x=0$ and$ y=0$
 Hence the answer is $x=0\quad & \quad y=0$

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Let $\begin{bmatrix} n \ k\end{bmatrix}$ represents the combination of 'n' things taken 'k' at a time, then the value of the sum $\begin{bmatrix} 99\ 97\end{bmatrix}+\begin{bmatrix} 98\ 96\end{bmatrix}+\begin{bmatrix} 97\ 95\end{bmatrix}+...+\begin{bmatrix} 3\ 1\end{bmatrix}+\begin{bmatrix} 2\ 0\end{bmatrix}$ equals?

  1. $\begin{bmatrix} 99\\ 97\end{bmatrix}$
  2. $\begin{bmatrix} 100\\ 98\end{bmatrix}$
  3. $\begin{bmatrix} 99\\ 98\end{bmatrix}$
  4. $\begin{bmatrix} 100\\ 97\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a sum of combinations. Using the identity C(n, k) + C(n, k-1) = C(n+1, k), the sum C(99, 97) + C(98, 96) + ... + C(2, 0) can be evaluated. This is a known property of Pascal's triangle diagonals.

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

The number of value of $x$ in the closed interval $[-4,-1]$, the matrix $\begin{bmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{bmatrix}$ is singular is 

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For a singular matrix the value of the determinant = 0 
$ [3(-1)(2)+(-1+x)(x+2)(x+3)+(2)(3)(-1)] $
$\times  [-(2)(-1)(x+3)-(-1+x)(3)(2)-(3)(x+2)(x+3)] = 0 $
$-6+x^3+4x^2+x-6-6+2x+6+6-6x-3x^2-15x-18 = 0 $
$ x^3+x^2-18x-24 = 0 $
$ x = -4 $
$x =(3-\sqrt{33})/2=-1.372 $
$x =(3+\sqrt{33})/2= 4.372 $

Value of x in the closed interval $[-4,-1]$ are $-4,   -1.372$