Mathematics

Linear Algebra

449 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

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 $-4,   -1.372$

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

The matrix $\begin{bmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \ 3 & 1 & 1 \end{bmatrix}$ is:

  1. nonsingular

  2. singular

  3. skew symmetric

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

$\begin{vmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \end{vmatrix}=1\begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix}-0\begin{vmatrix} 2 & 0 \ 3 & 1 \end{vmatrix}+1\begin{vmatrix} 2 & 1 \ 3 & 1 \end{vmatrix}$

$=1(1-0)-0+1(2-3)$
$=1-1$
$=0$
Therefore, it is a singular matrix.

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

If $\begin{bmatrix} 1 & 2 & x \  4 & -1 & 7  \  2 & 4 & 6  \end{bmatrix}$ is a singular matrix, then $x=$

  1. $0$
  2. $1$
  3. $-3$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{vmatrix} 1 & 2 & x \ 4 & -1 & 7 \ 2 & 4 & -6  \end{vmatrix}=0$

$\Rightarrow 1\begin{vmatrix} -1 & 7 \ 4 & -6 \end{vmatrix}-2\begin{vmatrix} 4 & 7 \ 2 & -6 \end{vmatrix}+x\begin{vmatrix} 4 & -1 \ 2 & 4 \end{vmatrix}=0$

$\Rightarrow (6-28)-2(-24-14)+x(16+2)=0$
$\Rightarrow -22+76+18x=0$
$\Rightarrow 18x=-54$

$\Rightarrow x=-3$

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

Consider the following statements:
1. The matrix
               $\begin{pmatrix} 1 & 2 & 1 \ a & 2a & 1 \ b & 2b & 1 \end{pmatrix}$ is singular.
2. The matrix
              $\begin{pmatrix} c & 2c & 1 \ a & 2a & 1 \ b & 2b & 1 \end{pmatrix}$ is non-singular.
Which of the above statements is/are correct?

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For matrix

  $\begin{vmatrix} 1 & 2 & 1 \ a & 2a & 1 \ b & 2b & 1 \end{vmatrix}$


$C _2\rightarrow C _2-2C _1 $

$\begin{vmatrix} 1 & 0 & 1 \\ a & 0 & 1 \\ b & 0 & 1 \end{vmatrix}$ The determinant is $0$. So the matrix is singular.

For matrix,
$ \begin{vmatrix} c & 2c & 1 \\ a & 2a & 1 \\ b & 2b & 1 \end{vmatrix}$

$C _2\rightarrow C _2-2C _1 $

$\begin{vmatrix} c & 0 & 1 \\ a & 0 & 1 \\ b & 0 & 1 \end{vmatrix}$ The determinant is $0$. So the matrix is singular.

Thus, only $1$ is true.
Hence, option A is correct.

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

Let $A$ be a square matrix all of whose entries are integers. Then which one of the following is true?

  1. If $det(A)=\pm 1$, then ${A}^{-1}$ exists but all its entries are not necessarily integers.
  2. If $det(A)=\pm 1$, then ${A}^{-1}$ exists and all its entries are non integers
  3. If $det(A)=\pm 1$, then ${A}^{-1}$ exists and all its entries are integers
  4. If $det(A)=\pm 1$, then ${A}^{-1}$ need not exist
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a matrix A with integer entries, the inverse is given by (1/det(A)) * adj(A). If det(A) = +/- 1, then 1/det(A) is +/- 1. Since the adjugate matrix consists of determinants of integer submatrices, it also contains only integers. Thus, A^-1 has only integer entries.

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

If $A$ and $B$ are two non-zero square matrices of the same order such that the product $AB=0$, then

  1. both $A$ and $B$ must be singular

  2. exactly one of them must be singular

  3. both of them are non singular

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

$AB=O$
taking determinant on both sides
$|AB|=|A||B|=0$
$\Rightarrow$ Either $A$ or $B$ should be a singular matrix.
Hence, option D.

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

If $A$ is a nonsingular matrix satisfying $AB=BA+A$ then

  1. $\left|B\right|=\left|I+B\right|$
  2. $\left|B\right|=\left|2I+B\right|$
  3. $\left|B\right|=\left|B-I\right|$
  4. $\left|B\right|=\left|B-2I\right|$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given AB = BA + A, we can rewrite this as AB - BA = A. This is a commutation relation. Using properties of determinants and the given condition, it can be shown that |B| = |I + B|.