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If $[x]$ denotes the greatest integer $\leq x$ , then the system

of linear equations $[sin\theta]x+[-cos\theta]y=0$

$[cot\theta]x+y=0$

• Option 1)

Have infinitely many solutions if $\theta\epsilon (\frac{\pi}{2},\frac{2\pi}{3})$ and has a unique solution if $\theta\epsilon (\pi,\frac{7\pi}{6})$.

• Option 2)

has a unique solution if $\theta\epsilon(\frac{\pi}{2},\frac{2\pi}{3})\cup (\pi,\frac{7\pi}{6})$

• Option 3)

has a unique solution if $\theta\epsilon(\frac{\pi}{2},\frac{2\pi}{3})$ and have infinitely many solutions if $\theta\epsilon (\pi,\frac{7\pi}{6})$

• Option 4)

infinitely many solutions if $\theta\epsilon(\frac{\pi}{2},\frac{2\pi}{3})\cup$$(\pi,\frac{7\pi}{6})$

linear equations  &                              For infinite many solution, i.e. ...................(1) * when                         so,     * when                         so,     *when   *when                          Option 1) Have infinitely many solutions if  and has a unique solution if . Option 2) has a unique solution if  Option 3) has a unique solution if  and have infinitely many...

A value of $\theta \epsilon (0,\pi /3)$, for which

$\begin{vmatrix} 1+cos^{2}\theta &sin^{2}\theta &4cos6\theta \\ cos^{2}\theta& 1+sin^{2}\theta &4cos6\theta \\ cos^{2}\theta& sin^{2}\theta & 1+4cos6\theta \end{vmatrix}=0$, is:

• Option 1)

$\frac{\pi }{9}$

• Option 2)

$\frac{\pi }{18}$

• Option 3)

$\frac{7\pi }{24}$

• Option 4)

$\frac{7\pi }{36}$

Option 1)       Option 2) Option 3) Option 4)

If $B=\begin{bmatrix} 5 &2\alpha &1 \\ 0 &2 &1 \\ \alpha &3 &-1 \end{bmatrix}$ is the inverse of a $3\times 3$ matrix A, then the sum of all values of $\alpha$ for which det $\left ( A \right )+1=0,$ is :



• Option 1)

$0$

• Option 2)

$1$

• Option 3)

$2$

• Option 4)

$-1$

Option 1) Option 2) Option 3) Option 4)

If A is a syymmetric matrix and B is a skew-symmetrix matrix such that $A+B=\begin{bmatrix} 2 &3 \\5 &-1 \end{bmatrix}$, then AB is equal to :

• Option 1)

$\begin{bmatrix} -4 &-2 \\ -1 &4 \end{bmatrix}$

• Option 2)

$\begin{bmatrix} 4 &-2 \\ -1 & -4 \end{bmatrix}$

• Option 3)

$\begin{bmatrix} 4 &-2 \\ 1 &-4 \end{bmatrix}$

• Option 4)

$\begin{bmatrix} -4 &2 \\ 1 & 4 \end{bmatrix}$

A is symmetric matrix  B is skew-symmetrix (1) + (2)  Option 1) Option 2) Option 3) Option 4)

The sum of the real roots of the equation

$\begin{vmatrix} x & -6 &-1 \\ 2 &-3x &x-3 \\ -3& 2x &x+2 \end{vmatrix}=0$, is equal to :

• Option 1)

6

• Option 2)

0

• Option 3)

1

• Option 4)

-4

=>  =>  Root of equation (-3,1,2) So, Sum of real root of equation = -3+1+2=0 So, option (2) is correct.Option 1)6Option 2)0Option 3)1Option 4)-4

Let $\lambda$ be a real number for which the system of linear equations

$x+y+z=6$

$4x+\lambda y-\lambda z=\lambda -2$

$3x+2 y-4 z= -5$

has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation :

• Option 1)

$\lambda^{2}+3\lambda-4=0$

• Option 2)

$\lambda^{2}-3\lambda-4=0$

• Option 3)

$\lambda^{2}+\lambda-6=0$

• Option 4)

$\lambda^{2}-\lambda-6=0$

linear equations             Now, using cramers law for infinite solution   all will be zero      Now put  in options and check for the correct one  (1) (2) (3) (4) So, option (4) is correct.     Option 1) Option 2) Option 3) Option 4)

If the system of linear equations

x + y + z = 5

x + 2y + 2z = 6

x + 3y + $\lambda$ z = $\mu$ , ($\lambda$,$\mu$ $\epsilon R$ ) , has infinitely

many solutions, then the value of $\lambda$ + $\mu$ is :

• Option 1)

12

• Option 2)

9

• Option 3)

7

• Option 4)

10

x + 3y +  z -  = p ( x + y + z - 5) + q ( x + 2y + 2z - 6 )  On comparing the coefficients  p + q = 1   and    p + 2q = 3 => ( p , q ) = ( -1 , 2 ) Hence, x + 3y +  z -  = x + 3y + 3z - 7  =>  =>  So, option (4) is correct. Option 1) 12 Option 2) 9 Option 3) 7 Option 4) 10

If $\bigtriangleup _{1}=\begin{vmatrix} x &\sin \theta &\cos \theta \\ -\sin \theta &-x &1 \\ \cos \theta & 1 & x \end{vmatrix}$  and

$\bigtriangleup _{2}=\begin{vmatrix} x &\sin 2\theta &\cos 2\theta \\ -\sin 2\theta &-x &1 \\ \cos2 \theta & 1 & x \end{vmatrix}$ , $x\neq0$ ; then

for all $\theta \epsilon (0,\frac{\pi}{2}) :$

• Option 1)

$\triangle_{1}-\triangle_{2}=-2x^{3}$

• Option 2)

$\triangle_{1}-\triangle_{2}=x(cos2\theta-cos4\theta)$

• Option 3)

$\triangle_{1}+\triangle_{2}=-2(x^{3}+x-1)$

• Option 4)

$\triangle_{1}+\triangle_{2}=-2x^{3}$

So,  So, option (4) is correct.Option 1)Option 2)Option 3)Option 4)

If the system of equations  $2x+3y-z=0,x+ky-2z=0\:\:and\:\:2x-y+z=0$ has a non-trivial solution $\left ( x,y,z \right )$  , then $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k$   is equal to :

• Option 1)

$\frac{3}{4}$

• Option 2)

$\frac{1}{2}$

• Option 3)

$-\frac{1}{4}$

• Option 4)

$-4$

for non-trivial solution  A=0   so       Option 1) Option 2) Option 3) Option 4)

The total number of matrices

$A=\begin{pmatrix} 0 &2y &1 \\ 2x&y & -1\\ 2x & -y& 1 \end{pmatrix}$  ,$\left ( x,y\epsilon R,\:x\neq y \right )$ for which $A^{T}A=3I_{3}$  is :

• Option 1)

$2$

• Option 2)

$3$

• Option 3)

$6$

• Option 4)

$4$

,     4 matrices    Option 1) Option 2) Option 3) Option 4)

If

$\begin{bmatrix} 1& 1\\ 0& 1 \end{bmatrix}$$\begin{bmatrix} 1& 2\\ 0& 1 \end{bmatrix}$$\begin{bmatrix} 1 & 3\\ 0 & 1 \end{bmatrix}.$.................$\begin{bmatrix} 1 & n-1\\ 0& 1 \end{bmatrix}=$$\begin{bmatrix} 1 & 78\\ 0& 1 \end{bmatrix}$ ,

then the inverse of $\begin{bmatrix} 1 & n\\ 0 & 1 \end{bmatrix}$ is :

• Option 1)

$\begin{bmatrix} 1 & 0\\ 12 & 1 \end{bmatrix}$

• Option 2)

$\begin{bmatrix} 1& -13\\ 0& 1\end{bmatrix}$

• Option 3)

$\begin{bmatrix} 1 & -12\\ 0 & 1 \end{bmatrix}$

• Option 4)

$\begin{bmatrix} 1 & 0\\ 13 &1 \end{bmatrix}$

and  and so on So,...........                  Inverse of   is   Option 1)    Option 2)   Option 3)   Option 4)

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+x+1=0.$ Then for $y\neq 0$ in $R$,$\begin{vmatrix} y+1 & \alpha & \beta \\ \alpha & y+\beta & 1\\ \beta & 1& y+\alpha \end{vmatrix}$ is equal to :

• Option 1)

$y\left ( y^{2}-1 \right )$

• Option 2)

$y\left ( y^{2}-3 \right )$

• Option 3)

$y^{3}$

• Option 4)

$y^{3}-1$

are roots of      Option 1)Option 2)Option 3)Option 4)

If the system of linear equations

$x-2y+kz=1$

$2x+y+z=2$

$3x-y-kz=3$

has a solution $(z,y,z), z\neq 0$ then $(x,y)$ lies on the straight line whose equation is :

• Option 1)

$3x-4y-1=0$

• Option 2)

$4x-3y-4=0$

• Option 3)

$4x-3y-1=0$

• Option 4)

$3x-4y-4=0$

add  and    Option 1) Option 2) Option 3) Option 4)

Let the numbers $2,b,c$ be in an A.P. and $A=\begin{vmatrix} 1 &1 &1 \\ 2 &b &c \\ 4 &b^{2} &c^{2} \end{vmatrix}.$. If $\det (A)\epsilon \left [ 2,16 \right ],$ then c lies in the interval :

• Option 1)

$[2,3)$

• Option 2)

$\left ( 2+2^{\frac{3}{4}},4 \right )$

• Option 3)

$\left [ 4,6 \right ]$

• Option 4)

$\left [ 3,2+2^{\frac{3}{4}} \right ]$

are in A.P Option 1) Option 2) Option 3)   Option 4)

The greatest value of $c\; \epsilon \; \mathbb{R}$ for which the system of linear equations

$x-cy-cz=0$

$cx-y+cz=0$

$cx+cy-z=0$

has a non-trivial solution, is :

• Option 1)

$\frac{1}{2}$

• Option 2)

$0$

• Option 3)

$2$

• Option 4)

$-1$

For non trivial solution                                 Option 1)Option 2)Option 3)  Option 4)

Let $A=\begin{pmatrix} \cos \alpha - &\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix},(\alpha \: \epsilon \: \mathbb{R})$ such that $A^{32}=\begin{pmatrix} 0 &-1 \\ 1 &0 \end{pmatrix}.$ Then a value of $\alpha$ is:

• Option 1)

$\frac{\pi}{16}$

• Option 2)

$0$

• Option 3)

$\frac{\pi}{32}$

• Option 4)

$\frac{\pi}{64}$

Similarly we can observe that Option 1) Option 2) Option 3)   Option 4)

Let $A=\begin{pmatrix} 0 & 2q & r\\ p & q & -r\\ p& -q & r \end{pmatrix}.\: If\: \: AA^{T}=I_{3},\: Then \: \left | p \right |\: is\: :$

• Option 1)

$\frac{1}{\sqrt2}$

• Option 2)

$\frac{1}{\sqrt6}$

• Option 3)

$\frac{1}{\sqrt3}$

• Option 4)

$\frac{1}{\sqrt5}$

Transpose of a Matrix - The matrix obtained from any given matrix , by interchanging its rows and columns. - wherein     Multiplication of matrices -   -     and , Hence, Option 1)  Option 2)  Option 3)  Option 4)

If $\begin{vmatrix} a-b-c & 2a & 2a\\ 2b & b-c-a & 2b\\ 2c & 2c & c-a-b \end{vmatrix}= (a+b+c) (x+a+b+c)^{2}, x\neq 0 \: and\:$

$a+b+c\neq 0$ , then $x$ is equal to :

• Option 1)

$-2(a+b+c)$

• Option 2)

abc

• Option 3)

$-(a+b+c)$

• Option 4)

$2(a+b+c)$

Value of determinants of order 3 - -                          Option 1)Option 2)abcOption 3)Option 4)

Let A and B be two invertible matrices of order 3 x 3. If  $det (ABA^{T})=8$ and  $det (AB^{-1})=8$, then $det (BA^{-1}B^{T})$ is equal to:

• Option 1)

$\frac{1}{16}$

• Option 2)

$\frac{1}{4}$

• Option 3)

1

• Option 4)

16

Property of inverse of a matrix - Every invertible matrix possesses a unique inverse  -     Reversal law - - wherein  and  are invertible matrices of same order      Property of Transpose - - wherein  being scalar ;  is transpose of A       and            and       Option 1)Option 2)Option 3)1Option 4)16

The set of all values of $\lambda$ for which the system of linear equations

$x-2y-2z=\lambda x$

$x+2y+z=\lambda y$

$-x-y=\lambda z$

has a non-trivial solution :

• Option 1)

contains more than two elements

• Option 2)

is a singleton

• Option 3)

is an empty set

• Option 4)

contains exactly two elements

Solution of a homogeneous system of linear equations - Let     If  is singular then the system of equations will have infinitely many solutions -     Cramer's rule for solving system of linear equations - When   and  , then  the system of equations has infinite solutions. - wherein and   are obtained by replacing column 1,2,3 of  by   column For non-trivial solution ,  Option...
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