Tag: inverse of a matrix and linear equations

Questions Related to inverse of a matrix and linear equations

$x + 2y + 3z = 0$ 

$2x + 3y + 4z = 0$
$7x + 13y + 19z =0$
The system of equations have non trivial solutions

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. True

  2. False

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Correct Option: A

$2x - 3y + z = 0$ 

$x + 2y - 3z =0$
$4x - y - 2z = 0$
The system of equations have a non trivial solution

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. True

  2. False

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Correct Option: B

If the lines $L _{1}:\lambda ^{2}x-y-1=0$ $L _{2}:x-\lambda ^{2}y+1=0$ $L _{3}:x+y-\lambda ^{2}=0$ pass through the same point the value(s) of $\lambda$ equals

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $1$

  2. $\sqrt{2}$

  3. $2$

  4. $0$

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Correct Option: B
Explanation:

The given equations passes through same point.So, they are concurrent lines

$\Rightarrow \left| \begin{matrix} { \lambda  }^{ 2 } & -1 & -1 \ 1 & -{ \lambda  }^{ 2 } & 1 \ 1 & 1 & -{ \lambda  }^{ 2 } \end{matrix} \right| =0$

$\Rightarrow { \lambda  }^{ 6 }-3{ \lambda  }^{ 2 }-2=0$

By doing synthetic division we get,
$(\lambda^4-2\lambda^2+1)(\lambda^2-2)=0$
$(\lambda^2-1)^2(\lambda^2-2)=0$
$\lambda=\pm \sqrt 2 or \lambda=\pm1$
But here $\lambda=\pm1\  does\  not\  satisfies\  ,hence\  \lambda=\sqrt2$
Option B satisfies above equation

Three digits numbers $ 7x,36y$  and  $12z$ where  $x , y , z$ are integers from  $0$  to  $9 ,$  are divisible by a fixed constant $k.$  Then the determinant  $\left| \begin{array} { l l l } { x } & { 3 } & { 1 } \ { 7 } & { 6 } & { z } \ { 1 } & { y } & { 2 } \end{array} \right|$ $\ +48$ must be divisible by 

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $k$

  2. $k ^ { 2 }$

  3. $k ^ { 3 }$

  4. $k ^ { 4}$

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Correct Option: A
Explanation:

Since $7x,36y,12z$ are divisible by $k$

Let us Assume $x=ka,y=kb,z=kc$
So the det $\implies \left| \begin {array}{c c c} ka&3&1\7&6&kc\1&kb&2 \end{array} \right|$
$\implies 12ka-k^3abc+3kc+7kb-48$
$\implies From \  Question, 12ka-k^3abc+3kc+7kb-48+48$
Hence A

Numbers of ways in which 75600 can be resolved as product of two divisors which are relatively prime ?

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. 44

  2. $8$

  3. $9$

  4. $16$

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Correct Option: B
Explanation:

First we will have to find the prime factors of $75600$

Prime factorization of $75600=2\times 2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 7$
$\Rightarrow 75600=2^4\times 3^3 \times 5^2\times 7$
The number of ways in which a composite number N can be resolved as product of two divisors which are relatively prime.
$=2^{n-1}$ where n is number of different factors of N
$=2^{n-1}$
$=2^{4-1}$
$=2^3$
$=8$ ways
Hence, the answer is $8.$

Number of values of $a$ for which the lines $2x+y-1=0, ax+3y-3=0, 3x+2y-2=0$ are concurrent is

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. 0

  2. 1

  3. 2

  4. $\infty$

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Correct Option: D
Explanation:

Here coefficient matrix,  $\Delta = \begin{vmatrix} 2 & 1 & -1 \ a & 3 & -3 \ 3 & 2 & -2 \end{vmatrix}$
Using $C _2 \to C _2+C _3$
$\Delta = \begin{vmatrix} 2 & 0 & -1 \ a & 0 & 3 \ 3 & 0 & -2 \end{vmatrix}=0$
Clearly $\Delta=0$, Hence given lines are concurrent for all values of $a$

If the lines $\displaystyle y-x=5,3x+4y=1$ and $\displaystyle y=mx+3$ are concurrent then the value of m is

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $\displaystyle \frac{19}{5}$

  2. $\displaystyle 1$

  3. $\displaystyle \frac{5}{19}$

  4. none of these

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Correct Option: C
Explanation:

Given lines $\displaystyle y-x=5,3x+4y=1$ and $\displaystyle y=mx+3$
For concurrency,
$\begin{vmatrix} -1 & 1 & 5 \ 3 & 4 & 1 \ -m & 1 & 3 \end{vmatrix}=0$
$\Rightarrow -5+19m=0$
$\Rightarrow \displaystyle m =\frac{5}{19}$

Three lines $px + qy + r = 0, qx + ry + p = 0$ and $rx + py + q = 0$ are concurrent if

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $p + q + r = 0$

  2. $p^2 + q^2 + r^2 = pr + qr + pq$

  3. $p^3 + q^3 + r^3 = 3pqr$

  4. none of these

Show answer
Correct Option: A,B,C
Explanation:

These lines are concurrent when area formed from these lines is zero
$\begin{vmatrix} p & q & r \ q & r & p \ r & p & q \end{vmatrix}=0$
Applying ${ R } _{ 1 }\rightarrow { R } _{ 1 }+{ R } _{ 2 }+{ R } _{ 3 }$
$\Rightarrow \begin{vmatrix} p+q+r & p+q+r & p+q+r \ q & r & p \ r & p & q \end{vmatrix}=0\ \Rightarrow \left( p+q+r \right) \begin{vmatrix} 1 & 1 & 1 \ q & r & p \ r & p & q \end{vmatrix}=0$
Again applying ${ C } _{ 2 }\rightarrow { C } _{ 2 }-{ C } _{ 1 },{ C } _{ 3 }\rightarrow { C } _{ 3 }-{ C } _{ 1 }$
$\Rightarrow \left( p+q+r \right) \begin{vmatrix} 1 & 0 & 0 \ q & r-q & p-q \ r & p-r & q-r \end{vmatrix}\ \Rightarrow \left( p+q+r \right) \left( \left( r-q \right) \left( q-r \right) -\left( p-r \right) \left( p-q \right)  \right) =0$<br>$\Rightarrow p^3+q^3+r^3-3pqr =0 $
Hence, options A, B and C are correct.

If the lines $2\mathrm{x}-\mathrm{a}\mathrm{y}+1 =0,\ 3\mathrm{x}-\mathrm{b}\mathrm{y}+1 =0,\ 4\mathrm{x}-\mathrm{c}\mathrm{y}+1 =0$ are concurrent then $\mathrm{a}$, b,c are in ?

.

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. G.P.

  2. A.P.

  3. H.P.

  4. A.G.P.

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Correct Option: B
Explanation:

$2x-ay+1 =0,\ 3x-by+1 =0,\ 4x-cy+1 =0$


Given lines are concurrent

$\Rightarrow \begin{vmatrix} 2 & -a & 1 \ 3 & -b & 1 \ 4 & -c & 1 \end{vmatrix}=0$

$\Rightarrow  2b-a-c=0$

$\Rightarrow 2b=a+c$

Hence, a,b,c are in A.P.

The points $(k, 2-2k)$, $(-k+1, 2k)$ and $(-4-k, 6-2k)$ are collinear for

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. all values of k

  2. $k=-1$

  3. $k=1/2$

  4. no value of k.

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Correct Option: B,C
Explanation:

The given points are collinear if 


$\displaystyle \begin{vmatrix} k  & 2-2k  & 1 \ -k+1  & 2k  & 1 \ -4-k & 6-2k  & 1  \end{vmatrix}=0$

$\Rightarrow  
\begin{vmatrix} k  & 2-2k  & 1 \ -2k+1  & 4k-2  & 0 \ -4-2k & 4  & 0  \end{vmatrix}=0$       $ \left [ R _{2}\rightarrow R _{2}-R _{1}, R _{3}\rightarrow R _{3}-R _{1} \right ]$


$\Rightarrow 4(-2k+1)-(-4-2k)(4k-2)=0$

$\Rightarrow (1-2k)(4-8-4k)=0$

$\Rightarrow (1-2k)(k+1)=0$

$\Rightarrow k=-1 \ or \ k=1/2$