# ISI2016-DCG-11

79 views

Let two systems of linear equations be defined as follows:

$\begin{array}{lll} & x+y & =1 \\ P: & 3x+3y & =3 \\ & 5x+5y & =5 \end{array}$ and $\begin{array}{lll} & x+y & =3 \\ Q: & 2x+2y & =4 \\ & 5x+5y & =12 \end{array}$. Then,

1. $P$ and $Q$ are inconsistent
2. $P$ and $Q$ are consistent
3. $P$ is consistent but $Q$ is inconsistent
4. None of the above

recategorized

P

$x+y = 1$ ....$eq(i)$

$3x+3y = 3$....$eq(ii)$

$5x+5y = 5$...$eq(iii)$

As we can see

$2*eq(i) + eq(ii) = eq(iii)$ this rule is satisfied by both the LHS and RHS parts of the equations.

$\implies$ they can have infinite solution

$\implies$ They are having consistent solutions.

Q

$x+y = 3$ ....$eq(i)$

$2x+2y = 4$....$eq(ii)$

$5x+5y = 12$...$eq(iii)$

As we can see

$eq(i) + 2*eq(ii) = eq(iii)$ this rule is satisfied by the LHS part and not by the RHS parts of the equations.(i.e. $3+4*2\neq 12$)

$\implies$ They have no solution

$\implies$ They are having inconsistent solution.

$\therefore$ Option $C.$ is the correct answer

## Related questions

1
69 views
The value of $\begin{vmatrix} 1+a& 1& 1& 1\\ 1&1+b &1 &1 \\ 1&1 &1+c &1 \\ 1&1 &1 &1+d \end{vmatrix}$ is $abcd(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})$ $abcd(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})$ $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ None of these
If $f(x)=\begin{bmatrix}\cos\:x & -\sin\:x & 0 \\ \sin\:x & \cos\:x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ then the value of $\big(f(x)\big)^2$ is $f(x)$ $f(2x)$ $2f(x)$ None of these
The value of $\:\:\begin{vmatrix} 1&\log_{x}y &\log_{x}z \\ \log_{y}x &1 &\log_{y}z \\\log_{z}x & \log_{z}y&1 \end{vmatrix}\:\:$ is $0$ $1$ $-1$ None of these
Let $A$ be an $n\times n$ matrix such that $\mid\: A^{2}\mid=1.\:\: \mid A\:\mid$ stands for determinant of matrix $A.$ Then $\mid\:(A)\mid=1$ $\mid\:(A)\mid=0\:\text{or}\:1$ $\mid\:(A)\mid=-1,0\:\text{or}\:1$ $\mid\:(A)\mid=-1\:\text{or}\:1$