Question

Let $S$ be the set of all column matrices $\left[\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right]$
such that $b_1,b_2,b_3\in R$ and the system of equations (in real variables)
$$\begin{array}{c}-x+2y+5z=b_1\\ 2x-4y+3z=b_2\\ x-2y+2z=b_3\end{array}$$
has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $\left[\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right]\in S$?

• A. $x+2y+3z=b_1,4y+5z=b_2\text{and}x+2y+6z=b_3$
• B. $x+y+3z=b_1,5x+2y+6z=b_2\text{and}-2x-y-3z=b_3$
• C. $-x+2y-5z=b_1,2x-4y+10z=b_2\text{and}x-2y+5z=b_3$
• D. $x+2y+5z=b_1,2x+3z=b_2\text{and}x+4y-5z=b_3$

Answer 1

Answer: (A), (C), (D)

$x+2y+5z=b_1,2x+3z=b_2\text{and}x+4y-5z=b_3$

$\begin{array}{c}-x+2y+5z=b_1..........\left(1\right)\\ 2x-4y+3z=b_2............\left(2\right)\\ x-2y+2z=b_3............\left(3\right)\end{array}$

$\Delta =\left|\begin{array}{ccc}-1& 2& 5\\ 2& -4& 3\\ 1& -2& 2\end{array}\right|=0$
No two planes are parallel so infinite solutions.
They are family of planes so,
$P_3=\lambda P_1+\mu P_{2}1=-\lambda +2\mu .......\left(i\right)-2=2\lambda -4\mu .......\left(ii\right)2=5\lambda +3\mu .......\left(iii\right)$
Using equation $\left(i\right)$ and $\left(iii\right)$,
$\lambda =\frac{1}{13};\mu =\frac{7}{13}\Rightarrow 13b_3=b_1+7b_2......\left(1\right)$

Now checking $\Delta$ for all the option,
Option (a)
$\Delta =\left|\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 2& 6\end{array}\right|=12\Rightarrow \Delta \ne 0$

Option (b)
$\Delta =\left|\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right|=0$
So
$P_3=\lambda P_1+\mu P_2-2=\lambda +5\mu .......\left(iv\right)-1=\lambda +2\mu .......\left(v\right)-3=3\lambda +6\mu .......\left(vi\right)$
Using equation $\left(iv\right)$ and $\left(v\right)$,
$\lambda =-\frac{1}{3};\mu =-\frac{1}{3}$
So $-\frac{b_1}{3}-\frac{b_2}{3}=b_3\Rightarrow -3b_3=b_1+b_2$
Which is not equal to equation $\left(1\right)$
No solutions.

Option (c)
$\Delta =\left|\begin{array}{ccc}-1& 2& -5\\ 2& -4& 10\\ 1& -2& 5\end{array}\right|=0$
Directly from the equation we can wrie
$b_1=-b_3$ and $b_2=2b_3$
Putting this in equation $\left(1\right)$
$14b_3-b_3=13b_3$
satisfies the equation, so infinite solution.

Option (d)
$\Delta =\left|\begin{array}{ccc}1& 2& 5\\ 2& 0& 3\\ 1& 4& -5\end{array}\right|=54\ne 0$