Matrices and Determinants 4 Question 23
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24. Let $S$ be the set of all column matrices $\begin{aligned} & b _1 \ & b _2 \ & b _3\end{aligned}$ such that $b _1$,
======= ####24. Let $S$ be the set of all column matrices $\left|\begin{aligned} & b _1 \\ & b _2 \\ & b _3\end{aligned}\right|$ such that $b _1$,
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed $b _2, b _3 \in R$ and the system of equations (in real variables)
$ \left|\begin{array}{r} -x+2 y+5 z=b _1 \\ 2 x-4 y+3 z=b _2 \\ x-2 y+2 z=b _3 \end{array}\right| $
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{aligned} & b _1 \\ & b _2 \\ & b _3\end{aligned}\right|$$\in S$
(a) $x+2 y+3 z=b _1, 4 y+5 z=b _2$ and $x+2 y+6 z=b _3$
(b) $x+y+3 z=b _1, 5 x+2 y+6 z=b _2$ and $-2 x-y-3 z=b _3$
(c) $-x+2 y-5 z=b _1, 2 x-4 y+10 z=b _2$ and $x-2 y+5 z=b _3$
(d) $x+2 y+5 z=b _1, 2 x+3 z=b _2$ and $x+4 y-5 z=b _3$
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Answer:
Correct Answer: 24. (a,d)
Solution:
- We have,
$ \left|\begin{aligned} -x+2 y+5 z & =b _1 \\ 2 x-4 y+3 z & =b _2 \\ x-2 y+2 z & =b _3 \end{aligned}\right| $
has at least one solution.
$ \begin{aligned} & \therefore \quad D=\left|\begin{array}{ccc} -1 & 2 & 5 \\ 2 & -4 & 3 \\ 1 & -2 & 2 \end{array}\right| \\ & \text { and } \quad D _1=D _2=D _3=0 \\ & \Rightarrow \quad D _1=\left|\begin{array}{ccc} b _1 & 2 & 5 \\ b _2 & -4 & 3 \\ b _3 & -2 & 2 \end{array}\right| \\\ & =-2 b _1-14 b _2+26 b _3=0 \\ & \Rightarrow \quad b _1+7 b _2=13 b _3 \\ & \text { (a) } D=\left|\begin{array}{lll} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 2 & 6 \end{array}\right|=1(24-10)+1(10-12) \\ & =14-2=12 \neq 0 \end{aligned} $
Here, $D \neq 0 \Rightarrow$ unique solution for any $b _1, b _2, b _3$.
(b) $D=\left|\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right|$
$ =1(-6+6)-1(-15+12)+3(-5+4)=0 $
For atleast one solution
$ \begin{aligned} & D _1=D _2=D _3=0 \\ & \text { Now, } D _1=\left|\begin{array}{ccc} b _1 & 1 & 3 \\ b _2 & 2 & 6 \\ b _3 & -1 & -3 \end{array}\right| \\ &=b _1(-6+6)-b _2(-3+3)+b _3(6-6) \\ &=0 \\ & D _2=\left|\begin{array}{ccc} 1 & b _1 & 3 \\ 5 & b _2 & 6 \\ -2 & b _3 & -3 \end{array}\right| \\ &=-b _1(-15+12)+b _2(-3+6)-b _3(6-15) \\ &= 3 b _1+3 b _2+9 b _3=0 \Rightarrow b _1+b _2+3 b _3=0 \end{aligned} $
not satisfies the Eq. (i)
It has no solution.
(c) $D=\left|\begin{array}{ccc}-1 & 2 & -5 \\ 2 & -4 & 10 \\ 1 & -2 & 5\end{array}\right|$
$ =-1(-20+20)-2(10-10)-5(-4+4) $
$ =0 $
Here, $b _2=-2 b _1$ and $b _3=-b _1$ satisfies the Eq. (i) Planes are parallel.
(d) $\begin{aligned} D & =\left|\begin{array}{ccc}1 & 2 & 5 \\ 2 & 0 & 3 \\ 1 & 4 & -5\end{array}\right|=1(0-12)-2(-10-3)+5(8-0) \\ & =54\end{aligned}$
$ D \neq 0 $
It has unique solution for any $b _1, b _2, b _3$.
