SATHEE: Matrices and Determinants 4 Question 23

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 $| \begin{aligned} b_{1} \\ b_{2} \\ b_{3} \end{aligned} |$ such that $b_{1}$ ,

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed $b_{2}, b_{3} \in R$ and the system of equations (in real variables)

$| \begin{array}{r} - x + 2 y + 5 z = b_{1} \\ 2 x - 4 y + 3 z = b_{2} \\ x - 2 y + 2 z = b_{3} \end{array} |$

has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $| \begin{aligned} b_{1} \\ b_{2} \\ b_{3} \end{aligned} |$ $\in S$

(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 $- 2 x - y - 3 z = b_{3}$

(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,

$| \begin{aligned} - x + 2 y + 5 z & = b_{1} \\ 2 x - 4 y + 3 z & = b_{2} \\ x - 2 y + 2 z & = b_{3} \end{aligned} |$

has at least one solution.

$\begin{aligned} ∴ D = | \begin{array}{ccc} - 1 & 2 & 5 \\ 2 & - 4 & 3 \\ 1 & - 2 & 2 \end{array} | \\ and D_{1} = D_{2} = D_{3} = 0 \\ \Rightarrow D_{1} = | \begin{array}{ccc} b_{1} & 2 & 5 \\ b_{2} & - 4 & 3 \\ b_{3} & - 2 & 2 \end{array} | \\ = - 2 b_{1} - 14 b_{2} + 26 b_{3} = 0 \\ \Rightarrow b_{1} + 7 b_{2} = 13 b_{3} \\ (a) D = | \begin{array}{lll} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 2 & 6 \end{array} | = 1 (24 - 10) + 1 (10 - 12) \\ = 14 - 2 = 12 \neq 0 \end{aligned}$

Here, $D \neq 0 \Rightarrow$ unique solution for any $b_{1}, b_{2}, b_{3}$ .

(b) $D = | \begin{array}{ccc} 1 & 1 & 3 \\ 5 & 2 & 6 \\ - 2 & - 1 & - 3 \end{array} |$

$= 1 (- 6 + 6) - 1 (- 15 + 12) + 3 (- 5 + 4) = 0$

For atleast one solution

$\begin{aligned} D_{1} = D_{2} = D_{3} = 0 \\ Now, D_{1} = | \begin{array}{ccc} b_{1} & 1 & 3 \\ b_{2} & 2 & 6 \\ b_{3} & - 1 & - 3 \end{array} | \\ = b_{1} (- 6 + 6) - b_{2} (- 3 + 3) + b_{3} (6 - 6) \\ = 0 \\ D_{2} = | \begin{array}{ccc} 1 & b_{1} & 3 \\ 5 & b_{2} & 6 \\ - 2 & b_{3} & - 3 \end{array} | \\ = - b_{1} (- 15 + 12) + b_{2} (- 3 + 6) - b_{3} (6 - 15) \\ = 3 b_{1} + 3 b_{2} + 9 b_{3} = 0 \Rightarrow b_{1} + b_{2} + 3 b_{3} = 0 \end{aligned}$

not satisfies the Eq. (i)

It has no solution.

$= - 1 (- 20 + 20) - 2 (10 - 10) - 5 (- 4 + 4)$

$= 0$

Here, $b_{2} = - 2 b_{1}$ and $b_{3} = - b_{1}$ satisfies the Eq. (i) Planes are parallel.

(d) $\begin{aligned} D & = | \begin{array}{ccc} 1 & 2 & 5 \\ 2 & 0 & 3 \\ 1 & 4 & - 5 \end{array} | = 1 (0 - 12) - 2 (- 10 - 3) + 5 (8 - 0) \\ = 54 \end{aligned}$