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Let {S}be the set of all column matrices {[{b}_{{1}}{b}_{{2}}{

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Let $S$ be the set of all column matrices $[b_{1} b_{2} b_{3}]$ such that $b_{1}, b_{2}, b_{3} \in R$ and the system of equations (in real variable) $- x + 2 y + 5 z = b_{1}$ $2 x - 4 y + 3 z = b_{2}$ $x - 2 y + 2 z = b_{3}$ has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $[b_{1} b_{2} b_{3}] \in S$ ?(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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Emily Discussed

Let

S

be the set of all column matrices

[b_{1} b_{2} b_{3}]

such that

b_{1}, b_{2}, b_{3} \in R

and the system of equations (in real variable)

- x + 2 y + 5 z = b_{1}

2 x - 4 y + 3 z = b_{2}

x - 2 y + 2 z = b_{3}

has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each

[b_{1} b_{2} b_{3}] \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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Question Text	Let $S$ be the set of all column matrices $[b_{1} b_{2} b_{3}]$ such that $b_{1}, b_{2}, b_{3} \in R$ and the system of equations (in real variable) $- x + 2 y + 5 z = b_{1}$ $2 x - 4 y + 3 z = b_{2}$ $x - 2 y + 2 z = b_{3}$ has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $[b_{1} b_{2} b_{3}] \in S$ ?(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 $x + 4 y - 5 z = b_{3}$
Answer Type	Video solution: 1
Upvotes	150