SATHEE: Matrices and Determinants 4 Question 29

Matrices and Determinants 4 Question 29

30. Show that the system of equations, $3 x - y + 4 z = 3$ , $x + 2 y - 3 z = - 2$ and $6 x + 5 y + λ z = - 3$ has atleast one solution for any real number $λ \neq - 5$ . Find the set of solutions, if $λ = - 5$ .

$(1983, 5 M)$

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Answer:

Correct Answer: 30. ( $k = 0$ , the given system has infinitely many solutions)

Solution:

The given system of equations

$\begin{matrix} 3 x - y + 4 z = 3 \\ x + 2 y - 3 z = - 2 \\ 6 x + 5 y + λ z = - 3 \end{matrix}$

has atleast one solution, if $Δ \neq 0$

$∴ Δ = | \begin{array}{ccc} 3 & - 1 & 4 \\ 1 & 2 & - 3 \\ 6 & 5 & λ \end{array} | \neq 0$

$\begin{array}{lc} \Rightarrow & 3 (2 λ + 15) + 1 (λ + 18) + 4 (5 - 12) \neq 0 \\ \Rightarrow & 7 (λ + 5) \neq 0 \\ \Rightarrow & λ \neq - 5 \end{array}$

Let $z = - k$ , then equations become

and $\begin{matrix} 3 x - y = 3 - 4 k \\ x + 2 y = 3 k - 2 \end{matrix}$

On solving, we get

$x = \frac{4 - 5 k}{7}, y = \frac{13 k - 9}{7}, z = k$

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Matrices and Determinants 4 Question 29

30. Show that the system of equations, $3 x - y + 4 z = 3$ , $x + 2 y - 3 z = - 2$ and $6 x + 5 y + λ z = - 3$ has atleast one solution for any real number $λ \neq - 5$ . Find the set of solutions, if $λ = - 5$ .

Answer:

इंजीनियरिंग की तैयारी

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एनसीईआरटी पुस्तकें और समाधान

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नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Matrices and Determinants 4 Question 29

30. Show that the system of equations, 3x−y+4z=3, x+2y−3z=−2 and 6x+5y+λz=−3 has atleast one solution for any real number λ≠−5. Find the set of solutions, if λ=−5.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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30. Show that the system of equations, $3 x - y + 4 z = 3$ , $x + 2 y - 3 z = - 2$ and $6 x + 5 y + λ z = - 3$ has atleast one solution for any real number $λ \neq - 5$ . Find the set of solutions, if $λ = - 5$ .