If A- a1 b1 c1- a2 b2 c2- a3 b3 c3 - and -A- 0- then the system of equations a1x - b1y - c1 z -0- a2 x - b2 y -c2 z -0 and a3 x - b3 y -c3z-0 has

The correct option is A only one solutionNote that A is invertible as |A | ≠ 0 .Now, nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nb ...

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Byju's Answer

Standard XII

Mathematics

Determinant

If A=[ a1 b...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ⎤ ⎥ ⎦$ and $| A | \neq 0$ , then the system of equations $a_{1} x + b_{1} y + c_{1} z = 0, a_{2} x + b_{2} y + c_{2} z = 0$ and $a_{3} x + b_{3} y + c_{3} z = 0$ has

only one solution

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infinite number of solutions

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no solution

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more than one but finite number of solutions.

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Solution

The correct option is A only one solution
Note that $A$ is invertible as $| A | \neq 0$ .
Now, $A X = 0 \Rightarrow A^{- 1} (A X) = 0$
$\Rightarrow (A^{- 1} A) X = 0 o r I X = 0 o r X = 0$
$∴ x = y = z = 0$ is the only solution of the system of equations.
Hence, option A.

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Similar questions

Q. If

A = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & a_{3} \end{matrix} ∣ ∣ ∣ ∣ \neq 0,

then the system of equations

a_{1} x + b_{1} y + c_{1} z = 0, a_{2} x + b_{2} y + c_{2} z = 0

and

a_{3} x + b_{3} y + c_{3} z = 0

has:

Q. Consider the system of equations

a_{1} x + b_{1} y + c_{1} z = 0

a_{2} x + b_{2} y + c_{2} z = 0

a_{3} x + b_{3} y + c_{3} z = 0

i f ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{3} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0,

then the system has

Q. Eliminate x, y, z from the equations

a_{1} x + b_{1} y + c_{1} z = 0 . . . (1),

a_{2} x + b_{2} y + c_{2} z = 0 . . . (2),

a_{3} x + b_{3} y + c_{3} z = 0 . . . (3)

Three lines given by $L_{1} : a_{1} x + b_{1} y + c_{1} = 0, L_{2} : a_{2} x + b_{2} y + c_{2} = 0 and L_{3} : a_{3} x + b_{3} y + c_{3} = 0$ are always concurrent given that

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

Q. Consider the system of equations:
a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z = 0,
if

|\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}|

= 0, then the system has
(a) more than two solutions
(b) one trivial and one non-trivial solutions
(c) no solution
(d) only trivial solution (0, 0, 0)

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If A=⎡⎢⎣a1b1c1a2b2c2a3b3c3⎤⎥⎦ and |A|≠0, then the system of equations a1x+b1y+c1z=0,a2x+b2y+c2z=0 and a3x+b3y+c3z=0 has

The correct option is A only one solutionNote that A is invertible as |A|≠0.Now, AX=0⇒A−1(AX)=0⇒(A−1A)X=0orIX=0orX=0∴x=y=z=0 is the only solution of the system of equations.Hence, option A.

If $A = ⎡ ⎢ ⎣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ⎤ ⎥ ⎦$ and $| A | \neq 0$ , then the system of equations $a_{1} x + b_{1} y + c_{1} z = 0, a_{2} x + b_{2} y + c_{2} z = 0$ and $a_{3} x + b_{3} y + c_{3} z = 0$ has

The correct option is A only one solution
Note that $A$ is invertible as $| A | \neq 0$ .
Now, $A X = 0 \Rightarrow A^{- 1} (A X) = 0$
$\Rightarrow (A^{- 1} A) X = 0 o r I X = 0 o r X = 0$
$∴ x = y = z = 0$ is the only solution of the system of equations.
Hence, option A.