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Chapter 10: Problem 280

Solve \(x+3 y-2 z=0\) \(2 x-y+4 z=0\) \(x-11 y+14 z=0\)

Short Answer

Expert verified
After performing the above steps, the ratio of \(x : y : z\) turns out to be \(12 : \frac{8}{7} : 1\). In other words, \(x = 12k, y = \frac{8k}{7}, z = k\) for an arbitrary number \(k\).

Step by step solution

01

Equating the first and second equation

Subtract the first equation \(x+3y-2z=0\) from the second equation \(2x-y+4z=0\), This gives \(x - 4y + 6z = 0\)
02

Equating the first and third equation

Subtract the first equation \(x+3y-2z=0\) from the third equation \(x-11y+14z=0\), This gives \( -14y + 16z = 0\)
03

Solving for y

From \(-14y + 16z = 0\), we can derive the value of \(y\) as \(y = \frac{8}{7}z\)
04

Substituting y in Step 1 equation

Substitute value of \(y\) from Step 3 into the equation derived in Step 1 i.e \(x - 4y + 6z = 0\) to get value of \(x\) = \(12z\)
05

Substituting x and y in original equation

Substitute \(x = 12z\) and \(y = \frac{8}{7}z\) in any of the original equations to find the ratio of \(z\)

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Most popular questions from this chapter

EVALUATING DETERMINANTS. $$ \left|\begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 3 & 6 & 10 & 15 \\ 1 & 4 & 10 & 20 & 35 \\ 1 & 5 & 15 & 35 & 69 \end{array}\right| $$

EVALUATING DETERMINANTS. $$ \left|\begin{array}{llll} 1 & a & a^{2} & a^{3}+b c d \\ 1 & b & b^{2} & b^{3}+c d a \\ 1 & c & c^{2} & c^{3}+a b d \\ 1 & d & d^{2} & d^{3}+a b c \end{array}\right| $$

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{ccc} a x-b y-c z & a y+b x & c x+a z \\ a y+b x & b y-c z-a x & b z+c y \\ c x+a z & b z+c y & c z-a x-b y \end{array}\right|=\left(x^{2}+y^{2}+z^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)(a x+b y+c z) $$

Find all values of \(k\) for which the following system possesses a non-trivial solution \(x+k y+3 z=0\) \(k x+2 y+2 z=0\) \(2 x+3 y+4 z=0\)

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{ccc} y+z & x & x \\ y & z+x & y \\ z & z & x+y \end{array}\right|=4 x y z $$
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