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I need to find all values $\lambda$ at which the vector $(7,-2, \lambda)$ is linearly expressed through vectors $(2,3,5), (3,7,8), (1,-6,1)$

I can't go to the next step after building a system:

$2x_{1} + 3x_{2} + x_{3} = 7$

$3x_{1} + 7x_{2} + 8x_{3} = -2$

$5x_{1} + 8x_{2} + x_{3} = \lambda$

manabou11
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  • The way you've set up the system seems wrong. The coefficients of $x_3$ ought to reflect the components of the third vector $(1,-6,1)$, right? – hardmath May 29 '20 at 02:40

2 Answers2

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Reduce the augmented matrix: $$ \left[\begin{array}{ccc|c} 2&3&1&7\\ 3&7&-6&-2\\ 5&8&1&\lambda \end{array}\right] \sim \left[\begin{array}{ccc|c} 1&0&5&11\\ 0&1&-3&-5\\ 0&0&0&15 - \lambda \end{array}\right]. $$

Therefore, the only possible $\lambda$ that works is $\lambda = 15$, and it does in fact work: if $\lambda = 15$, then $(7,-2,15) = (-5z + 11)(2,3,5)+(3z-5)(3,7,8)+z(1,-6,1)$ where $z$ can be any number.

Why does this work? Reducing a matrix doesn't change the set of solutions, so this justifies the reduction. The system is inconsistent if the augmented matrix has a row of the form $\left[\begin{array}{cccc|c}0 & 0 & \cdots & 0 & x \end{array}\right]$ for some $x \neq 0$. So, we must have $\lambda = 15$. For $\lambda = 15$, the matrix has the last row as zero, with one free variable. For a vector $(x,y,z)$ to be a solution of the system means that $x + 5z = 11$ and $y -3z = -5$. This can be rewritten as $x = -5z + 11$ and $y = 3z - 5$. In other words, $z$ is free to be any number and $z$ determines $x$ and $y$ by those two equations. So any vector of the form $v = (-5z+11, 3z-5, z)$ is a solution. By definition of matrix-vector multiplication, $Av$ is a linear combination of the columns of $A$ with the coefficients coming from $v$. Therefore, we can write out the equation $Av = (7,-2,15)$ as $(-5z + 11)(2,3,5)+(3z-5)(3,7,8)+z(1,-6,1) = (7,-2,15)$.

twosigma
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  • could you please describe reducing in more details? – manabou11 May 29 '20 at 23:46
  • @manabou11 Do you know Gaussian elimination? – twosigma May 30 '20 at 00:04
  • yeah but it doesn't work out for me, could you show a logic of your process? – manabou11 May 30 '20 at 00:29
  • You can try the following and let me know what you get: add $-R_1$ to $R_2$; then add $-2R_2$ to $R_1$ and $-5R_2$ to $R_3$; then divide $R_1$ by $-5$ and divide $R_3$ by $-12$; then add $-R_1$ to $R_3$; then divide $R_3$ by $12$; then add $-4R_1$ to $R_2$; then swap $R_1$ and $R_2$. – twosigma May 30 '20 at 00:39
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To make the calculus easy, you find that your vectors family's rank is $2$ while the cardinal is $3$. Then you can choose any $2$ of your vectors, since none of them is equal to zero, or directly proportional to each other. Then with a little system of $2$ equations, for example: $2a + 3b = 7$ and $3a + 7b = -2$, you find the values of $a = 11$ and $b = -5$, which forces your $\lambda$ being equal to $15$. So you only have one value :P.

Hope it had been useful.

csch2
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Tortilla
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  • Nice way to simplify the calculations. – twosigma May 29 '20 at 00:58
  • To expand on this answer: we can check that the span of the three vectors has dimension 2, and no two vectors are in the span of each other. This means any 2 of the 3 vectors span the column space of the $3 \times 3$ matrix whose columns are those 3 vectors. For example, take the vectors $(2,3,5)$ and $(3,7,8)$. So, the vector $(7,-2,\lambda)$ is in the span of the original 3 vectors if and only if it is in the span of these 2 vectors. We can represent this as a system of 3 equations in 2 unknowns. But we can look at the sub-system consisting of the first two equations; because... – twosigma May 29 '20 at 01:43
  • ...if all 3 equations are satisfied, then certainly the first two are, i.e. $2a + 3b = 7$ and $3a + 7b = -2$, from which we necessarily conclude that $a = 11$ and $b = -5$. Plugging this into the 3rd equation forces $\lambda$ to be $15$. – twosigma May 29 '20 at 01:45