Solve for the missing vector in a cross product

Question

How can I answer this question?

Solve for all vectors $A$ so that $( 1, 2, 1) \times A = (3, 1, -5)$.

Answer 1

Let $A=\left(\begin{array}{c} a\\ b\\ c \end{array}\right)$ so$$\left(\begin{array}{c} 3\\ 1\\ -5 \end{array}\right)=\left(\begin{array}{c} 1\\ 2\\ 1 \end{array}\right)\times\left(\begin{array}{c} a\\ b\\ c \end{array}\right)=\left(\begin{array}{c} 2c-b\\ a-c\\ b-2a \end{array}\right),$$which has general solution $a=1+c,\,b=2a-5=-3+2c$, i.e.$$A=\left(\begin{array}{c} 1\\ -3\\ 0 \end{array}\right)+c\left(\begin{array}{c} 1\\ 2\\ 1 \end{array}\right).$$

Answer 2

Write the cross product using the matrix notation. https://en.wikipedia.org/wiki/Cross_product#Matrix_notation

Solve the system you obtain!

Answer 3

If you know how to solve simultaneous equation (either by substitution or elimination) you can find the components of the vector A after applying the definition of the cross product of the two vectors.

$ $$ \begin{pmatrix} i & j & k \\ 1 & 2 & 1 \\ a_1 & a_2 & a_3 \\ \end{pmatrix} $$ $

$(2\times a_3 - a_2) \hat i - (a_3 - a_1)\hat j + (a_2 - 2 \times a_1)\hat k = (3) \hat i + (1)\hat j - (5)\hat k $

compare/equate the components in the above equations.

from equation 1 $a_3 = \frac{3 + a_2}{2} $

substitute this into 2nd equation, $ a_3 = a_1 - 1$

$\frac{3 + a_2}{2} = a_1 - 1$

solve the above equation with the third equation $ a_2 - 2\times a_1 = -5$ as simultaneous equations in two variables and you will encounter the problem of

$2a_1 - 2 = 2a_1 - 2$ so that there is no solution. This is the same for $a_2$ and $a_1$

I could write the above system of linear equations in matrix form as

$ $$ \begin{pmatrix} 0 & -1 & 2 \\ 1 & 0 & -1 \\ -2 & 1 & 0 \\ \end{pmatrix} $$ $ $ $$ \begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ \end{pmatrix} $$ $ $=$ $ $$ \begin{pmatrix} 3 \\ 1 \\ -5 \\ \end{pmatrix} $$ $

If I try to invert the 3x3 matrix on the left I find that it is singular (non-invertible) which suggests that there is no vector $A$ that can satisfy this linear system of equations. The vector $<3,1,-5>$ is outside the column space of the matrix so it cannot be reached.

No such vector A exists.