Question

Let A be a 3×3 matrix and suppose we know that −2a1+3a2−5a3=0 where a1,a2 and a3 are the columns of A. Write a non-trivial solution to the system Ax=0

Answer 1

One of the obvious non-trivial solutions is
`(x_1, x_2, x_3)=(-2, 3, -5)`.

Explanation:

Suppose the matrix A is as follows:

`A=\left[\begin{array}{ccc}a_(11)&a_(12)&a_(13)\\a_(21)&a_(22)&3_(23)\\a_(31)&a_(32)&a_(33)\end{array}\right]`

The observed system
`Ax=0` after multiplying looks like this

`Ax=0 \iff \left[\begin{array}{ccc}a_(11)&a_(12)&a_(13)\\a_(21)&a_(22)&a_(23)\\a_(31)&a_(32)&a_(33)\end{array}\right] \cdot \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] =0 \iff \\ \\a_(11)x_1+a_(12)x_2+a_(13)x_3=0\\a_(21)x_1+a_(22)x_2+a_(23)x_3=0\\a_(31)x_1+a_(32)x_2+a_(33)x_3=0\\\\`

Since we now that
`-2A_1+3A_2-5A_3=0`, where
`A_i\ ,\ i=1, 2, 3` are the columns of the matrix A, we actually know this:

`-2\cdot \left[\begin{array}{ccc}a_(11)\\a_(21)\\a_(31)\end{array}\right] +3\cdot \left[\begin{array}{ccc}a_(12)\\a_(22)\\a_(32)\end{array}\right] -5\cdot \left[\begin{array}{ccc}a_(13)\\a_(23)\\a_(33)\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right]`

Once we multiply and sum up these 3 by 1 matrices, we get that these equations hold:

`-2a_(11)+3a_(12)-5a_(13)=0\\-2a_(21)+3a_(22)-5a_(23)=0\\-2a_(31)+3a_(32)-5a_(33)=0`

This actually means that the solution to the previously observed system of equations (or equivalently, our system
`Ax=0`) has a non-trivial solution
`(x_1, x_2, x_3)=(-2, 3, -5)`.