Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Could someone please explain to me how they found $x_1$? By my calculations I got $x_2$ & $x_3=0$ and $0=0$.

enter image description here

share|improve this question
    
From the matrix equation we get $0 \cdot x_1 = 0$, so the equation will hold for any $x_1$. –  please delete me Jun 5 '11 at 20:16

2 Answers 2

up vote 2 down vote accepted

The system says that $x_2$ must equal $0$, and $x_3$ must equal $0$; however, there are no constraints on $x_1$ (the third equation you got, which is "$0=0$", is always satisfied, so it imposes no conditions whatsoever on the solutions).

That means that for $\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)$ to be a solution, you need $x_2=0$, $x_3=0$, but $x_1$ can be anything; that is, the solutions are all vectors of the form $$\left(\begin{array}{c}t\\0\\0\end{array}\right),$$ with $t$ arbitrary.

share|improve this answer

Well, if you carry out the matrix multiplication, you get: $$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_2 \\ x_3 \\ 0 \end{pmatrix} $$ for this to be equal to the zero vector, you indeed need $x_2 = 0$ and $x_3 = 0$. So far so good.

Now, notice that $x_1$ is not in any element of the vector on the right hand side, so there are no constraints on $x_1$. Therefore, $x_1$ can be any number, here called $t$.

You can check your solution; do the multplication $$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} t \\ 0 \\ 0 \end{pmatrix} $$ and see what you get.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.