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I need to find $\det X$ where $$8GX=XX^T,\quad G=\left(\begin{matrix}5 & 4\\3 & 2\\\end{matrix}\right).$$

My answer is that the determinant of $X$ is $-128$ and that is correct but there is one more value of $\det X$ that can solve the equation and I belive that it is 0 but how do I write it a matrix that is equal to 0?

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  • You may simply let $X=0$. – 23rd Jun 10 '13 at 11:02
  • @Landscape How do you mean? – user1838781 Jun 10 '13 at 11:04
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    I think you want to find some $X$ with $\det X=0$ which satisfies the equation. Then you may choose $X=0$. – 23rd Jun 10 '13 at 11:06
  • @user1551 No I have asked it one more time due to I don't get any answers – user1838781 Jun 10 '13 at 11:18
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    @user1838781 What do you mean? I have given, in my answer to your previous question, two solutions $X$ such that $\det(X)=0$, namely $X=0$ and $X=\lambda uu^T$, where $(\lambda,u)$ is an eigenpair of $A$ and $|u|=1$. Why isn't it an answer? – user1551 Jun 10 '13 at 11:21
  • @user1551 Sure it is an answer but it is really hard to understand for someone that is not that good at mathematical vocbles – user1838781 Jun 10 '13 at 11:29
  • @user1838781 No offense, but if you can't even understand why $0$ and $\lambda uu^T$ are solutions, you should really brush up your basic knowledge in linear algebra. By the way, I didn't downvote your question. – user1551 Jun 10 '13 at 12:15

1 Answers1

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Note that you're asked to find $\det (X)$ and not $X$.

Your reasoning probably was as follows:

$$\begin{align} 8GX=XX^T &\implies \det (8GX)=\det (XX^T)\\ &\implies 8^2\det (G)\det(X)=\det(X)\det(X^T)\\ &\implies 64\cdot (-2)\det (X)=(\det(X))^2\\ &\implies \det (X)(\det(X)+128)=0\end{align}$$

from where you can simply conclude that $\det (X)\in \{0,-128\}$ and that's it.

Your actual question was how do I write a matrix that is equal to zero? Well, there's only one null matrix of each order and since by hypothesis $X$ is a $2\times 2$ square matrix, the answer to this question is $\begin{pmatrix} 0 &0\\ 0 & 0\end{pmatrix}$.

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