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Given matrics $A = \begin{bmatrix}1 && 5 && 2 \\ -1 && 0 && 1 \\ 3 && 2 && 4\end{bmatrix}$ and $B = \begin{bmatrix}6 && 1 && 3 \\ -1 && 1 && 2 \\ 4 && 1 && 3\end{bmatrix}$, find $-3\mathop{Tr}(A-3B)$.

I am not exactly sure what is $-3\mathop{Tr}(\cdot)$, but I did till $(A-3B)$ part.

Where $3B = \begin{bmatrix}18 && 3 && 9 \\ -3 && 3 && 6 \\ 12 && 3 && 9\end{bmatrix}$, and $A-3B$ is $\begin{bmatrix}-17 && 2 && -7 \\ 2 && -3 && -5 \\ -9 && -1 && -5\end{bmatrix}$

How do I apply the $-3\mathop{Tr}(\cdot)$ to $(A-3B)$?

Thanks for the help!

Rajat
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Qing Yong
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2 Answers2

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I suppose that you know that the trace of a matrix is the sum of the elements on the principal diagonal. You have correctly found $A-3B$, so: trace$(A-3B)=-25$ and $-3 \mbox{trace} (A-3B)= -3\times (-25)$ .

Emilio Novati
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Hint: for a matrix $A\in \mathbb R^{n\times n}$ the trace of $A$ is a real number. Thus $3\operatorname{trace}(A-3B)$ is just...can you take it from here?

Hirshy
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