3.2.1: Exercises 3.2

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In Exercises \(\PageIndex{1}\) - \(\PageIndex{15}\), find the trace of the given matrix.

Exercise \(\PageIndex{1}\)

\(\left[\begin{array}{cc}{1}&{-5}\\{9}&{5}\end{array}\right]\)

Answer: \(6\)

Exercise \(\PageIndex{2}\)

\(\left[\begin{array}{cc}{-3}&{-10}\\{-6}&{4}\end{array}\right]\)

Answer: \(1\)

Exercise \(\PageIndex{3}\)

\(\left[\begin{array}{cc}{7}&{5}\\{-5}&{-4}\end{array}\right]\)

Answer: \(3\)

Exercise \(\PageIndex{4}\)

\(\left[\begin{array}{cc}{-6}&{0}\\{-10}&{9}\end{array}\right]\)

Answer: \(3\)

Exercise \(\PageIndex{5}\)

\(\left[\begin{array}{ccc}{-4}&{1}&{1}\\{-2}&{0}&{0}\\{-1}&{-2}&{-5}\end{array}\right]\)

Answer: \(-9\)

Exercise \(\PageIndex{6}\)

\(\left[\begin{array}{ccc}{0}&{-3}&{1}\\{5}&{-5}&{5}\\{-4}&{1}&{0}\end{array}\right]\)

Answer: \(-5\)

Exercise \(\PageIndex{7}\)

\(\left[\begin{array}{ccc}{-2}&{-3}&{5}\\{5}&{2}&{0}\\{-1}&{-3}&{1}\end{array}\right]\)

Answer: \(1\)

Exercise \(\PageIndex{8}\)

\(\left[\begin{array}{ccc}{4}&{2}&{-1}\\{-4}&{1}&{4}\\{0}&{-5}&{5}\end{array}\right]\)

Answer: \(10\)

Exercise \(\PageIndex{9}\)

\(\left[\begin{array}{ccc}{2}&{6}&{4}\\{-1}&{8}&{-10}\end{array}\right]\)

Answer: Not defined; the matrix must be square.

Exercise \(\PageIndex{10}\)

\(\left[\begin{array}{cc}{6}&{5}\\{2}&{10}\\{3}&{3}\end{array}\right]\)

Answer: Not defined; the matrix must be square.

Exercise \(\PageIndex{11}\)

\(\left[\begin{array}{cccc}{-10}&{6}&{-7}&{-9}\\{-2}&{1}&{6}&{-9}\\{0}&{4}&{-4}&{0}\\{-3}&{-9}&{3}&{-10}\end{array}\right]\)

Answer: \(-23\)

Exercise \(\PageIndex{12}\)

\(\left[\begin{array}{cccc}{5}&{2}&{2}&{2}\\{-7}&{4}&{-7}&{-3}\\{9}&{-9}&{-7}&{2}\\{-4}&{8}&{-8}&{-2}\end{array}\right]\)

Answer: \(0\)

Exercise \(\PageIndex{13}\)

\(I_{4}\)

Answer: \(4\)

Exercise \(\PageIndex{14}\)

\(I_{n}\)

Answer: \(n\)

Exercise \(\PageIndex{15}\)

A matrix \(A\) that is skew symmetric.

Answer: \(0\)

In Exercises \(\PageIndex{16}\) - \(\PageIndex{19}\), verify Theorem 3.2.1 by:

Showing that \(\text{tr}(A)+\text{tr}(B)=\text{tr}(A+B)\) and
Showing that \(\text{tr}(AB)=\text{tr}(BA)\).

Exercise \(\PageIndex{16}\)

\(A=\left[\begin{array}{cc}{1}&{-1}\\{9}&{-6}\end{array}\right],\quad B=\left[\begin{array}{cc}{-1}&{0}\\{-6}&{3}\end{array}\right]\)

Answer

\(\text{tr}(A)=-5;\:\text{tr}(B)=-4;\:\text{tr}(A+B)=-9\)
\(\text{tr}(AB)=23=\text{tr}(BA)\)

Exercise \(\PageIndex{17}\)

\(A=\left[\begin{array}{cc}{0}&{-8}\\{1}&{8}\end{array}\right],\quad B=\left[\begin{array}{cc}{-4}&{5}\\{-4}&{2}\end{array}\right]\)

Answer

\(\text{tr}(A)=8;\:\text{tr}(B)=-2;\:\text{tr}(A+B)=6\)
\(\text{tr}(AB)=53=\text{tr}(BA)\)

Exercise \(\PageIndex{18}\)

\(A=\left[\begin{array}{ccc}{-8}&{-10}&{10}\\{10}&{5}&{-6}\\{-10}&{1}&{3}\end{array}\right],\quad B=\left[\begin{array}{ccc}{-10}&{-4}&{-3}\\{-4}&{-5}&{4}\\{3}&{7}&{3}\end{array}\right]\)

Answer

\(\text{tr}(A)=0;\:\text{tr}(B)=-12;\:\text{tr}(A+B)=-12\)
\(\text{tr}(AB)=86=\text{tr}(BA)\)

Exercise \(\PageIndex{19}\)

\(A=\left[\begin{array}{ccc}{-10}&{7}&{5}\\{7}&{7}&{-5}\\{8}&{-9}&{2}\end{array}\right],\quad B=\left[\begin{array}{ccc}{-3}&{-4}&{9}\\{4}&{-1}&{-9}\\{-7}&{-8}&{10}\end{array}\right]\)

Answer

\(\text{tr}(A)=-1;\:\text{tr}(B)=6;\:\text{tr}(A+B)=5\)
\(\text{tr}(AB)=201=\text{tr}(BA)\)