\(\displaystyle \begin{align*}&\text{Tr, that is to say the trace, of a square nxn matrix is the its diagonal elements:}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{For the matrix }A=\begin{bmatrix}1&6&-4&13\\9&19&1&-7\\-16&5&11&-3\\-17&-10&-14&-9\end{bmatrix}\\&\text{We find: }\\&Tr(A)=(1)+(19)+(11)+(-9)=22\end{align*}\)

\(\displaystyle \begin{align*}&\text{Tr, that is to say the trace, of a square nxn matrix is the its diagonal elements:}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{For the matrix }A=\begin{bmatrix}6&19&-11&7&-9&7&8&-18\\-10&-11&7&14&-6&12&7&-20\\4&-5&17&-20&-2&-3&-2&11\\-7&12&-1&-19&-13&9&-1&-14\\-7&4&-13&10&-11&17&-9&11\\-13&-9&-17&3&8&2&-3&6\\6&7&6&18&-12&9&-11&-16\\4&-2&-2&7&11&-6&7&-3\end{bmatrix}\\&\text{We find: }\\&Tr(A)=(6)+(-11)+(17)+(-19)+(-11)+(2)+(-11)+(-3)=-30\end{align*}\)

\(\displaystyle \begin{align*}&\text{The trace of a square nxn matrix is the sum of its diagonal elements:}\\&\sum_{i=1}^{n}A_{i,i}\\&\text{For the matrix }\begin{bmatrix}15&-10&-7&-16&18\\6&-1&6&2&6\\2&9&1&20&-12\\-16&-16&-18&-4&-2\\-6&11&5&11&18\end{bmatrix}\\&\text{This becomes: }\\&(15)+(-1)+(1)+(-4)+(18)=29\end{align*}\)

\(\displaystyle \begin{align*}&\text{Tr in the context of linear algebra represents the trace.}\\&\text{We can find this quantity for a square nxn matrix by summing the elements of its diagonal:}\\&\sum_{i=1}^{n}A_{i,i}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{Doing so we find for}\begin{bmatrix}1&15&-1&-4&7\\10&1&-6&-14&4\\-10&-19&10&-11&-2\\8&-6&10&-4&8\\8&-2&-20&-7&-3\end{bmatrix}\\&\text{We find:}\\&(1)+(1)+(10)+(-4)+(-3)=5\end{align*}\)

\(\displaystyle \begin{align*}&\text{Tr, that is to say the trace, of a square nxn matrix is the its diagonal elements:}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{For the matrix }A=\begin{bmatrix}13&10&-5&-12\\12&18&-7&7\\-3&14&11&-14\\15&20&1&16\end{bmatrix}\\&\text{We find: }\\&Tr(A)=(13)+(18)+(11)+(16)=58\end{align*}\)

\(\displaystyle \begin{align*}&\text{Tr, that is to say the trace, of a square nxn matrix is the its diagonal elements:}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{For the matrix }A=\begin{bmatrix}12&-7&1&-17&-16&-15&7\\0&-13&0&-14&-18&14&2\\18&8&3&13&16&20&-20\\15&5&20&1&-1&12&-11\\0&16&3&14&10&4&-10\\7&-17&5&7&9&16&20\\11&3&18&3&-20&-16&15\end{bmatrix}\\&\text{We find: }\\&Tr(A)=(12)+(-13)+(3)+(1)+(10)+(16)+(15)=44\end{align*}\)

\(\displaystyle A\) is a diagonal matrix - its only nonzero elements are on the main diagonal, from upper left to lower right - so its square can be taken by simply squaring the diagonal elements. Since

\(\displaystyle A = \begin{bmatrix} 3&0 & 0\\ 0& 6& 0\\ 0&0 &x \end{bmatrix}\)

it follows that

\(\displaystyle A ^{2}= \begin{bmatrix} 3^{2}&0 & 0\\ 0& 6^{2}& 0\\ 0&0 &x^{2} \end{bmatrix} = \begin{bmatrix} 9&0 & 0\\ 0& 36 & 0\\ 0&0 &x^{2} \end{bmatrix}\)

The trace of a matrix is equal to the sum of the elements in its main diagonal, so the trace of \(\displaystyle A^{2}\) is

\(\displaystyle Tr(A^{2}) = 9 + 36 + x^{2}\)

\(\displaystyle Tr(A^{2}) = x^{2} + 45\)

Since the trace is given to be 50, set this equal to 50 and solve for \(\displaystyle x\):

\(\displaystyle x^{2} + 45= 50\)

\(\displaystyle x^{2} =5\)

\(\displaystyle x= \pm \sqrt{5}\)

The trace of a matrix is equal to the sum of the elements in its main diagonal - the elements going from upper left to lower right. Therefore,

\(\displaystyle \textrm{Tr} (A) = 3+a+a+3\)

\(\displaystyle \textrm{Tr} (A) = 2a+6\)

Since the trace of \(\displaystyle A\) is equal to 10, set \(\displaystyle 2a+6\) equal to this and solve for \(\displaystyle a\):

\(\displaystyle 2a+6 = 10\)

\(\displaystyle 2a=4\)

\(\displaystyle a = 2\)

The trace of a matrix is equal to the sum of the elements in its main diagonal - the elements going from upper left to lower right. Therefore,

\(\displaystyle \textrm{Tr}(A) = a+a+2+5 = 2a + 7\)

\(\displaystyle A\) and \(\displaystyle B\) are both diagonal matrices - their only nonzero elements are on their main diagonals, from upper left to lower right - so their product can be found by multiplying their corresponding diagonal elements:

\(\displaystyle AB= \begin{bmatrix} 2&0 &0 &0 \\ 0& x& 0& 0\\ 0& 0&6 &0 \\ 0&0 &0 & -4 \end{bmatrix} \begin{bmatrix} -4&0 &0 &0 \\ 0& 2& 0& 0\\ 0& 0&x &0 \\ 0&0 &0 & -3 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} 2(-4)&0 &0 &0 \\ 0& x(2)& 0& 0\\ 0& 0&6(x) &0 \\ 0&0 &0 & -4(-3) \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} -8&0 &0 &0 \\ 0&2x & 0& 0\\ 0& 0&6x&0 \\ 0&0 &0 & 12\end{bmatrix}\)

The trace of a matrix is equal to the sum of the elements in its main diagonal, so the trace of \(\displaystyle AB\) is

\(\displaystyle \textrm{Tr}(AB) = -8 + 2x + 6x + 12 = 8x+4\)