\(\displaystyle \begin{align*}&\text{Observation of the matrix given will reveal that it is symmetric.}\\&\text{In other words, it is equal to its transpose. If a matrix is}\\&\text{symmetric, then it is also positive semi-definite if all of}\\&\text{its eigenvalues are non-negative. So then, let us compute these}\\&\text{eigenvalues:}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&16\lambda + \lambda ^{2} - 381\\&\lambda_{1}=-29.10;\lambda_{2}=13.10\\&\text{The matrix is not positive semi-definite.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&(-1)\cdot (7\lambda ^{2} - 845\lambda + \lambda ^{3} + 7671)\\&\lambda_{1}=-36.20;\lambda_{2}=13.50;\lambda_{3}=15.70\\&\text{The matrix is not positive semi-definite.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}13&-4\\-4&3\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}13 - \lambda&-4\\-4&3 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\lambda ^{2} - 16\lambda + 23\\&\lambda_{1}=1.60;\lambda_{2}=14.40\\&\text{The matrix is positive semi-definite.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Observation of the matrix given will reveal that it is symmetric.}\\&\text{In other words, it is equal to its transpose. If a matrix is}\\&\text{symmetric, then it is also positive semi-definite if all of}\\&\text{its eigenvalues are non-negative. So then, let us compute these}\\&\text{eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}-9&-15&8\\-15&12&6\\8&6&6\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}- \lambda - 9&-15&8\\-15&12 - \lambda&6\\8&6&6 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&(-1)\cdot (\lambda ^{3} - 9\lambda ^{2} - 415\lambda + 3882)\\&\lambda_{1}=-20.49;\lambda_{2}=9.45;\lambda_{3}=20.04\\&\text{The matrix is not positive semi-definite.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}12&-5\\-5&9\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}12 - \lambda&-5\\-5&9 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\lambda ^{2} - 21\lambda + 83\\&\lambda_{1}=5.28;\lambda_{2}=15.72\\&\text{The matrix is positive semi-definite.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}4&14\\14&9\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}4 - \lambda&14\\14&9 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\lambda ^{2} - 13\lambda - 160=0\\&\lambda_{1}=-7.72;\lambda_{2}=20.72\\&\text{The matrix is not positive semi-definite.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}6&5&12\\5&19&3\\12&3&7\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}6 - \lambda&5&12\\5&19 - \lambda&3\\12&3&7 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&(-1)\cdot (111\lambda - 32\lambda ^{2} + \lambda ^{3} + 1807)=0\\&\lambda_{1}=-5.61;\lambda_{2}=13.18;\lambda_{3}=24.43\\&\text{The matrix is not positive semi-definite.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Observation of the matrix given will reveal that it is symmetric.}\\&\text{In other words, it is equal to its transpose. If a matrix is}\\&\text{symmetric, then it is also positive semi-definite if all of}\\&\text{its eigenvalues are non-negative. So then, let us compute these}\\&\text{eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}-14&2\\2&-15\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}- \lambda - 14&2\\2&- \lambda - 15\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&29\lambda + \lambda ^{2} + 206=0\\&\lambda_{1}=-16.56;\lambda_{2}=-12.44\\&\text{The matrix is not positive semi-definite.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}15&-15\\-15&20\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}15 - \lambda&-15\\-15&20 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\lambda ^{2} - 35\lambda + 75=0\\&\lambda_{1}=2.29;\lambda_{2}=32.71\\&\text{The matrix is positive semi-definite.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}17&-5&9\\-5&18&-5\\9&-5&18\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}17 - \lambda&-5&9\\-5&18 - \lambda&-5\\9&-5&18 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&(-1)\cdot (805\lambda - 53\lambda ^{2} + \lambda ^{3} - 3625)=0\\&\lambda_{1}=8.48;\lambda_{2}=14.01;\lambda_{3}=30.51\\&\text{The matrix is positive semi-definite.}\end{align*}\)