The matrix A=\(\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)=A’=-A’.Hence, a null matrix is both symmetric and skew-symmetric.

The number of rows (m) and number of columns (n) in given matrix A=\(\begin{bmatrix}3&5\\7&9\end{bmatrix}\) is 2. Therefore, order of matrix is 2×2(m×n).

A matrix is said to be skew-symmetric if it is equal to negative of its transpose i.e. A=-A’.

A matrix is called a scalar matrix if elements along diagonal of matrix are equal and are non-zero i.e. a_ij=k for i=j and a_ij=0 for i?j.
Therefore, matrix A=\(\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}\) is a scalar matrix.

The given matrix is A=\(\begin{bmatrix}8&5\\2&8\end{bmatrix}\)
Applying elementary operation R₁→R₁+2R₂, we get
\(\begin{bmatrix}8+2(2)&5+2(8)\\2&8\end{bmatrix}\)=\(\begin{bmatrix}12&21\\2&8\end{bmatrix}\).

Given that, 2\(\begin{bmatrix}5&x\\y-4&6\end{bmatrix}\)+\(\begin{bmatrix}-4&1\\3&2\end{bmatrix}\)=\(\begin{bmatrix}6&3\\10&14\end{bmatrix}\)
?\(\begin{bmatrix}2(5)-4&2x+1\\2(y-4)+3&2(6)+2\end{bmatrix}\)=\(\begin{bmatrix}6&3\\10&14\end{bmatrix}\)
Comparing two matrices, 2x+1=3, 2y-8=10
Solving two equations we get, x=1, y=9.

The given matrix A=\(\begin{bmatrix}2&3\\1&9\\5&2\end{bmatrix}\) has 3 rows and 2 columns. Therefore, order of matrix is 3×2.

The given statement is false. A matrix is non-invertible if it has all zeroes in one or more rows on L.H.S. This is because after applying all elementary operations on matrix, we should get an identity matrix on L.H.S. to obtain an inverse of given matrix, which is not possible if we obtain all zeroes in one or more rows.

Consider matrix A=\(\begin{bmatrix}5&1&3\\4&2&6\\5&4&2\end{bmatrix}\)
Using elementary row operations, we write A=IA
\(\begin{bmatrix}5&1&3\\4&2&6\\5&4&2\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)A
Applying R₂→5R₂-4R₁ and R₃→R₃-R₁
\(\begin{bmatrix}5&1&3\\0&6&18\\0&3&-1\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\-4&5&0\\-1&0&1\end{bmatrix}\)A
Applying R₁→R₂-6R₁ and R₃→R₂-2R₃
\(\begin{bmatrix}-30&0&0\\0&6&18\\0&0&20\end{bmatrix}\)=\(\begin{bmatrix}-10&5&0\\-4&5&0\\-2&5&-2\end{bmatrix}\)A
Applying R₂→20R₂-18R₃
\(\begin{bmatrix}-30&0&0\\0&120&0\\0&0&20\end{bmatrix}\)=\(\begin{bmatrix}-10&5&0\\-44&10&36\\-2&5&-2\end{bmatrix}\)A
Applying R₁→R₁/(-30), R₂→R₂/120, R₃→R₃/20
\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}\)A
A^-1=\(\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}\).

Minor matrix is not a type of matrix. Scalar, diagonal, symmetric are various type of matrices.