Q.Classify the matrix $\begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$ as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular, a symmetric or a skew-symmetric matrix.

The matrix has all off-diagonal entries zero (so it IS diagonal, upper-triangular, and lower-triangular), but its diagonal entries $1,0,5$ are unequal, so it is NOT a scalar matrix.

Given matrix: $\begin{bmatrix}1&0&0\\0&0&0\\0&0&5\end{bmatrix}$.

• Upper triangular: all entries below the main diagonal are 0. True here.
• Lower triangular: all entries above the main diagonal are 0. True here.
• Diagonal matrix: all off-diagonal entries are 0 (both triangular). True here. …

A diagonal matrix is a square matrix with all off-diagonal entries zero.

State the definition of an identity matrix.

A square matrix $A=[a_{ij}]_{n\times n}$ is called an identity matrix if

$a_{ij}=\begin{cases}1, & i=j\\ 0, & i e j\end{cases}$

i.e. all diagonal entries are $1$ and all non-diagonal entries are $0$. It is denoted $I_n$ (or simply $I$). For example,

…

A matrix is singular exactly when $\det A = 0$. For $A = \begin{bmatrix} 2 & 3 \\ a & 6 \end{bmatrix}$ this forces $12 - 3a = 0$, i.e. $a = 4$.

For a $2\times 2$ matrix $\begin{bmatrix} p & q \\ r & s \end{bmatrix}$ the determinant is $ps - qr$.

Here

$\det A = (2)(6) - (3)(a) = 12 - 3a.$

…

The matrix has zero everywhere off the main diagonal, making it diagonal (and so also upper- and lower-triangular), but its diagonal entries $1,0,5$ are unequal, so it is NOT a scalar matrix.

The matrix is $\begin{bmatrix}1&0&0\\0&0&0\\0&0&5\end{bmatrix}$.

A matrix is a diagonal matrix if every off-diagonal entry is 0 — true here, since all entries except the $(1,1)$, $(2,2)$, $(3,3)$ positions are already 0. This makes (d) true.

A diagonal matrix is automatically both upper triangular (nothing below the diagonal) and lower triangular (nothing above the diagonal), making (a) and (b) both true.

…

definition recall

The identity matrix $I_n$ of order $n$ is the square matrix in which every diagonal element is $1$ and every off-diagonal element is $0$, i.e. $I=[a_{ij}]$ where $a_{ij}=1$ if $i=j$ and $a_{ij}=0$ if $i\ne j$. It satisfies $AI=IA=A$ for any …