Q.(a) Show that $\begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{pmatrix}$ is a non-singular matrix.

Compute each $2\times2$ determinant; only (b) is non-zero, giving $2$.

For a $2\times2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc$.

• (a) $\begin{vmatrix} x & 1 \\ x^2+x & x+1 \end{vmatrix}=x(x+1)-1\cdot(x^2+x)=x^2+x-x^2-x=0$.
• (b) $\begin{vmatrix} x+3 & x+1 \\ x+4 & x+2 \end{vmatrix}=(x+3)(x+2)-(x+1)(x+4)=(x^2+5x+6)-(x^2+5x+4)=2$.
• (c) $\begin{vmatrix} x & 1 \\ 2x & 2 \end{vmatrix}=2x-2x=0$ (second row is $2\times$ first). …

One-word term: Singular matrix.

For a square matrix $A$, if $|A|=0$ the matrix is singular (it has no inverse). If $|A|\neq 0$ it is non-singular/invertible. This is a stan …

Correction: "third" should be second order.

The order of a determinant equals the number of its rows (= columns). $\begin{vmatrix} a & d \\ b & c \end{vmatrix}$ is $2\times2$, hence a second order determinant …

Determinant $\neq 0 \Rightarrow$ non-singular matrix.

A square matrix $A$ is classified by the value of its determinant $|A|$:

• If $|A|=0$, the matrix is singular (it has no inverse). …

$\begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix}=10$.

For a $2\times 2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc$.

Here $a=2,\ b=3,\ c=0,\ d=5$:

…

Vertical bars + $3\times 3$ size ⇒ option (a).

Two things identify a third-order determinant:

1. It must be enclosed in vertical bars $\begin{vmatrix}\;\end{vmatrix}$ — options (b), (c), (d) use square brackets, which denote matrices, not determinants. …

Correct option: (b) 4.

A determinant of order $n$ corresponds to an $n \times n$ arrangement and so contains $n^2$ elements. For order two, …