====== Question 3, Exercise 2.3 ======
Solutions of Question 3 of Exercise 2.3 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 3(i)=====
Determine which of the matrix $\left[\begin{array}{ccc}3 & 1 & 2 \\ 2 & 3 & 1 \\ -4 & 1 & -3\end{array}\right]$ is singular and which are non-singular.

** Solution. **

\begin{align*}
A &= \left[\begin{array}{ccc} 3 & 1 & 2 \\ 2 & 3 & 1 \\ -4 & 1 & -3\end{array}\right]\end{align*}
The determinant of a \(3 \times 3\) matrix is calculated as follows:
\begin{align*}
|A| &= 3(3 \cdot (-3) - 1 \cdot 1) - 1(2 \cdot (-3) - 1 \cdot (-4)) + 2(2 \cdot 1 - 3 \cdot (-4)) \\
&= 3(-9 - 1) - 1(-6 + 4) + 2(2 + 12) \\
&= 3(-10) - 1(-2) + 2(14) \\
&= -30 + 2 + 28 \\
&= 0
\end{align*}
Since $|A| = 0$, the matrix $A$ is a singular matrix.

=====Question 3(ii)=====
Determine which of the matrix $\left[\begin{array}{ccc}3 & -1 & 2 \\ 2 & 0 & 1 \\ -1 & 5 & 1\end{array}\right]$ is singular and which are non-singular.

** Solution. **

\begin{align*}
A &= \left[\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 0 & 1 \\ -1 & 5 & 1\end{array}\right]\\
|A| &= 3(0 \cdot 1 - 1 \cdot 5) - (-1)(2 \cdot 1 - 1 \cdot (-1)) + 2(2 \cdot 5 - 0 \cdot (-1)) \\
&= 3(0 - 5) - (-1)(2 + 1) + 2(10 - 0) \\
&= 3(-5) + 1(3) + 2(10) \\
&= -15 + 3 + 20 \\
&= 8\end{align*}
Since $|A| = 8 \neq 0$, the matrix $A$ is a non-singular matrix.

=====Question 3(iii)=====
Determine which of the matrix $\left[\begin{array}{ccc}3 i & 1 & 2 \\ -4 & 1 & i \\ 2 & 0 & 1\end{array}\right]$ is singular and which are non-singular.

** Solution. **

\begin{align*}
A &= \left[\begin{array}{ccc}3i & 1 & 2 \\ -4 & 1 & i \\ 2 & 0 & 1\end{array}\right]\\
|A| &= 3i \cdot 1 - 1 \cdot (-4 - 2i) + 2 \cdot (-2) \\
&= 3i - (-4 - 2i) + 2 \cdot (-2) \\
&= 3i + 4 + 2i - 4 \\
&= 5i
\end{align*}
Since $|A| = 5i \neq 0$, the matrix $A$ is a non-singular matrix.

=====Question 3(iv)=====
Determine which of the matrix $\left[\begin{array}{ccc}2 & -i & 1 \\ i & 3 & -2 \\ -2+i & i+3 & -3\end{array}\right]$ is singular and which are non-singular.
 
** Solution. **

\begin{align*}
A &= \left[\begin{array}{ccc}2 & -i & 1 \\ i & 3 & -2 \\ -2+i & i+3 & -3\end{array}\right]\\
|A| &= 2 \cdot (-3 + 2i) - (-i) \cdot (-4 - i) + 1 \cdot 5 \\
&= 2(-3 + 2i) - (-i)(-4 - i) + 5 \\
&= -6 + 4i - (4i + i^2) + 5 \\
&= -6 + 4i - 4i - (-1) + 5 \\
&= -6 + 1 + 5 \\
&= 0
\end{align*}
Since $|A| = 0$, the matrix $A$ is a singular matrix.

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