====== Question 1, Exercise 2.3 ======
Solutions of Question 1 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 1(i)=====
Evaluate the determinant of the matrix $\left[\begin{array}{ccc}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2\end{array}\right]$.

** Solution. **
Let
\begin{align*}
A &= \left[\begin{array}{ccc}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2\end{array}\right]\\
|A|&=2(-2-2)-3(2-8)+1(1+4)\\
\implies |A|&=-8+18+5\\
\implies |A|&=15
\end{align*}
	


=====Question 1(ii)=====
Evaluate the determinant of the matrix $\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$.

** Solution. **

Let \begin{align*}
A&= \left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]\\
|A| &= cos \theta (cos \theta -0)+ sin \theta(sin \theta -0)+0\\
\implies |A| &= cos^2 \theta + sin^2 \theta\\
\implies |A| &= 1
\end{align*}


=====Question 1(iii)=====
Evaluate the determinant of the matrix $\left[\begin{array}{ccc}i & 3 & -2 i \\ 1 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$.

** Solution. **

\begin{align*}
A &= \left[\begin{array}{ccc} i & 3 & -2i \\ 1 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] \\
|A| &= i \left(3 \cdot 2 - 4 \cdot 1\right) - 3 \left(1 \cdot 2 - 4 \cdot 0\right) + (-2i) \left(1 \cdot 1 - 3 \cdot 0\right) \\
&= i (6 - 4) - 3 (2 - 0) + (-2i) (1 - 0) \\
&= i (2) - 3 (2) + (-2i) (1) \\
&= 2i - 6 - 2i \\
&= -6\end{align*}
Therefore, the determinant of the matrix \(\left[\begin{array}{ccc} i & 3 & -2i \\ 1 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]\) is:
$$|A| = -6$$

=====Question 1(iv)=====
Evaluate the determinant of the matrix $\left[\begin{array}{ccc}2+i & 1 & i \\ 0 & 2 & 1 \\ -3 i & 1 & 6\end{array}\right]$.
 
** Solution. **

\begin{align*}
A &= \left[\begin{array}{ccc} 2+i & 1 & i \\ 0 & 2 & 1 \\ -3i & 1 & 6 \end{array}\right] \\
|A| &= (2+i) \left( 2 \cdot 6 - 1 \cdot 1 \right) - 1 \left( 0 \cdot 6 - 1 \cdot (-3i) \right) + i \left( 0 \cdot 1 - 2 \cdot (-3i) \right) \\
&= (2+i) \left( 12 - 1 \right) - 1 \left( 0 + 3i \right) + i \left( 0 + 6i \right) \\
&= (2+i) \cdot 11 - 3i + 6i^2 \\
&= 22 + 11i - 3i + 6(-1) \quad \text{(since \(i^2 = -1\))} \\
&= 22 + 8i - 6 \\
&= 16 + 8i
\end{align*}
Therefore, the determinant of the matrix \(\left[\begin{array}{ccc} 2+i & 1 & i \\ 0 & 2 & 1 \\ -3i & 1 & 6 \end{array}\right]\) is: $$ |A| = 16 + 8i$$

	


====Go to ====

<text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-3-p2|Question 2 >]]</btn></text>