====== Question 2, Exercise 2.3 ======
Solutions of Question 2 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 2(i)=====
 Evaluate the determinant of the following matrix $\left[\begin{array}{lll}3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0\end{array}\right]$ using cofactor method.\\

** Solution. **

The elements of \(R_1\) are \(a_{11} = 3\), \(a_{12} = 2\), and \(a_{13} = 3\). Now we find their corresponding cofactors.
\begin{align*}
A &= \left[\begin{array}{ccc} 3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0 \end{array}\right]\\
& A_{11} = (-1)^{1+1} \left|\begin{array}{cc}
5 & 1 \\
1 & 0
\end{array}\right| = (-1)^{2} (5 \cdot 0 - 1 \cdot 1) = (1) (-1) = -1 \\
& A_{12} = (-1)^{1+2} \left|\begin{array}{cc}
4 & 1 \\
2 & 0
\end{array}\right| = (-1)^{3} (4 \cdot 0 - 1 \cdot 2) = (-1) (-2) = 2 \\
& A_{13} = (-1)^{1+3} \left|\begin{array}{cc}
4 & 5 \\
2 & 1
\end{array}\right| = (-1)^{4} (4 \cdot 1 - 5 \cdot 2) = (1) (4 - 10) = -6
\end{align*}
Now, we use these cofactors to find the determinant:
\begin{align*}
\det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\
&= 3(-1) + 2(2) + 3(-6) \\
&= -3 + 4 - 18 \\
&= -17
\end{align*}
Thus, the determinant of the matrix \(\left[\begin{array}{ccc} 3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0 \end{array}\right]\) is:
$-17$

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

** Solution. **

The elements of \(R_1\) are \(a_{11} = 2\), \(a_{12} = 3\), and \(a_{13} = -1\). Now we find their corresponding cofactors.
\begin{align*}
A &= \left[\begin{array}{ccc} 2 & 3 & -1 \\ -1 & 0 & 2 \\ 3 & 1 & 4 \end{array}\right]\\
& A_{11} = (-1)^{1+1} \left|\begin{array}{cc}
0 & 2 \\
1 & 4
\end{array}\right| = (-1)^{2} (0 \cdot 4 - 2 \cdot 1) = (1) (0 - 2) = -2 \\
& A_{12} = (-1)^{1+2} \left|\begin{array}{cc}
-1 & 2 \\
3 & 4
\end{array}\right| = (-1)^{3} (-1 \cdot 4 - 2 \cdot 3) = (-1) (-4 - 6) = 10 \\
& A_{13} = (-1)^{1+3} \left|\begin{array}{cc}
-1 & 0 \\
3 & 1
\end{array}\right| = (-1)^{4} (-1 \cdot 1 - 0 \cdot 3) = (1) (-1) = -1
\end{align*}
Now, we use these cofactors to find the determinant:
\begin{align*}
\det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\
&= 2(-2) + 3(10) + (-1)(-1) \\
&= -4 + 30 + 1 \\
&= 27
\end{align*}
Thus, the determinant of the matrix \(\left[\begin{array}{ccc} 2 & 3 & -1 \\ -1 & 0 & 2 \\ 3 & 1 & 4 \end{array}\right]\) is:$27$

=====Question 2(iii)=====
 Evaluate the determinant of the following matrix $\left[\begin{array}{ccc}2 i & 6 & 1 \\ 1 & -i & 2 \\ 0 & 1 & 3 i\end{array}\right]$ using cofactor method.

** Solution. **

The elements of \(R_1\) are \(a_{11} = 2i\), \(a_{12} = 6\), and \(a_{13} = 1\).
	
\begin{align*}
A &= \left[\begin{array}{ccc}2i & 6 & 1 \\ 1 & -i & 2 \\ 0 & 1 & 3i\end{array}\right]\\
& A_{11} = (-1)^{1+1} \left|\begin{array}{cc}
-i & 2 \\
1 & 3i
\end{array}\right| = (-1)^{2} (-i \cdot 3i - 2 \cdot 1) = (1) (-3i^2 - 2) = -3(-1) - 2 = 3 - 2 = 1 \\
& A_{12} = (-1)^{1+2} \left|\begin{array}{cc}
1 & 2 \\
0 & 3i
\end{array}\right| = (-1)^{3} (1 \cdot 3i - 2 \cdot 0) = (-1) (3i - 0) = -3i \\
& A_{13} = (-1)^{1+3} \left|\begin{array}{cc}
1 & -i \\
0 & 1
\end{array}\right| = (-1)^{4} (1 \cdot 1 - (-i) \cdot 0) = (1) (1 - 0) = 1
\end{align*}
Now, we use these cofactors to find the determinant:
\begin{align*}
\det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\
&= 2i(1) + 6(-3i) + 1(1) \\
&= 2i - 18i + 1 \\
&= -16i + 1
\end{align*}
Thus, the determinant of the matrix \(\left[\begin{array}{ccc}2i & 6 & 1 \\ 1 & -i & 2 \\ 0 & 1 & 3i\end{array}\right]\) is:$1 - 16i$



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

** Solution. **

The elements of $R_1$ are $a_{11} = 1-i$, $a_{12} = 2$, and $a_{13} = 1+i$.	
\begin{align*}A &= \left[\begin{array}{ccc}1-i & 2 & 1+i \\ 3 & 1 & 4 \\ 0 & 2 & 3\end{array}\right]\\
& A_{11} = (-1)^{1+1} \left|\begin{array}{cc}
1 & 4 \\
2 & 3
\end{array}\right| = (-1)^{2} (1 \cdot 3 - 4 \cdot 2) = (1) (3 - 8) = -5 \\
& A_{12} = (-1)^{1+2} \left|\begin{array}{cc}
3 & 4 \\
0 & 3
\end{array}\right| = (-1)^{3} (3 \cdot 3 - 4 \cdot 0) = (-1) (9 - 0) = -9 \\
& A_{13} = (-1)^{1+3} \left|\begin{array}{cc}
3 & 1 \\
0 & 2
\end{array}\right| = (-1)^{4} (3 \cdot 2 - 1 \cdot 0) = (1) (6 - 0) = 6
\end{align*}
Now, we use these cofactors to find the determinant:
\begin{align*}
\det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\
&= (1-i)(-5) + 2(-9) + (1+i)(6) \\
&= -5 + 5i - 18 + 6 + 6i \\
&= -17 + 11i
\end{align*}
Thus, the determinant of the matrix \(\left[\begin{array}{ccc}1-i & 2 & 1+i \\ 3 & 1 & 4 \\ 0 & 2 & 3\end{array}\right]\) is:
$-17 + 11i$

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