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.

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*}

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*}

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: $$|A| = -6$$

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: $$ |A| = 16 + 8i$$

Question 2 >