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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