Question 4, Exercise 2.3
Solutions of Question 4 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 4(i)
Find the value of $\lambda$, so that the given matrix is singular $\left[\begin{array}{lll}\lambda & 1 & 3 \\ 2 & 1 & 8 \\ 0 & 3 & 1\end{array}\right]$.
Solution.
\begin{align*} A &= \left[\begin{array}{ccc} \lambda & 1 & 3 \\ 2 & 1 & 8 \\ 0 & 3 & 1 \end{array}\right]\\ |A| &= \lambda \cdot (-23) - 1 \cdot 2 + 3 \cdot 6 \\ &= -23\lambda - 2 + 18 \\ &= -23\lambda + 16 \end{align*} $A$ is singular, Then \begin{align*} |A|&=0\\ -23\lambda + 16 &= 0\\ -23\lambda &= -16 \\ \lambda &= \dfrac{16}{23} \end{align*} Thus, $\lambda = \dfrac{16}{23}$.
Question 4(ii)
Find the value of $\lambda$, so that the given matrix is singular $\left[\begin{array}{lll}\lambda & 2 & 0 \\ 2 & 1 & 3 \\ \lambda & 2 & 1\end{array}\right]$.
Solution.
\begin{align*} A &= \left[\begin{array}{ccc} \lambda & 2 & 0 \\ 2 & 1 & 3 \\ \lambda & 2 & 1 \end{array}\right]\\ |A| &= \lambda \cdot (-5) - 2 \cdot (2 - 3\lambda) \\ &= -5\lambda - 2(2 - 3\lambda) \\ &= -5\lambda - 4 + 6\lambda \\ &= \lambda - 4 \end{align*} We know that the matrix $A$ is singular, then: \begin{align*} |A|&=0\\ \lambda - 4 = 0\\ \lambda = 4 \end{align*}
Question 4(iii)
Find the value of $\lambda$, so that the given matrix is singular $\left[\begin{array}{lll}\lambda & i & 1 \\ 2 & 1 & 3 \\ 3 & 1 & 2\end{array}\right]$.
Solution.
Do yourself
Question 4(iv)
Find the value of $\lambda$, so that the given matrix is singular $\left[\begin{array}{ccc}2+i & 1 & 6 \\ 2 & \lambda & 1 \\ 3 & 0 & 2\end{array}\right]$.
Solution.
\begin{align*} A = \left[\begin{array}{ccc} 2+i & 1 & 6 \\ 2 & \lambda & 1 \\ 3 & 0 & 2 \end{array}\right]\\ |A| &= (2+i) \cdot 2\lambda - 1 \cdot 1 + 6 \cdot (-3\lambda) \\ &= (2+i) \cdot 2\lambda - 1 - 18\lambda \\ &= 4\lambda + 2i\lambda - 1 - 18\lambda \\ &= (4\lambda - 18\lambda + 2i\lambda - 1) \\ &= -14\lambda + 2i\lambda - 1 \\ &= (-14 + 2i)\lambda - 1 \end{align*}
| A |
\begin{align*} (-14 + 2i)\lambda - 1 &= 0\\ (-14 + 2i)\lambda &= 1\\ \lambda &= \dfrac{1}{-14 + 2i}\\ \implies \lambda &= \dfrac{1}{-14 + 2i} \cdot \dfrac{-14 - 2i}{-14 - 2i}\\ & = \dfrac{-14 - 2i}{(-14)^2 - (2i)^2}\\ & = \dfrac{-14 - 2i}{196 + 4}\\ & = \dfrac{-14 - 2i}{200}\\ & = \dfrac{-14 - 2i}{200}\\ &= -\dfrac{7}{100} - \dfrac{i}{100} \end{align*} Hence $\lambda = -\dfrac{7}{100} - \dfrac{i}{100}$
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