# 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.

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, $\lambda = \dfrac{16}{23}$.

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*}

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

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 $\lambda = -\dfrac{7}{100} - \dfrac{i}{100}$