====== 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| is singular, then
\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 $\lambda = -\dfrac{7}{100} - \dfrac{i}{100}$
====Go to ====
[[math-11-nbf:sol:unit02:ex2-3-p3|< Question 3]]
[[math-11-nbf:sol:unit02:ex2-3-p5|Question 5 >]]