NCERT Solutions for Class 12 Maths Chapter 4 Determinants: Download Exercise Solutions PDF

Exercise - 4.1

Q1.Evaluate the determinants in Exercises 1 and 2.

$\left|\begin{array}{cc}\hfill 2\hfill & \hfill 4\hfill \\ \hfill -5\hfill & \hfill -1\hfill \end{array}\right|$

A.1. $\left|\begin{array}{cc}\hfill 2\hfill & \hfill 4\hfill \\ \hfill -5\hfill & \hfill -1\hfill \end{array}\right|=$ 2×(−1) – 4×(−5) = −2+20 = 18

Q2.(i) $\left|\begin{array}{cc}\hfill cos\theta \hfill & \hfill -sin\theta \hfill \\ \hfill sin\theta \hfill & \hfill cos\theta \hfill \end{array}\right|$ (ii) $\left|\begin{array}{cc}\hfill x^2-x+1\hfill & \hfill x-1\hfill \\ \hfill x+1\hfill & \hfill x+1\hfill \end{array}\right|$

A.2. (i) $\left|\begin{array}{cc}\hfill cos\theta \hfill & \hfill -sin\theta \hfill \\ \hfill sin\theta \hfill & \hfill cos\theta \hfill \end{array}\right|$

$=cos\theta \times cos\theta -(-sin\theta )\times sin\theta$

= cos2+sin2

=1

(ii)  $\left|\begin{array}{cc}\hfill x^2-x+1\hfill & \hfill x-1\hfill \\ \hfill x+1\hfill & \hfill x+1\hfill \end{array}\right|$

$=(x^2-x+1)(x+1)-(x-1)(x+1)$

= $x^3$ + $x^2$ - $x^2$ - $x$ + $x$ +1 – x² + 1

= x³ – x²+2

Q3.If $A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right]$ , then show that | 2A | = 4 | A |

A.3.Given, A= $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right]$

So, 2A= 2 $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right]$

= $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 4\hfill \\ \hfill 8\hfill & \hfill 4\hfill \end{array}\right]$ .

L.H.S. = $\left|2A\right|$  = $\left|\begin{array}{cc}\hfill 2\hfill & \hfill 4\hfill \\ \hfill 8\hfill & \hfill 4\hfill \end{array}\right|$ = 2×4 – 4×8 = 8−32 = −24

R.H.S. = 4|A| = 4 $\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right|$ = 4(1×2 – 2×4)

= 4[2 − 8]

= 4[−6] = - 24.

LHS = RHS

Q4.If A= $\left[\begin{array}{ccc}\begin{array}{l}1\\ 0\\ 0\end{array}& \begin{array}{l}0\\ 1\\ 0\end{array}& \begin{array}{l}1\\ 2\\ 4\end{array}\end{array}\right]$ ,then show that | 3 A | = 27 | A |

A.4. Given, A= $\left[\begin{array}{ccc}\begin{array}{l}1\\ 0\\ 0\end{array}& \begin{array}{l}0\\ 1\\ 0\end{array}& \begin{array}{l}1\\ 2\\ 4\end{array}\end{array}\right]$

Then, |A|= $\left[\begin{array}{ccc}\begin{array}{l}1\\ 0\\ 0\end{array}& \begin{array}{l}0\\ 1\\ 0\end{array}& \begin{array}{l}1\\ 2\\ 4\end{array}\end{array}\right]$

= 1  $\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 4\hfill \end{array}\right|$ − 0 $\left|\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 4\hfill \end{array}\right|$  + 0 $\left|\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right|$

= 4×1 – 2×0

= 4

And 3A= $3\left[\begin{array}{ccc}\begin{array}{l}1\\ 0\\ 0\end{array}& \begin{array}{l}0\\ 1\\ 0\end{array}& \begin{array}{l}1\\ 2\\ 4\end{array}\end{array}\right]$  =

LHS= |3A|=

=

=3[36 – 6×0]

= 108

RHS = 27. |A| = 27 4 = 108

LHS = RHS

 

$\left|\begin{array}{ccc}\begin{array}{l}3\\ 0\\ 3\end{array}& \begin{array}{l}-1\\ 0\\ -5\end{array}& \begin{array}{l}-2\\ -1\\ 0\end{array}\end{array}\right|$