In this page, you can find the complete solutions of the second exercise of  Matrices and Determinants chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, circle is the 5th chapter and has five exercises only. Out of which, this is the solution of the second exercise.

Check: Grade 11 Matrices and Determinants Exercise 1 Complete Solutions


Check: Grade 11 Matrices and Determinants Exercise 3 Complete Solutions


Check: Grade 11 Circle Exercise 1 Complete Solutions

Grade 11 Circle Exercise 1 Solutions | Basic Mathematics Grade XI by Sukunda Pustak Bhawan

Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:

Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal


Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here.


3. Without expanding the determinants, show that the value of each of the following determinants is zero.

i) $\left | \displaylines{6&1&9 \\ 2&4&7 \\ 18&3&27} \right | $

Solution:
To prove: $\left | \displaylines{6&1&9 \\ 2&4&7 \\ 18&3&27} \right | = 0$

LHS
$ = \left | \displaylines{6&1&9 \\ 2&4&7 \\ 18&3&27} \right |$

Taking 3 common from R3
$= 3 \left | \displaylines{6 & 1 & 9 \\ 2&4&7 \\ 6 &1 &9} \right |$

Since R1 and R3 are identical, value of given determinant is zero.

$= 3 * 0$
$= 0$
RHS

4. Evaluate:

i) $\left | \displaylines{51 & 61 & 71 \\ 5 & 6& 7 \\ 1 & 1 & 1}\right |$

Solution:
Given,
$\left | \displaylines{51 & 61 & 71 \\ 5 & 6& 7 \\ 1 & 1 & 1}\right |$

Expanding along R1
$= a11 * C11 + a12 * C12 + a13 * C13$

$= 51 * (-1)^{1+1} \left | \displaylines{6 & 7 \\ 1 & 1} \right | + 61  (-1)^{1+2} \left | \displaylines{5 & 7 \\ 1 & 1} \right | + 71  (-1)^{1+3} \left | \displaylines{5 & 6 \\ 1 & 1} \right | $

$= 51 * (6 - 7) - 61 * (5 - 7) + 71 * ( 5 - 6)$

$= -51 + 122 - 71$
$= -122 + 122$
$= 0$

Also,
Given,
$\left | \displaylines{51 & 61 & 71 \\ 5 & 6& 7 \\ 1 & 1 & 1}\right |$

Applying R1 $\to$ R1 - R3
$= \left | \displaylines{51-1 & 61-1 & 71-1 \\ 5 & 6& 7 \\ 1 & 1 & 1}\right |$

$= \left | \displaylines{ 50 & 60 & 70 \\  5 & 6& 7 \\ 1 & 1 & 1}\right |$

Taking 10 common from R1
$= 10 \left | \displaylines{ 5 & 6 & 7 \\ 5 & 6& 7 \\ 1 & 1 & 1}\right |$

Since R1 and R2 are identical, value of determinant is zero.

$= 10 * 0$
$= 0$


6. Show that

i) $\left | \displaylines{ 1 & 1 & 1 \\ a & b & c \\ bc & ca & ab }\right | = (a-b) (b-c)(c-a)$

Solution:

LHS

$= \left | \displaylines{ 1 & 1 & 1 \\ a & b & c \\ bc & ca & ab }\right |$

Applying C1 $\to$ C1 - C2 and C2 $\to$ C2 - C3

$= \left | \displaylines{ 1 -1 & 1-1 & 1 \\ a- b & b-c & c \\ bc - ca & ca - ab & ab } \right |$

$= \left | \displaylines{ 0 & 0 & 1 \\ a - b & b - c & c \\ -c(a -b) & -a (b-c) & ab} \right |$

Taking (a-b) common from C1 and (b-c) common from C2

$=(a-b) (b-c) \left | \displaylines{ 0 & 0 & 1 \\ 1 & 1 & c \\ - c & - a & ab } \right |$

Expanding along R1

$= (a-b)(b-c) * 1 \left | \displaylines{ 1 & 1 \\ -c & -a } \right |$

$= (a-b)(b-c) * (-a + c)$

$= (a-b)(b-c)(c-a)$

RHS



About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)