RD Sharma Class 12 Exercise 5.5 Determinants Solutions Maths

RD Sharma Class 12 Exercise 5.5 Determinant Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 07:11 PM IST

The CBSE board students use the RD Sharma Solutions to prepare for their public examinations. Therefore, it is recommended to use these books to understand the concepts better in a particular subject. Significantly, the students who find mathematics challenging to solve the sums in the Determinants chapter use the RD Sharma Class 12th exercise 5.5 book. Although not every student gets the opportunity to go for daily mathematics tuition, these books are a boon. Moreover, in RD Sharma solution books, the students can clear their doubts and verify their answers.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 5 Determinants - Other Exercise

Determinants Excercise: 5.5

Determinant exercise 5.5 question 1

$x = k, y = k, z = k, where \: k \in R$
Hint: The homogeneous system of linear equations has non-trivial solutions when the determinant is zero.
Given: $x + y - 2z =0$

$2x + y - 3z = 0$

$5x + 4y - 9z = 0$
Solution:
The above linear equations can be represented in matrix form i.e.
$A = \begin{bmatrix} 1& 1& -2\\ 2& 1& -3\\ 5& 4& -9 \end{bmatrix}$ , $X= \begin{bmatrix} x\\ y\\ z\end{bmatrix}$ , $B= \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$
Now, $A = \begin{vmatrix} 1 & 1&-2 \\ 2& 1& -3\\ 5& 4& -9 \end{vmatrix}$
$= 1(-9 + 12) -1(-18 + 15) - 2(8 -5)$
$=3 + 3 - 6$
$=0$
So, the given system has non-trivial solutions.
Consider first two equations and put z=k.
$x+ y - 2z =0$
$2x + y -3z = 0$
Solving these equations by Crammer’s rule:
$x = \frac{D_{1}}{D}=\frac{\begin{vmatrix} 2k&1 \\ 3k& 1 \end{vmatrix}}{\begin{vmatrix} 1 & 1\\ 2& 1 \end{vmatrix}} = \frac{-k}{-1}=k$

$y = \frac{D_{2}}{D}=\frac{\begin{vmatrix} 1&2k \\ 2& 3k \end{vmatrix}}{\begin{vmatrix} 1 & 1\\ 2& 1 \end{vmatrix}} = \frac{-k}{-1}=k$
Thus, $x = k, y = k, z = k$ satisfy the third equation.
Hence, $x = k, y = k, z = k$ where $k \in R$

determinants exercise 5.5 question 3

Answer: $x = -7k, y = 8k, z = 13k$ , where $k \in R$
Hint: : If $\left | A \right |= 0$ , then the system has an infinite number of solutions.
Given:
$3x+y+z=0$
$x - 4y + 3z = 0$
$2 x + 5y - 2z = 0$
Solution:
The above linear equations can be represented in matrix form i.e.
$A = \begin{bmatrix} 3 & 1 & 1\\ 1& -4 & 3\\ 2& 5& -2 \end{bmatrix}, X = \begin{bmatrix} x\\ y\\ z\end{bmatrix}, B = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$
Now,
$\left | A \right | = \begin{vmatrix} 3 & 1& 1\\ 1 & -4 &3 \\ 2& 5& -2 \end{vmatrix}$
$= 3(8 -15) -1(-2 -6) + 1(5+8)$
$= 3(-7) + 8 +15$
$= - 21 + 21$
$= 0$
So, the system of equations has non – trivial solutions.
Consider, first two equation and put z = k
$3 x + y = -k$
$x - 4y = -3k$
Now, By Crammer’s rule:
$x = \frac{D_{1}}{D} = \frac{\begin{vmatrix} -k & 1\\ -3k & -4 \end{vmatrix}}{\begin{vmatrix} 3&1 \\ 1& -4 \end{vmatrix}} = \frac{7k}{-13} = \frac{-7k}{13}$
$y = \frac{D_{2}}{D} = \frac{\begin{vmatrix} 3 & -k\\ 1 & -3k \end{vmatrix}}{\begin{vmatrix} 3&1 \\ 1& -4 \end{vmatrix}} = \frac{-8k}{-13} = \frac{8k}{13}$
$x=\frac{-7k}{13}, y = \frac{8k}{13}$ and $z=k$
or $x = -7k$ , $y = 8k$ and $z = 13k$ , where $k \in R$
It will satisfy the third equation.
Hence, $x = -7k$ , $y = 8k$ and $z = 13k$ , where $k \in R$

Determinants exercise 5.5 question 4

Answer: $\lambda = 2, x = k, y =k/2, z = -k,$ where $k \in R$
Hint: The system has non-trivial solution implies that the determinant is zero.
Given: $2\lambda x - 2y + 3z =0$
$x x + \lambda y + 2z = 0$
$2 x + \lambda z = 0 \lambda$
Solution:
Firstly, form 3x3 matrix from the given equations:
Let A = [ $2 \lambda$ , 2, 3 ]
x $\lambda$ 2
[ 2 0 $\lambda$ ]
We know that, $|A| = 0$
Expanding through first row,
$2\lambda (\lambda ^2 - 0) + 2 (\lambda - 4) + 3 (0 - 2 \lambda ) = 0$
$2\lambda ^3 + 2\lambda - 8 - 6\lambda = 0$
$2\lambda ^3 - 4\lambda - 8 = 0$
$\lambda ^3 - 2\lambda - 4= 0$
Put $\lambda = 2:$
$= (2)^{3} - 2 (2) - 4$
$= 8 - 4 - 4$
$= 0$
As it satisfies the given equation, $\therefore \lambda = 2.$
So, the given linear equation transforms into:
$4x-2y+3z=0$
$x+2y+2z=0$
$2 x+2z=0$
Now, the above equations can be represented in matrix form i.e.
$A = \begin{bmatrix} 4 & -2 & 3\\ 1& 2& 2\\ 2& 0& 2\end{bmatrix}, X = \begin{bmatrix} x\\ y\\ z\end{bmatrix}, B =\begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$
or $AX=B$
Now, Put z = k from first two equations, we get:
$4 x - 2y = -3k$
$x + 2y = -2k$
Now, By Crammer’s rule:
$x = \frac{D_{1}}{D} = \frac{\begin{vmatrix} -3k &-2 \\ -2k& 2\end{vmatrix}}{\begin{vmatrix} 4 & -2\\ 1 & 2\end{vmatrix}} =\frac{10k}{10} = k$
$y = \frac{D_{2}}{D} = \frac{\begin{vmatrix} 4-3k & 1-2k \end{vmatrix}}{\begin{vmatrix} 4-2 & 12 \end{vmatrix}}=\frac{-5k}{10} = \frac{-k}{2}$

It also satisfies the third equation.
Hence, $x = k$ , $y = \frac{-k}{2}$ and $z = -k$ where $k \in R$

_{Determinant exercise 5.5 question 5}

Answer: $ab+bc+ca=abc$
Hint: The system has non-trivial solution implies that the determinant is zero.
Given: $(a-1) x = y + z$
$(b - 1) y = z + x$
$(c - 1) z = x + y$
Solution:
Take x, y and z to the LHS:
$(a - 1)x - y - z = 0$
$x - (b - 1) y + z = 0$
$x + y - (c - 1) z = 0$
Let $A = [a-1 -1 -1 -1 b-1 -1 -1 -1 c-1 ]$
We know that, $|A|=0$
$=>(a-1)[(b-1)(c-1)-1]+1[-(c-1)-1]-1[1+b-1] = 0$
$=>(a-1)[bc-c-b+1-1]+1[-c+1-1]-1[b] = 0$
$=>(a-1)[bc-c-b]-c-b = 0$
$=>abc-ac-ab-bc+c+b-c-b=0$
$=>abc-ac-ab-bc+c+b-c-b$
$=>ab+bc+ca=abc$
Hence proved.

The Class 12, Mathematics chapter 5, Determinants, consists of five exercises in total. RD Sharma class 12th exercise 5.5, includes the left-out topics from the previous exercises in the same chapter. Concepts like Homogeneous linear equations and non-trivial solutions are present in this exercise. It contains around five questions, including the subparts. If they find it difficult or need a good solutions guide to verify the answers, the RD Sharma Class 12 Chapter 5 Exercise 5.5 is the best option.

A committee of experts provides all the answers given in this book. There are many methods by which a sum is solved. This is to allow the students to select the way that they find easy and go ahead with it. The pattern of the book follows the NCERT syllabus making an opportunity for the CBSE students to follow it. Using the Class 12 RD Sharma Chapter 5 Exercise 5.5 Solution guide to doing their homework will make the students fluent regarding the concept. Students who aim to score high marks in the tests and exams must stick to the RD Sharma Class 12th exercise 5.5 solutions.

The main advantage is that these RD Sharma Class 12th exercise 5.5 books are available free of cost. The students can visit the Career 360 website and download the solutions without paying even a single rupee. For instance, visit the Career 360 official site and search for the RD Sharma Class 12 Solutions Chapter 5 ex 5.5 to specifically download the required portion. Many sets of students have already benefited from the RD Sharma solutions books. Now, it is time for the next set of students to shine brighter and gain knowledge.

Download the RD Sharma Class 12 Solutions Determinants Ex 5.5 to fully understand the fifth chapter in mathematics. With constant practice, you will find yourself crossing your benchmark score.

Chapter-wise RD Sharma Class 12 Solutions

Frequently Asked Questions (FAQs)

1. What solution book do Class 12 CBSE students prefer to use as a reference?

Most of the CBSE school students prefer the RD Sharma solution books to look into the answers and clarify their doubts.

2. Is it enough if a student works out the sums for chapter 5 as given in the RD Sharma book?

Yes, It is more than enough to practice the sums as given in the RD Sharma Class 12 Chapter 5 Exercise 5.5 solutions to understand the Determinants concepts.

3. Are the RD Sharma solutions for the topper or the average learner students?

With the solutions of educational experts, the RD Sharma solution books consist of various methods of solving a mathematical function. Therefore, students of every category can depend on RD Sharma books for their doubt clarification.

4. Can everyone access the RD Sharma books?

The Career360 site allows everyone to access and download the RD Sharma solution books free of cost. Therefore, there is no need to pay any monetary charge to access this set of keys at this site.

5. Where can I find the right solutions for the class 12 Determinants chapter?

You can get help from the RD Sharma Class 12th exercise 5.5 solutions book to understand the concepts better.