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NCERT Solutions for Exercise 4.2 Class 12 Maths Chapter 4 - Determinants

NCERT Solutions for Exercise 4.2 Class 12 Maths Chapter 4 - Determinants

Edited By Ramraj Saini | Updated on Dec 03, 2023 04:00 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.2

NCERT Solutions for Exercise 4.2 Class 12 Maths Chapter 4 Determinants are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In this article, you will get NCERT solutions for Class 12 Maths chapter 4 exercise 4.2. These Exercise 4.2 Class 12 Maths solutions are consist of questions related to properties of determinants. Properties of determinants make it easy for us to finding determinants without complicated calculations. There are 6 properties of determinants related to operation on the determinants given in the NCERT textbook before the Class 12 Maths ch 4 ex 4.2. You are advised to go through the proof of these properties given in the textbook to get a better understanding. There are some examples given after each property which will also help you to get conceptual clarity.

12th class Maths exercise 4.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Determinants Exercise:4.2

Question:1 Using the property of determinants and without expanding, prove that

\begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}=0

Answer:

We can split it in manner like;

\begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}= \begin{vmatrix} x &a &x \\ y & b &y \\ z &c &z \end{vmatrix} + \begin{vmatrix} x &a & a\\ y &b &b \\ z&c & c \end{vmatrix}

So, we know the identity that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Clearly, expanded determinants have identical columns.

\therefore 0 + 0 = 0

Hence the sum is zero.

Question:2 Using the property of determinants and without expanding, prove that

\begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0

Answer:


Given determinant \triangle =\begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0

Applying the rows addition R_{1} \rightarrow R_{1}+R_{2} then we have;

\triangle =\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\-(a-c) &-(b-a) &-(c-b) \end{vmatrix}=0

=-\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\(a-c) &(b-a) &(c-b) \end{vmatrix}=0

So, we have two rows R_{1} and R_{2} identical hence we can say that the value of determinant = 0

Therefore \triangle = 0.

Question:3 Using the property of determinants and without expanding, prove that

\begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix}=0

Answer:

Given determinant \dpi{100} \begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix}

So, we can split it in two addition determinants:

\begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix} = \begin{vmatrix} 2 &7 &63+2 \\ 3& 8 &72+3 \\ 5& 9 & 81+5 \end{vmatrix}

\begin{vmatrix} 2 &7 &63+2 \\ 3& 8 &72+3 \\ 5& 9 & 81+5 \end{vmatrix} = \begin{vmatrix} 2 & 7 &2 \\ 3& 8& 3\\ 5 & 9 & 5 \end{vmatrix} + \begin{vmatrix} 2 & 7 &63 \\ 3& 8 &72 \\ 5 & 9 & 81 \end{vmatrix}

\begin{vmatrix} 2 & 7 &2 \\ 3& 8& 3\\ 5 & 9 & 5 \end{vmatrix} = 0 [\because Here two columns are identical ]

and \begin{vmatrix} 2 & 7 &63 \\ 3& 8 &72 \\ 5 & 9 & 81 \end{vmatrix} = \begin{vmatrix} 2 & 7 &9(7) \\ 3& 8 &9(8) \\ 5 &9 & 9(9) \end{vmatrix} = 9 \begin{vmatrix} 2 & 7 &7 \\ 3& 8& 8\\ 5& 9&9 \end{vmatrix} [\because Here two columns are identical ]

= 0

Therefore we have the value of determinant = 0.

Question:4 Using the property of determinants and without expanding, prove that

\begin{vmatrix}1 &bc &a(b+c) \\1 &ca &b(c+a) \\1 &ab & c(a+b) \end{vmatrix}=0

Answer:

We have determinant:

\triangle = \begin{vmatrix} 1 &bc &a(b+c) \\ 1& ca &b(c+a) \\ 1& ab &c(a+b) \end{vmatrix}

Applying C_{3} \rightarrow C_{2} + C_{3} we have then;

\triangle = \begin{vmatrix} 1 &bc & ab+bc+ca \\ 1& ca &ab+bc+ca \\ 1& ab &ab+bc+ca \end{vmatrix}

So, here column 3 and column 1 are proportional.

Therefore, \triangle = 0.

Question:5 Using the property of determinants and without expanding, prove that

\begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}=2\begin{vmatrix} a &p &x \\ b &q &y \\ c &r & z \end{vmatrix}

Answer:

Given determinant :

\triangle= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}

Splitting the third row; we get,

= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a &p & x \end{vmatrix} + \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ b &q & y \end{vmatrix} = \triangle_{1} + \triangle_{2}\ (assume\ that).

Then we have,

\triangle_{1} = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a &p & x \end{vmatrix}

On Applying row transformation R_{2} \rightarrow R_{2} - R_{3} and then R_{1} \rightarrow R_{1} - R_{2};

we get, \triangle_{1} = \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}

Applying Rows exchange transformation R_{1} \leftrightarrow R_{2} and R_{2} \leftrightarrow R_{3}, we have:

\triangle_{1} =(-1)^2 \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}= \begin{vmatrix} a & p & x\\ b & q&y \\ c& r & z \end{vmatrix}

also \triangle_{2} = \begin{vmatrix} b+c & q+r & y+z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}

On applying rows transformation, R_{1} \rightarrow R_{1} - R_{3} and then R_{2} \rightarrow R_{2} - R_{1}

\triangle_{2} = \begin{vmatrix} c & r & z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix} and then \triangle_{2} = \begin{vmatrix} c & r & z \\ a&p &x \\ b & q & y \end{vmatrix}

Then applying rows exchange transformation;

R_{1} \leftrightarrow R_{2} and then R_{2} \leftrightarrow R_{3}. we have then;

\triangle_{2} =(-1)^2 \begin{vmatrix} a & p & x \\ b&q &y \\ c & r & z \end{vmatrix}

So, we now calculate the sum = \triangle_{1} + \triangle _{2}

\triangle_{1} + \triangle _{2} = 2 \begin{vmatrix} a &p &x \\ b& q& y\\ c & r& z \end{vmatrix}

Hence proved.

Question:6 Using the property of determinants and without expanding, prove that

\begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}=0

Answer:

We have given determinant

\triangle = \begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}

Applying transformation, \dpi{100} R_{1} \rightarrow cR_{1} we have then,

\triangle = \frac{1}{c}\begin{vmatrix} 0 &ac &-bc \\-a &0 & -c\\b &c &0 \end{vmatrix}

We can make the first row identical to the third row so,

Taking another row transformation: R_{1} \rightarrow R_{1}-bR_{2} we have,

\triangle = \frac{1}{c}\begin{vmatrix} ab &ac &0 \\-a &0 & -c\\b &c &0 \end{vmatrix} = \frac{a}{c} \begin{vmatrix} b &c &0 \\-a &0 & -c\\b &c &0 \end{vmatrix}

So, determinant has two rows R_{1}\ and\ R_{3} identical.

Hence \triangle = 0.

Question:7 Using the property of determinants and without expanding, prove that

\begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}=4a^2b^2c^2

Answer:

Given determinant : \dpi{100} \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}

\triangle = \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}

As we can easily take out the common factors a,b,c from rows R_{1},R_{2},R_{3} respectively.

So, get then:

=abc \begin{vmatrix} -a &b &c \\ a &-b &c \\ a & b & -c \end{vmatrix}

Now, taking common factors a,b,c from the columns C_{1},C_{2},C_{3} respectively.

=a^2b^2c^2 \begin{vmatrix} -1 &1 &1 \\ 1 &-1 &1 \\ 1 & 1 & -1 \end{vmatrix}

Now, applying rows transformations R_{1} \rightarrow R_{1} + R_{2} and then R_{3} \rightarrow R_{2} + R_{3} we have;

\triangle = a^2b^2c^2\begin{vmatrix} 0 &0 &2 \\ 1&-1 &1 \\ 2& 0 &0 \end{vmatrix}

Expanding to get R.H.S.

\triangle = a^2b^2c^2 \left ( 2\begin{vmatrix} 1 &-1 \\ 2& 0 \end{vmatrix} \right ) = 2a^2b^2c^2(0+2) =4a^2b^2c^2

Question:8(i) By using properties of determinants, show that:

\begin{vmatrix} 1 &a &a^2 \\ 1 &b &b^2 \\ 1 &c &c^2 \end{vmatrix}=(a-b)(b-c)(c-a)
Answer:

We have the determinant \dpi{100} \begin{vmatrix} 1 &a &a^2 \\ 1 &b &b^2 \\ 1 &c &c^2 \end{vmatrix}

Applying the row transformations R_{1} \rightarrow R_{1} -R_{2} and then R_{2} \rightarrow R_{2} -R_{3} we have:

\triangle = \begin{vmatrix} 0 &a-b &a^2-b^2 \\ 0 &b-c &b^2-c^2 \\ 1 &c &c^2 \end{vmatrix}

= \begin{vmatrix} 0 &a-b &(a-b)(a+b) \\ 0 &b-c &(b-c)(b+c) \\ 1 &c &c^2 \end{vmatrix} = (a-b)(b-c)\begin{vmatrix} 0 &1 &(a+b) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix}

Now, applying R_{1} \rightarrow R_{1} -R_{2} we have:

= (a-b)(b-c)\begin{vmatrix} 0 &0 &(a-c) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix} or = (a-b)(b-c)(a-c)\begin{vmatrix} 0 &0 &1 \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix} =(a-b)(b-c)(a-c)\begin{vmatrix} 0 &1 \\ 1 & c \end{vmatrix}

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

Hence proved.

Question:8(ii) By using properties of determinants, show that:

\dpi{100} \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)

Answer:

Given determinant :

\begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix},

Applying column transformation C_{1} \rightarrow C_{1}-C_{3} and then C_{2} \rightarrow C_{2}-C_{3}

We get,

\triangle =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ a^3-c^3 &b^3-c^3 & c^3 \end{vmatrix}

=\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ (a-c)(a^2+ac+c^2) &(b-c)(b^2+bc+c^2) & c^3 \end{vmatrix}

=(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 1& 1 & c \\ (a^2+ac+c^2) &(b^2+bc+c^2) & c^3 \end{vmatrix}

Now, applying column transformation C_{1} \rightarrow C_{1} - C_{2}, we have:

=(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a^2-b^2+ac-bc) &(b^2+bc+c^2) & c^3 \end{vmatrix}

=(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a-b)(a+b+c) &(b^2+bc+c^2) & c^3 \end{vmatrix}

=(a-c)(b-c)(a-b)(a+b+c)\begin{vmatrix} 0&1 \\ 1& c \end{vmatrix}

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

Hence proved.

Question:9 By using properties of determinants, show that:

\begin{vmatrix} x & x^2 & yz\\ y & y^2 &zx \\ z & z^2 & xy \end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)

Answer:

We have the determinant:

\triangle = \begin{vmatrix} x & x^2 & yz\\ y & y^2 &zx \\ z & z^2 & xy \end{vmatrix}

Applying the row transformations R_{1} \rightarrow R_{1}- R_{3} and then R_{2} \rightarrow R_{2}- R_{3}, we have;

\triangle = \begin{vmatrix} x-z & x^2-z^2 & yz-xy\\ y-z & y^2-z^2 &zx-xy \\ z & z^2 & xy \end{vmatrix}

= \begin{vmatrix} x-z & (x-z)(x+z) & y(z-x)\\ y-z & (y-z)(y+z) &x(z-y) \\ z & z^2 & xy \end{vmatrix}

= (x-z)(y-z)\begin{vmatrix} 1 & (x+z) & -y\\ 1 & (y+z) &-x \\ z & z^2 & xy \end{vmatrix}

Now, applying R_{1} \rightarrow R_{1} - R_{2}; we have

= (x-z)(y-z)\begin{vmatrix} 0 & (x-y) & (x-y)\\ 1 & (y+z) &-x \\ z & z^2 & xy \end{vmatrix}

= (x-z)(y-z)(x-y)\begin{vmatrix} 0 & 1 & 1\\ 1 & (y+z) &-x \\ z & z^2 & xy \end{vmatrix}

Now, expanding the remaining determinant;

= (x-z)(y-z)(x-y) \left [ (xy+zx) + (z^2 - zy-z^2) \right]

= -(x-z)(y-z)(x-y) \left [ xy+zx + zy \right]

= (x-y)(y-z)(z-x) \left [ xy+zx + zy \right]

Hence proved.

Question:10(i) By using properties of determinants, show that:

\begin{vmatrix} x+4 &2x &2x \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}=(5x+4)(4-x)

Answer:

Given determinant:

\begin{vmatrix} x+4 &2x &2x \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}

Applying row transformation: R_{1} \rightarrow R_{1} + R_{2} + R_{3} then we have;

\triangle = \begin{vmatrix} 5x+4 &5x+4 &5x+4 \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}

Taking a common factor: 5x+4

= (5x+4)\begin{vmatrix} 1 &1 &1 \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}

Now, applying column transformations C_{1} \rightarrow C_{1}- C_{2} and C_{2} \rightarrow C_{2}- C_{3}

= (5x+4)\begin{vmatrix} 0 &0 &1 \\ x-4 & 4-x & 2x\\ 0 & x-4 & x+4 \end{vmatrix}

= (5x+4)(4-x)(4-x)\begin{vmatrix} 0 &0 &1 \\ 1 & 1 & 2x\\ 0 & 1 & x+4 \end{vmatrix}

= (5x+4)(4-x)^2

Question:10(ii) By using properties of determinants, show that:

\begin{vmatrix} y+k & y & y\\ y & y+k &y \\ y & y & y+k \end{vmatrix}=k^2(3y+k)

Answer:

Given determinant:

\triangle = \begin{vmatrix} y+k & y & y\\ y & y+k &y \\ y & y & y+k \end{vmatrix}

Applying row transformation R_{1} \rightarrow R_{1} +R_{2}+R_{3} we get;

= \begin{vmatrix} 3y+k & 3y+k & 3y+k\\ y & y+k &y \\ y & y & y+k \end{vmatrix}

=(3y+k) \begin{vmatrix}1 & 1 & 1\\ y & y+k &y \\ y & y & y+k \end{vmatrix} [taking common (3y + k) factor]

Now, applying column transformation C_{1} \rightarrow C_{1} - C_{2} and C_{2} \rightarrow C_{2} - C_{3}

=(3y+k) \begin{vmatrix}0 & 0 & 1\\ -k & k &y \\ 0 & -k & y+k \end{vmatrix}

=(3y+k)(k^2) \begin{vmatrix}0 & 0 & 1\\ -1 & 1 &y \\ 0 & -1 & y+k \end{vmatrix}

=k^2 (3y+k)

Hence proved.

Question:11(i) By using properties of determinants, show that:

\begin{vmatrix} a-b-c &2a &2a \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}=(a+b+c)^3

Answer:

Given determinant:

\triangle = \begin{vmatrix} a-b-c &2a &2a \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}

We apply row transformation: R_{1} \rightarrow R_{1}+R_{2}+R_{3} we have;

= \begin{vmatrix} a+b+c &a+b+c &a+b+c \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}

Taking common factor (a+b+c) out.

=(a+b+c) \begin{vmatrix} 1 &1 &1 \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}

Now, applying column tranformation C_{1} \rightarrow C_{1}- C_{2} and then C_{2} \rightarrow C_{2}- C_{3}

We have;

=(a+b+c) \begin{vmatrix} 0 &0 &1 \\ b+c+a &-b-c-a &2b \\ 0 &c+a+b &c-a-b \end{vmatrix}

=(a+b+c)(a+b+c)(a+b+c) \begin{vmatrix} 0 &0 &1 \\ 1 &-1 &2b \\ 0 &1 &c-a-b \end{vmatrix}

=(a+b+c)(a+b+c)(a+b+c) = (a+b+c)^3

Hence Proved.

Question:11(ii) By using properties of determinants, show that:

\begin{vmatrix} x+y+2z &x &y \\ z & y+z+2x & y\\ z & x &z+x+2y \end{vmatrix}=2(x+y+z)^3

Answer:

Given determinant

\triangle =\begin{vmatrix} x+y+2z &x &y \\ z & y+z+2x & y\\ z & x &z+x+2y \end{vmatrix}

Applying C_{1} \rightarrow C_{1}+C_{2}+C_{3} we get;

=\begin{vmatrix} 2(x+y+z) &x &y \\ 2(z+y+x) & y+z+2x & y\\ 2(z+y+x) & x &z+x+2y \end{vmatrix}

Taking 2(x+y+z) factor out, we get;

=2(x+y+z)\begin{vmatrix} 1 &x &y \\ 1 & y+z+2x & y\\ 1 & x &z+x+2y \end{vmatrix}

Now, applying row transformations, R_{1} \rightarrow R_{1} -R_{2} and then R_{2} \rightarrow R_{2} -R_{3}.

we get;

=2(x+y+z)\begin{vmatrix} 0 &-x-y-z &0 \\ 0 & y+z+x & -y-z-x\\ 1 & x &z+x+2y \end{vmatrix}

=2(x+y+z)^3\begin{vmatrix} 0 &-1 &0 \\ 0 & 1 & -1\\ 1 & x &z+x+2y \end{vmatrix}

=2(x+y+z)^3\begin{vmatrix} -1 &0 \\ 1& -1 \end{vmatrix} = 2(x+y+z)^3

Hence proved.

Question:12 By using properties of determinants, show that:

\begin{vmatrix} 1 &x &x^2 \\ x^2 &1 &x \\ x &x^2 &1 \end{vmatrix}=(1-x^3)^2

Answer:

Give determinant \begin{vmatrix} 1 &x &x^2 \\ x^2 &1 &x \\ x &x^2 &1 \end{vmatrix}

Applying column transformation C_{1} \rightarrow C_{1}+C_{2}+C_{3} we get;

\triangle = \begin{vmatrix} 1+x+x^2 &x &x^2 \\ x^2+1+x &1 &x \\ x+x^2+1 &x^2 &1 \end{vmatrix}

= (1+x+x^2)\begin{vmatrix} 1 &x &x^2 \\ 1 &1 &x \\ 1 &x^2 &1 \end{vmatrix} [after taking the (1+x+x2 ) factor common out.]

Now, applying row transformations, R_{1} \rightarrow R_{1}-R_{2} and then R_{2} \rightarrow R_{2}-R_{3}.

we have now,

= (1+x+x^2)\begin{vmatrix} 0 &x-1 &x^2-x \\ 0 &1-x^2 &x-1 \\ 1 &x^2 &1 \end{vmatrix}

= (1+x+x^2)\begin{vmatrix} x-1 &x^2-x \\ 1-x^2 &x-1 \end{vmatrix}

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

= (1+x+x^2)(x-1)(x^3-1) = (x^3-1)^2

As we know \left [\because (1+x+x^2)(x-1) = (x^3-1) \right ]

Hence proved.

Question:13 By using properties of determinants, show that:

\begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}=(1+a^2+b^2)^3

Answer:

We have determinant:

\triangle = \begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}

Applying row transformations, R_{1} \rightarrow R_{1} +bR_{3} and R_{2} \rightarrow R_{2} -aR_{3} then we have;

= \begin{vmatrix} 1+a^2+b^2 &0 &-b(1+a^2+b^2) \\ 0 &1+a^2+b^2 &a(1+a^2+b^2) \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}

taking common factor out of the determinant;

= (1+a^2+b^2)^2\begin{vmatrix} 1 &0 &-b \\ 0 &1 &a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}

Now expanding the remaining determinant we get;

= (1+a^2+b^2)^2\left [ (1)\begin{vmatrix} 1& a\\ -2a&1-a^2-b^2 \end{vmatrix} - b\begin{vmatrix} 0&1 \\ 2b&-2a \end{vmatrix}\right ]

= (1+a^2+b^2)^2\left [ 1-a^2-b^2+2a^2-b(-2b)\right ]

= (1+a^2+b^2)^2\left [ 1+a^2+b^2\right ] = (1+a^2+b^2)^3

Hence proved.

Question:14 By using properties of determinants, show that:

\begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}=1+a^2+b^2+c^2

Answer:

Given determinant:

\dpi{100} \begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}

Let \triangle = \begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}

Then we can clearly see that each column can be reduced by taking common factors like a,b, and c respectively from C1,C2,and C3.

We then get;

=abc \begin{vmatrix} \left ( a+\frac{1}{a} \right ) &a &a \\ b &(b+\frac{1}{b}) &b \\ c & c &(c+\frac{1}{c}) \end{vmatrix}

Now, applying column transformations: C_{1} \rightarrow C_{1} -C_{2} and C_{2} \rightarrow C_{2} -C_{3}

then we have;

=abc \begin{vmatrix} \left ( \frac{1}{a} \right ) &0 &a \\ -\frac{1}{b} &(\frac{1}{b}) &b \\ 0 & -\frac{1}{c} &(c+\frac{1}{c}) \end{vmatrix}

=abc\times \frac{1}{abc} \begin{vmatrix} 1 &0 &a^2 \\ -1 &1 &b^2 \\ 0 & -1 &(c^2+1) \end{vmatrix}

= \begin{vmatrix} 1 &0 &a^2 \\ -1 &1 &b^2 \\ 0 & -1 &(c^2+1) \end{vmatrix}

Now, expanding the remaining determinant:

\triangle = 1\begin{vmatrix} 1&b^2 \\ -1&(c^2+1) \end{vmatrix} + a^2\begin{vmatrix} -1&1 \\ 0& -1 \end{vmatrix}

= 1[(c^2+1)+b^2] + a^2(1)=a^2+b^2+c^2+1.

Hence proved.

Question:15 Choose the correct answer. Let A be a square matrix of order 3\times 3 , then |kA| is equal to

(A) k|A| (B) k^2|A| (C) k^3|A| (D) 3k|A|

Answer:

Assume a square matrix A of order of 3\times3.

A = \begin{bmatrix} a_1 & b_1&c_1 \\ a_2& b_2& c_2\\ a_3& b_3 & c_3 \end{bmatrix}

Then we have;

kA = \begin{bmatrix} ka_1 & kb_1&kc_1 \\ ka_2& kb_2& kc_2\\ ka_3& kb_3 & kc_3 \end{bmatrix}

(Taking the common factors k from each row.)

|kA| = \begin{vmatrix} ka_1 & kb_1&kc_1 \\ ka_2& kb_2& kc_2\\ ka_3& kb_3 & kc_3 \end{vmatrix} = k^3 \begin{vmatrix} a_1 & b_1&c_1 \\a_2& b_2& c_2\\ a_3& b_3 & c_3 \end{vmatrix}

= k^3 |A|

Therefore correct option is (C).

Question:16 Choose the correct answer.

Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these

Answer:

The answer is (C) Determinant is a number associated to a square matrix.

As we know that To every square matrix A = [a_{ij}]of order n, we can associate a number (real or complex) called determinant of the square matrix A, where a_{ij} = (i, j)^{th} element of A.

More About NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2

This article NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 is consists of questions related to properties of determinants. In Class 12th Maths chapter 4 exercise 4.2 there are 16 questions including 2 multiple choice type questions. There are 11 examples given in NCERT book before the exercise 4.2 Class 12 Maths. First, try to solve these examples given in the textbook. It will help you to get conceptual clarity and solving NCERT problems. NCERT syllabus Class 12th Maths chapter 4 exercise 4.2 questions are very important for the board exam as well as for the engineering competitive exams.

Also Read| Determinants Class 12 Chapter 4 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2

  • Class 12 Maths chapter 4 exercise 4.2 solutions are prepared by the subject matter experts who know how best to answer in order to perform well in the board exams.
  • Class 12th Maths chapter 4 exercise 4.2 questions are prepared in a very descriptive manner which you will get easily.
  • NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 are important in competitive exams like JEE, SRMJEE, etc.
  • As most of the time, one question from this exercise is asked in the board exam, so you are advised to be thorough with them.
  • You can use these NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 for reference.
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Key Features Of NCERT Solutions for Exercise 4.2 Class 12 Maths Chapter 4

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 4.2 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 4.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 4.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 4.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 4.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 4.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. How the value of determinants affect if the rows and columns of determinant are interchanged ?

The value of the determinant remains unchanged when the rows and columns of determinants are interchanged.  

2. How the value of determinants change if any two rows of a determinant are interchanged

If any two rows of a determinant are interchanged then the sign of the determinant change.

3. How the value of determinants change if any two columns of a determinant are interchanged

The sign of the determinant change when any two columns of a determinant are interchanged.

4. What is the value of determinant when the two rows of determinant are identical ?

The value of the determinant is zero if any two rows of a determinant are identical.

5. What is the value of determinant when the two columns of determinant are identical ?

The value of the determinant is zero if any two columns of a determinant are identical.

6. If A be a square matrix of order 2 then | kA| ?

If the order of the square matrix is 2 then |kA| = k^2|A|.

7. How many questions are there is the exercise 4.2 Class 12 Maths ?

There are 16 questions including two multiple choice questions are given in this exercise. All questions are useful to get a conceptual clarity. Following NCERT syllabus is beneficial for the CBSE board exam

8. where can I get NCERT solutions ?

By clicking on the link you will get  NCERT solutions. NCERT Solutions for Mathematics and Science are given chapter wise.

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Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.
  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.
  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Hi,

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hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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