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

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. Here you can get NCERT Solutions by clicking on this link.

Also, see

• Determinants Exercise 4.1
• Determinants Exercise 4.3
• Determinants Exercise 4.4
• Determinants Exercise 4.5
• Determinants Exercise 4.6
• Determinants Miscellaneous Exercise

## Determinants Exercise:4.2

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

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

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

$\dpi{100} \begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0$

Given determinant $\dpi{100} \triangle =\begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0$

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

$\dpi{100} \triangle =\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\-(a-c) &-(b-a) &-(c-b) \end{vmatrix}=0$

$\dpi{100} =-\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\(a-c) &(b-a) &(c-b) \end{vmatrix}=0$

So, we have two rows $\dpi{100} R_{1}$ and $\dpi{100} R_{2}$ identical hence we can say that the value of determinant = 0

Therefore $\dpi{100} \triangle = 0$.

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

$\dpi{100} \begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix}=0$

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:

$\dpi{100} \begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix} = \begin{vmatrix} 2 &7 &63+2 \\ 3& 8 &72+3 \\ 5& 9 & 81+5 \end{vmatrix}$

$\dpi{100} \begin{vmatrix} 2 &7 &63+2 \\ 3& 8 &72+3 \\ 5& 9 & 81+5 \end{vmatrix} = \begin{vmatrix} 2 & 7 &2 \\ 3& 8& 3\\ 5 & 9 & 5 \end{vmatrix} + \begin{vmatrix} 2 & 7 &63 \\ 3& 8 &72 \\ 5 & 9 & 81 \end{vmatrix}$

$\dpi{100} \begin{vmatrix} 2 & 7 &2 \\ 3& 8& 3\\ 5 & 9 & 5 \end{vmatrix} = 0$ [$\dpi{100} \because$ Here two columns are identical ]

and $\dpi{100} \begin{vmatrix} 2 & 7 &63 \\ 3& 8 &72 \\ 5 & 9 & 81 \end{vmatrix} = \begin{vmatrix} 2 & 7 &9(7) \\ 3& 8 &9(8) \\ 5 &9 & 9(9) \end{vmatrix} = 9 \begin{vmatrix} 2 & 7 &7 \\ 3& 8& 8\\ 5& 9&9 \end{vmatrix}$ [$\dpi{100} \because$ Here two columns are identical ]

$\dpi{100} = 0$

Therefore we have the value of determinant = 0.

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

$\dpi{100} \begin{vmatrix}1 &bc &a(b+c) \\1 &ca &b(c+a) \\1 &ab & c(a+b) \end{vmatrix}=0$

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

$\dpi{100} \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}=2\begin{vmatrix} a &p &x \\ b &q &y \\ c &r & z \end{vmatrix}$

Given determinant :

$\dpi{100} \triangle= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}$

Splitting the third row; we get,

$\dpi{100} = \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a &p & x \end{vmatrix} + \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ b &q & y \end{vmatrix} = \triangle_{1} + \triangle_{2}\ (assume\ that)$.

Then we have,

$\dpi{100} \triangle_{1} = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a &p & x \end{vmatrix}$

On Applying row transformation $\dpi{100} R_{2} \rightarrow R_{2} - R_{3}$ and then $\dpi{100} R_{1} \rightarrow R_{1} - R_{2}$;

we get, $\dpi{100} \triangle_{1} = \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}$

Applying Rows exchange transformation $\dpi{100} R_{1} \leftrightarrow R_{2}$ and $\dpi{100} R_{2} \leftrightarrow R_{3}$, we have:

$\dpi{100} \triangle_{1} =(-1)^2 \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}= \begin{vmatrix} a & p & x\\ b & q&y \\ c& r & z \end{vmatrix}$

also $\dpi{100} \triangle_{2} = \begin{vmatrix} b+c & q+r & y+z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$

On applying rows transformation, $\dpi{100} R_{1} \rightarrow R_{1} - R_{3}$ and then $\dpi{100} R_{2} \rightarrow R_{2} - R_{1}$

$\dpi{100} \triangle_{2} = \begin{vmatrix} c & r & z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$ and then $\dpi{100} \triangle_{2} = \begin{vmatrix} c & r & z \\ a&p &x \\ b & q & y \end{vmatrix}$

Then applying rows exchange transformation;

$\dpi{100} R_{1} \leftrightarrow R_{2}$ and then $\dpi{100} R_{2} \leftrightarrow R_{3}$. we have then;

$\dpi{100} \triangle_{2} =(-1)^2 \begin{vmatrix} a & p & x \\ b&q &y \\ c & r & z \end{vmatrix}$

So, we now calculate the sum = $\dpi{100} \triangle_{1} + \triangle _{2}$

$\dpi{100} \triangle_{1} + \triangle _{2} = 2 \begin{vmatrix} a &p &x \\ b& q& y\\ c & r& z \end{vmatrix}$

Hence proved.

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

$\dpi{100} \begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}=0$

We have given determinant

$\dpi{100} \triangle = \begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}$

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

$\dpi{100} \triangle = \frac{1}{c}\begin{vmatrix} 0 &ac &-bc \\-a &0 & -c\\b &c &0 \end{vmatrix}$

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

Taking another row transformation: $\dpi{100} R_{1} \rightarrow R_{1}-bR_{2}$ we have,

$\dpi{100} \triangle = \frac{1}{c}\begin{vmatrix} ab &ac &0 \\-a &0 & -c\\b &c &0 \end{vmatrix} = \frac{a}{c} \begin{vmatrix} b &c &0 \\-a &0 & -c\\b &c &0 \end{vmatrix}$

So, determinant has two rows $\dpi{100} R_{1}\ and\ R_{3}$ identical.

Hence $\dpi{100} \triangle = 0$.

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

$\dpi{100} \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}=4a^2b^2c^2$

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:

$\dpi{100} \begin{vmatrix} 1 &a &a^2 \\ 1 &b &b^2 \\ 1 &c &c^2 \end{vmatrix}=(a-b)(b-c)(c-a)$

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:

$\dpi{100} \triangle = \begin{vmatrix} 0 &a-b &a^2-b^2 \\ 0 &b-c &b^2-c^2 \\ 1 &c &c^2 \end{vmatrix}$

$\dpi{100} = \begin{vmatrix} 0 &a-b &(a-b)(a+b) \\ 0 &b-c &(b-c)(b+c) \\ 1 &c &c^2 \end{vmatrix} = (a-b)(b-c)\begin{vmatrix} 0 &1 &(a+b) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix}$

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

$\dpi{100} = (a-b)(b-c)\begin{vmatrix} 0 &0 &(a-c) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix}$ or $\dpi{100} = (a-b)(b-c)(a-c)\begin{vmatrix} 0 &0 &1 \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix} =(a-b)(b-c)(a-c)\begin{vmatrix} 0 &1 \\ 1 & c \end{vmatrix}$

$\dpi{100} = (a-b)(b-c)(c-a)$

Hence proved.

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

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

Given determinant :

$\dpi{100} \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix}$,

Applying column transformation $\dpi{100} C_{1} \rightarrow C_{1}-C_{3}$ and then $\dpi{100} C_{2} \rightarrow C_{2}-C_{3}$

We get,

$\dpi{100} \triangle =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ a^3-c^3 &b^3-c^3 & c^3 \end{vmatrix}$

$\dpi{100} =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ (a-c)(a^2+ac+c^2) &(b-c)(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 1& 1 & c \\ (a^2+ac+c^2) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

Now, applying column transformation $\dpi{100} C_{1} \rightarrow C_{1} - C_{2}$, we have:

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a^2-b^2+ac-bc) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a-b)(a+b+c) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)(a-b)(a+b+c)\begin{vmatrix} 0&1 \\ 1& c \end{vmatrix}$

$\dpi{100} =-(a-c)(b-c)(a-b)(a+b+c) = (a-b)(b-c)(c-a)(a+b+c)$

Hence proved.

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

$\dpi{100} \begin{vmatrix} x & x^2 & yz\\ y & y^2 &zx \\ z & z^2 & xy \end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$

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:

$\dpi{100} \begin{vmatrix} x+4 &2x &2x \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}=(5x+4)(4-x)$

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:

$\dpi{100} \begin{vmatrix} y+k & y & y\\ y & y+k &y \\ y & y & y+k \end{vmatrix}=k^2(3y+k)$

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:

$\dpi{100} \begin{vmatrix} a-b-c &2a &2a \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}=(a+b+c)^3$

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:

$\dpi{100} \begin{vmatrix} x+y+2z &x &y \\ z & y+z+2x & y\\ z & x &z+x+2y \end{vmatrix}=2(x+y+z)^3$

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:

$\dpi{100} \begin{vmatrix} 1 &x &x^2 \\ x^2 &1 &x \\ x &x^2 &1 \end{vmatrix}=(1-x^3)^2$

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:

$\dpi{100} \begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}=(1+a^2+b^2)^3$

We have determinant:

$\dpi{100} \triangle = \begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

Applying row transformations, $\dpi{100} R_{1} \rightarrow R_{1} +bR_{3}$ and $\dpi{100} R_{2} \rightarrow R_{2} -aR_{3}$ then we have;

$\dpi{100} = \begin{vmatrix} 1+a^2+b^2 &0 &-b(1+a^2+b^2) \\ 0 &1+a^2+b^2 &a(1+a^2+b^2) \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

taking common factor out of the determinant;

$\dpi{100} = (1+a^2+b^2)^2\begin{vmatrix} 1 &0 &-b \\ 0 &1 &a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

Now expanding the remaining determinant we get;

$\dpi{100} = (1+a^2+b^2)^2\left [ (1)\begin{vmatrix} 1& a\\ -2a&1-a^2-b^2 \end{vmatrix} - b\begin{vmatrix} 0&1 \\ 2b&-2a \end{vmatrix}\right ]$

$\dpi{100} = (1+a^2+b^2)^2\left [ 1-a^2-b^2+2a^2-b(-2b)\right ]$

$\dpi{100} = (1+a^2+b^2)^2\left [ 1+a^2+b^2\right ] = (1+a^2+b^2)^3$

Hence proved.

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

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

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) $3k|A|$

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).

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

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.

Also see-

• NCERT Solutions for Class 12 Maths Chapter 4

• NCERT Exemplar Solutions Class 12 Maths Chapter 4

## NCERT Solutions of Class 12 Subject Wise

• NCERT Solutions for Class 12 Maths

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Chemistry

• NCERT Solutions for Class 12 Biology

Subject Wise NCERT Exampler Solutions

• NCERT Exemplar Solutions for Class 12th Maths

• NCERT Exemplar Solutions for Class 12th Physics

• NCERT Exemplar Solutions for Class 12th Chemistry

• NCERT Exemplar Solutions for Class 12th Biology

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