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

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

$\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|=0$

Answer:

Given determinant $\mathrm{△}=\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|=0$

Applying the rows addition $R_1\to R_1+R_2$ then we have;

$\mathrm{△}=\left|\begin{array}{ccc}a-c& b-a& c-b\\ b-c& c-a& a-b\\ -(a-c)& -(b-a)& -(c-b)\end{array}\right|=0$

$=-\left|\begin{array}{ccc}a-c& b-a& c-b\\ b-c& c-a& a-b\\ (a-c)& (b-a)& (c-b)\end{array}\right|=0$

So, we have two rows $R_1$ and $R_2$ identical hence we can say that the value of determinant = 0

Therefore $\mathrm{△}=0$.

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

$\left|\begin{array}{ccc}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|=0$

Answer:

Given determinant $\text{dpi}100\left|\begin{array}{ccc}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|$

So, we can split it in two addition determinants:

$\left|\begin{array}{ccc}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|=\left|\begin{array}{ccc}2& 7& 63+2\\ 3& 8& 72+3\\ 5& 9& 81+5\end{array}\right|$

$\left|\begin{array}{ccc}2& 7& 63+2\\ 3& 8& 72+3\\ 5& 9& 81+5\end{array}\right|=\left|\begin{array}{ccc}2& 7& 2\\ 3& 8& 3\\ 5& 9& 5\end{array}\right|+\left|\begin{array}{ccc}2& 7& 63\\ 3& 8& 72\\ 5& 9& 81\end{array}\right|$

$\left|\begin{array}{ccc}2& 7& 2\\ 3& 8& 3\\ 5& 9& 5\end{array}\right|=0$ [$\because$ Here two columns are identical ]

and $\left|\begin{array}{ccc}2& 7& 63\\ 3& 8& 72\\ 5& 9& 81\end{array}\right|=\left|\begin{array}{ccc}2& 7& 9(7)\\ 3& 8& 9(8)\\ 5& 9& 9(9)\end{array}\right|=9\left|\begin{array}{ccc}2& 7& 7\\ 3& 8& 8\\ 5& 9& 9\end{array}\right|$ [$\because$ 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

$\left|\begin{array}{ccc}1& bc& a(b+c)\\ 1& ca& b(c+a)\\ 1& ab& c(a+b)\end{array}\right|=0$

Answer:

We have determinant:

$\mathrm{△}=\left|\begin{array}{ccc}1& bc& a(b+c)\\ 1& ca& b(c+a)\\ 1& ab& c(a+b)\end{array}\right|$

Applying $C_3\to C_2+C_3$ we have then;

$\mathrm{△}=\left|\begin{array}{ccc}1& bc& ab+bc+ca\\ 1& ca& ab+bc+ca\\ 1& ab& ab+bc+ca\end{array}\right|$

So, here column 3 and column 1 are proportional.

Therefore, $\mathrm{△}=0$.

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

$\left|\begin{array}{ccc}b+c& q+r& y+z\\ c+a& r+p& z+x\\ a+b& p+q& x+y\end{array}\right|=2\left|\begin{array}{ccc}a& p& x\\ b& q& y\\ c& r& z\end{array}\right|$

Answer:

Given determinant :

$\mathrm{△}=\left|\begin{array}{ccc}b+c& q+r& y+z\\ c+a& r+p& z+x\\ a+b& p+q& x+y\end{array}\right|$

Splitting the third row; we get,

$=\left|\begin{array}{ccc}b+c& q+r& y+z\\ c+a& r+p& z+x\\ a& p& x\end{array}\right|+\left|\begin{array}{ccc}b+c& q+r& y+z\\ c+a& r+p& z+x\\ b& q& y\end{array}\right|={\mathrm{△}}_1+{\mathrm{△}}_2(assumethat)$.

Then we have,

${\mathrm{△}}_1=\left|\begin{array}{ccc}b+c& q+r& y+z\\ c+a& r+p& z+x\\ a& p& x\end{array}\right|$

On Applying row transformation $R_2\to R_2-R_3$ and then $R_1\to R_1-R_2$;

we get, ${\mathrm{△}}_1=\left|\begin{array}{ccc}b& q& y\\ c& r& z\\ a& p& x\end{array}\right|$

Applying Rows exchange transformation $R_1\leftrightarrow R_2$ and $R_2\leftrightarrow R_3$, we have:

${\mathrm{△}}_1=(-1{)}^2\left|\begin{array}{ccc}b& q& y\\ c& r& z\\ a& p& x\end{array}\right|=\left|\begin{array}{ccc}a& p& x\\ b& q& y\\ c& r& z\end{array}\right|$

also ${\mathrm{△}}_2=\left|\begin{array}{ccc}b+c& q+r& y+z\\ c+a& r+p& z+x\\ b& q& y\end{array}\right|$

On applying rows transformation, $R_1\to R_1-R_3$ and then $R_2\to R_2-R_1$

${\mathrm{△}}_2=\left|\begin{array}{ccc}c& r& z\\ c+a& r+p& z+x\\ b& q& y\end{array}\right|$ and then ${\mathrm{△}}_2=\left|\begin{array}{ccc}c& r& z\\ a& p& x\\ b& q& y\end{array}\right|$

Then applying rows exchange transformation;

$R_1\leftrightarrow R_2$ and then $R_2\leftrightarrow R_3$. we have then;

${\mathrm{△}}_2=(-1{)}^2\left|\begin{array}{ccc}a& p& x\\ b& q& y\\ c& r& z\end{array}\right|$

So, we now calculate the sum = ${\mathrm{△}}_1+{\mathrm{△}}_2$

${\mathrm{△}}_1+{\mathrm{△}}_2=2\left|\begin{array}{ccc}a& p& x\\ b& q& y\\ c& r& z\end{array}\right|$

Hence proved.

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

$\left|\begin{array}{ccc}0& a& -b\\ -a& 0& -c\\ b& c& 0\end{array}\right|=0$

Answer:

We have given determinant

$\mathrm{△}=\left|\begin{array}{ccc}0& a& -b\\ -a& 0& -c\\ b& c& 0\end{array}\right|$

Applying transformation, $\text{dpi}100R_1\to c{R}_1$ we have then,

$\mathrm{△}=\frac{1}{c}\left|\begin{array}{ccc}0& ac& -bc\\ -a& 0& -c\\ b& c& 0\end{array}\right|$

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

Taking another row transformation: $R_1\to R_1-b{R}_2$ we have,

$\mathrm{△}=\frac{1}{c}\left|\begin{array}{ccc}ab& ac& 0\\ -a& 0& -c\\ b& c& 0\end{array}\right|=\frac{a}{c}\left|\begin{array}{ccc}b& c& 0\\ -a& 0& -c\\ b& c& 0\end{array}\right|$

So, determinant has two rows $R_{1}and{R}_3$ identical.

Hence $\mathrm{△}=0$.

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

$\left|\begin{array}{ccc}-a^2& ab& ac\\ ba& -b^2& bc\\ ca& cb& -c^2\end{array}\right|=4a^2{b}^2{c}^2$

Answer:

Given determinant : $\text{dpi}100\left|\begin{array}{ccc}-a^2& ab& ac\\ ba& -b^2& bc\\ ca& cb& -c^2\end{array}\right|$

$\mathrm{△}=\left|\begin{array}{ccc}-a^2& ab& ac\\ ba& -b^2& bc\\ ca& cb& -c^2\end{array}\right|$

As we can easily take out the common factors a,b,c from rows $R_1,R_2,R_3$ respectively.

So, get then:

$=abc\left|\begin{array}{ccc}-a& b& c\\ a& -b& c\\ a& b& -c\end{array}\right|$

Now, taking common factors a,b,c from the columns $C_1,C_2,C_3$ respectively.

$=a^2{b}^2{c}^2\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$

Now, applying rows transformations $R_1\to R_1+R_2$ and then $R_3\to R_2+R_3$ we have;

$\mathrm{△}=a^2{b}^2{c}^2\left|\begin{array}{ccc}0& 0& 2\\ 1& -1& 1\\ 2& 0& 0\end{array}\right|$

Expanding to get R.H.S.

$\mathrm{△}=a^2{b}^2{c}^2\left(2\left|\begin{array}{cc}1& -1\\ 2& 0\end{array}\right|\right)=2a^2{b}^2{c}^2(0+2)=4a^2{b}^2{c}^2$

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

$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)(c-a)$
Answer:

We have the determinant $\text{dpi}100\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$

Applying the row transformations $R_1\to R_1-R_2$ and then $R_2\to R_2-R_3$ we have:

$\mathrm{△}=\left|\begin{array}{ccc}0& a-b& a^2-b^2\\ 0& b-c& b^2-c^2\\ 1& c& c^2\end{array}\right|$

$=\left|\begin{array}{ccc}0& a-b& (a-b)(a+b)\\ 0& b-c& (b-c)(b+c)\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)\left|\begin{array}{ccc}0& 1& (a+b)\\ 0& 1& (b+c)\\ 1& c& c^2\end{array}\right|$

Now, applying $R_1\to R_1-R_2$ we have:

$=(a-b)(b-c)\left|\begin{array}{ccc}0& 0& (a-c)\\ 0& 1& (b+c)\\ 1& c& c^2\end{array}\right|$ or $=(a-b)(b-c)(a-c)\left|\begin{array}{ccc}0& 0& 1\\ 0& 1& (b+c)\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)(a-c)\left|\begin{array}{cc}0& 1\\ 1& c\end{array}\right|$

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

Hence proved.

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

$\text{dpi}100\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

Answer:

Given determinant :

$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|$,

Applying column transformation $C_1\to C_1-C_3$ and then $C_2\to C_2-C_3$

We get,

$\mathrm{△}=\left|\begin{array}{ccc}0& 0& 1\\ a-c& b-c& c\\ a^3-c^3& b^3-c^3& c^3\end{array}\right|$

$=\left|\begin{array}{ccc}0& 0& 1\\ a-c& b-c& c\\ (a-c)(a^2+ac+c^2)& (b-c)(b^2+bc+c^2)& c^3\end{array}\right|$

$=(a-c)(b-c)\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& c\\ (a^2+ac+c^2)& (b^2+bc+c^2)& c^3\end{array}\right|$

Now, applying column transformation $C_1\to C_1-C_2$, we have:

$=(a-c)(b-c)\left|\begin{array}{ccc}0& 0& 1\\ 0& 1& c\\ (a^2-b^2+ac-bc)& (b^2+bc+c^2)& c^3\end{array}\right|$

$=(a-c)(b-c)\left|\begin{array}{ccc}0& 0& 1\\ 0& 1& c\\ (a-b)(a+b+c)& (b^2+bc+c^2)& c^3\end{array}\right|$

$=(a-c)(b-c)(a-b)(a+b+c)\left|\begin{array}{cc}0& 1\\ 1& c\end{array}\right|$

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

Hence proved.

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

$\left|\begin{array}{ccc}x& x^2& yz\\ y& y^2& zx\\ z& z^2& xy\end{array}\right|=(x-y)(y-z)(z-x)(xy+yz+zx)$

Answer:

We have the determinant:

$\mathrm{△}=\left|\begin{array}{ccc}x& x^2& yz\\ y& y^2& zx\\ z& z^2& xy\end{array}\right|$

Applying the row transformations $R_1\to R_1-R_3$ and then $R_2\to R_2-R_3$, we have;

$\mathrm{△}=\left|\begin{array}{ccc}x-z& x^2-z^2& yz-xy\\ y-z& y^2-z^2& zx-xy\\ z& z^2& xy\end{array}\right|$

$=\left|\begin{array}{ccc}x-z& (x-z)(x+z)& y(z-x)\\ y-z& (y-z)(y+z)& x(z-y)\\ z& z^2& xy\end{array}\right|$

$=(x-z)(y-z)\left|\begin{array}{ccc}1& (x+z)& -y\\ 1& (y+z)& -x\\ z& z^2& xy\end{array}\right|$

Now, applying $R_1\to R_1-R_2$; we have

$=(x-z)(y-z)\left|\begin{array}{ccc}0& (x-y)& (x-y)\\ 1& (y+z)& -x\\ z& z^2& xy\end{array}\right|$

$=(x-z)(y-z)(x-y)\left|\begin{array}{ccc}0& 1& 1\\ 1& (y+z)& -x\\ z& z^2& xy\end{array}\right|$

Now, expanding the remaining determinant;

$=(x-z)(y-z)(x-y)[(xy+zx)+(z^2-zy-z^2)]$

$=-(x-z)(y-z)(x-y)[xy+zx+zy]$

$=(x-y)(y-z)(z-x)[xy+zx+zy]$

Hence proved.

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

$\left|\begin{array}{ccc}x+4& 2x& 2x\\ 2x& x+4& 2x\\ 2x& 2x& x+4\end{array}\right|=(5x+4)(4-x)$

Answer:

Given determinant:

$\left|\begin{array}{ccc}x+4& 2x& 2x\\ 2x& x+4& 2x\\ 2x& 2x& x+4\end{array}\right|$

Applying row transformation: $R_1\to R_1+R_2+R_3$ then we have;

$\mathrm{△}=\left|\begin{array}{ccc}5x+4& 5x+4& 5x+4\\ 2x& x+4& 2x\\ 2x& 2x& x+4\end{array}\right|$

Taking a common factor: 5x+4

$=(5x+4)\left|\begin{array}{ccc}1& 1& 1\\ 2x& x+4& 2x\\ 2x& 2x& x+4\end{array}\right|$

Now, applying column transformations $C_1\to C_1-C_2$ and $C_2\to C_2-C_3$

$=(5x+4)\left|\begin{array}{ccc}0& 0& 1\\ x-4& 4-x& 2x\\ 0& x-4& x+4\end{array}\right|$

$=(5x+4)(4-x)(4-x)\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& 2x\\ 0& 1& x+4\end{array}\right|$

$=(5x+4)(4-x{)}^2$

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

$\left|\begin{array}{ccc}y+k& y& y\\ y& y+k& y\\ y& y& y+k\end{array}\right|=k^2(3y+k)$

Answer:

Given determinant:

$\mathrm{△}=\left|\begin{array}{ccc}y+k& y& y\\ y& y+k& y\\ y& y& y+k\end{array}\right|$

Applying row transformation $R_1\to R_1+R_2+R_3$ we get;

$=\left|\begin{array}{ccc}3y+k& 3y+k& 3y+k\\ y& y+k& y\\ y& y& y+k\end{array}\right|$

$=(3y+k)\left|\begin{array}{ccc}1& 1& 1\\ y& y+k& y\\ y& y& y+k\end{array}\right|$ [taking common (3y + k) factor]

Now, applying column transformation $C_1\to C_1-C_2$ and $C_2\to C_2-C_3$

$=(3y+k)\left|\begin{array}{ccc}0& 0& 1\\ -k& k& y\\ 0& -k& y+k\end{array}\right|$

$=(3y+k)(k^2)\left|\begin{array}{ccc}0& 0& 1\\ -1& 1& y\\ 0& -1& y+k\end{array}\right|$

$=k^2(3y+k)$

Hence proved.

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

$\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|=(a+b+c{)}^3$

Answer:

Given determinant:

$\mathrm{△}=\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$

We apply row transformation: $R_1\to R_1+R_2+R_3$ we have;

$=\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$

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

$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$

Now, applying column tranformation $C_1\to C_1-C_2$ and then $C_2\to C_2-C_3$

We have;

$=(a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ b+c+a& -b-c-a& 2b\\ 0& c+a+b& c-a-b\end{array}\right|$

$=(a+b+c)(a+b+c)(a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ 1& -1& 2b\\ 0& 1& c-a-b\end{array}\right|$

$=(a+b+c)(a+b+c)(a+b+c)=(a+b+c{)}^3$

Hence Proved.

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

$\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|=2(x+y+z{)}^3$

Answer:

Given determinant

$\mathrm{△}=\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|$

Applying $C_1\to C_1+C_2+C_3$ we get;

$=\left|\begin{array}{ccc}2(x+y+z)& x& y\\ 2(z+y+x)& y+z+2x& y\\ 2(z+y+x)& x& z+x+2y\end{array}\right|$

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

$=2(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 1& y+z+2x& y\\ 1& x& z+x+2y\end{array}\right|$

Now, applying row transformations, $R_1\to R_1-R_2$ and then $R_2\to R_2-R_3$.

we get;

$=2(x+y+z)\left|\begin{array}{ccc}0& -x-y-z& 0\\ 0& y+z+x& -y-z-x\\ 1& x& z+x+2y\end{array}\right|$

$=2(x+y+z{)}^3\left|\begin{array}{ccc}0& -1& 0\\ 0& 1& -1\\ 1& x& z+x+2y\end{array}\right|$

$=2(x+y+z{)}^3\left|\begin{array}{cc}-1& 0\\ 1& -1\end{array}\right|=2(x+y+z{)}^3$

Hence proved.

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

$\left|\begin{array}{ccc}1& x& x^2\\ x^2& 1& x\\ x& x^2& 1\end{array}\right|=(1-x^3{)}^2$

Answer:

Give determinant $\left|\begin{array}{ccc}1& x& x^2\\ x^2& 1& x\\ x& x^2& 1\end{array}\right|$

Applying column transformation $C_1\to C_1+C_2+C_3$ we get;

$\mathrm{△}=\left|\begin{array}{ccc}1+x+x^2& x& x^2\\ x^2+1+x& 1& x\\ x+x^2+1& x^2& 1\end{array}\right|$

$=(1+x+x^2)\left|\begin{array}{ccc}1& x& x^2\\ 1& 1& x\\ 1& x^2& 1\end{array}\right|$ [after taking the (1+x+x² ) factor common out.]

Now, applying row transformations, $R_1\to R_1-R_2$ and then $R_2\to R_2-R_3$.

we have now,

$=(1+x+x^2)\left|\begin{array}{ccc}0& x-1& x^2-x\\ 0& 1-x^2& x-1\\ 1& x^2& 1\end{array}\right|$

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

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

$=(1+x+x^2)(x-1)(x^3-1)=(x^3-1{)}^2$

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

Hence proved.

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

$\left|\begin{array}{ccc}1+a^2-b^2& 2ab& -2b\\ 2ab& 1-a^2+b^2& 2a\\ 2b& -2a& 1-a^2-b^2\end{array}\right|=(1+a^2+b^2{)}^3$

Answer:

We have determinant:

$\mathrm{△}=\left|\begin{array}{ccc}1+a^2-b^2& 2ab& -2b\\ 2ab& 1-a^2+b^2& 2a\\ 2b& -2a& 1-a^2-b^2\end{array}\right|$

Applying row transformations, $R_1\to R_1+b{R}_3$ and $R_2\to R_2-a{R}_3$ then we have;