Ncert Solutions For Class 12 Maths Ex 4.2

Ncert Solutions For Class 12 Maths Chapter 4 Ex 4.2

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

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

Sol:

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

(Here, the two columns of the determinants are identical)

Q2: Using the property of determinants and without expanding, prove that:

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

Sol:

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

Applying $$R_{1} \rightarrow R_{1} + R_{2}$$, we have:

$$\boldsymbol{\Rightarrow }$$ $$\Delta = \begin{vmatrix} a – b & b – a & c – b \\ b – c & c – a & a – b \\ -(a – c) & -(b – a) & -(c – b) \end{vmatrix} = 0\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$– \begin{vmatrix} a – c & b – a & c – b \\ b – c & c – a & a – b \\ a – c & b – a & c – b \end{vmatrix} = 0$$

Here, the two rows R1 and R3 are identical.

Therefore, $$\bigtriangleup = 0$$

Q3: Using the property of determinants and without expanding, prove that:

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

Sol:

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

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

$$\\\boldsymbol{\Rightarrow }$$ $$\begin{vmatrix} 2 & 7 & 9(7) \\ 3 & 8 & 9(8) \\ 5 & 9 & 9(9) \end{vmatrix} + 0$$ (Two columns are identical)

$$\\9 \begin{vmatrix} 2 & 7 & 7 \\ 3 & 8 & 8 \\ 5 & 9 & 9 \end{vmatrix} + 0$$ (Two columns are identical)

= 0

Q4: Using the property of determinants and without expanding, prove that:

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

Sol:

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

By applying $$C_{3} \rightarrow C_{3} + C_{2}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\Delta = \begin{vmatrix} 1 & bc & ab + bc + ca \\ 1 & ca & ab + bc + ca \\ 1 & ab & ab + bc + ca \end{vmatrix} = 0\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$(ab + bc + ca ) \begin{vmatrix} 1 & bc & 1 \\ 1 & ca & 1 \\ 1 & ab & 1 \end{vmatrix} = 0$$

Here, two columns C1 and C3 are proportional.

$$\boldsymbol{\Rightarrow }$$ $$\Delta = 0$$

Q5: Using the property of determinants and without expanding, prove that:

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

Sol:

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

= $$\Delta _{1} + \Delta _{2}$$ ……………………….(i)

Now, $$\Delta _{1} = \begin{vmatrix} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a & p & x \end{vmatrix}$$

Applying $$R_{2}\rightarrow R_{2} – R_{3}$$, we have:

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

Applying $$R_{1}\rightarrow R_{1} – R_{2}$$, we have:

$$\Delta _{1} = \begin{vmatrix} b & q & y \\ c & r & z \\ a & p & x \end{vmatrix}\\$$

Applying $$R_{1} \leftrightarrow R_{3} \;\; and R_{2} \leftrightarrow R_{3}$$ we have:

$$\Delta _{1} = (-1)^{2} \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix}$$ …………………..(ii)

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

Applying $$R_{1} \rightarrow R_{1} – R_{3}$$, we have:

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

Applying $$R_{2} \rightarrow R_{2} – R_{1}$$, we have:

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

Applying $$R_{1} \leftrightarrow R_{2} \;\; and R_{2} \leftrightarrow R_{3}$$ we have:

$$\Delta _{2} = (-1)^{2} \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix}$$ ………………………(iii)

From (i), (ii) and (iii), we have:

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

Q6: By using properties of determinants, show that:

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

Sol:

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

Applying $$R_{1} = c R_{1}$$, we have:

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

Applying $$R_{1} \rightarrow R_{1} – b R_{2}$$, we have:

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

Here, the two rows $$R_{1} \;\; and \;\; R_{3}$$ are identical.

$$\boldsymbol{\Rightarrow }$$ $$\Delta = 0$$

Q7: By using the properties of determinants, show that:

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

Sol:

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

$$\\\begin{vmatrix} -a^{2} & ab & ac \\ ba & -b^{2} & bc \\ ca & cb & -c^{2} \end{vmatrix}$$ (Taking out factors a, b, c from R1 , R2 and R3)

$$\Delta = a^{2}b^{2}c^{2} \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix}$$ (Taking out factors a,b,c from C1, C2 and C3)

Applying $$R_{2} \rightarrow R_{2} + R_{1} \;\; and \;\; R_{3} \rightarrow R_{3} + R_{1}$$ we have:

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

$$\\\boldsymbol{\Rightarrow }$$ $$\Delta = a^{2}b^{2}c^{2} (-1) \begin{vmatrix} 0 & 2 \\ 2 & 0 \end{vmatrix}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$-a^{2} b^{2} c^{2} (0 – 4) = 4a^{2} b^{2} c^{2}$$

Q8: By using the properties of determinants, show that:

(i) $$\begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix} = (a – b) (b – c) (c – a)$$

(ii) $$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{vmatrix} = (a – b) (b – c) (c – a)(a + b + c)$$

Sol:

Let $$\Delta = \begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix}$$

Applying $$R_{1} \rightarrow R_{1} – R_{3}$$ and $$R_{2} \rightarrow R_{2} – R_{3}$$, we have:

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

Applying $$R_{1}\rightarrow R_{1}+R_{2}$$ we have:

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

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

Expanding along $$C_{1}$$ we have:

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

Hence, the given result is proved.

(ii) Let $$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{vmatrix}$$

Applying $$C_{1}\rightarrow C_{1}-C_{3} \;\; and C_{2}\rightarrow C_{2}- C_{3}$$ we have:

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

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

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

Applying $$C_{1}\rightarrow C_{1}+C_{2}$$ we have:

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

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

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

Expanding along $$C_{1}$$ we have:

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

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

Hence, the given result is proved.

Q9: By using the properties of determinants, show that:

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

Sol:

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

Applying $$R_{2}\rightarrow R_{2}-R_{1} \;\; and R_{3}\rightarrow R_{3}-R_{1}$$ we have:

$$\\\boldsymbol{\Rightarrow }$$ $$\Delta = \begin{vmatrix} x & x^{2} & yz \\ y-x & y^{2}-x^{2} & zx-yz \\ z-x & z^{2}-x^{2} xy-yz & -y(z-x) \end{vmatrix}\\$$

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

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

Applying $$R_{3}\rightarrow R_{3}+R_{2}$$ we have:

$$\boldsymbol{\Rightarrow }$$ $$\Delta = (x-y)(z-x)\begin{vmatrix} x & x^{2} & yz \\ -1 & -x-y & z \\ 0 & z-y & z-y \end{vmatrix}\\$$

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

Expanding along $$R_{3}$$ we have:

$$\boldsymbol{\Rightarrow }$$ $$\Delta = \left [ (x-y)(z-x)(z-y) \right ] \left [ (-1) \begin{vmatrix} x & yz \\ -1 & z \end{vmatrix} + 1 \begin{vmatrix} x & x^{2} \\ -1 & -x-y \end{vmatrix} \right ]\\$$

= (x-y)(z-x)(z-y)[(-xz-zy)$$+(x^{2}-xy+x^{2})$$]

= -(x-y)(y-z)(z-x)(xy+yz+zx)

Hence, the given result is proved.

Q10: By using properties of determinants, show that:

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

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

Sol:

(i) $$\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} \\$$

Applying $$R_{1}\rightarrow R_{1}+R_{2}+R_{3}$$ we have:

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

Applying $$C_{2}\rightarrow C_{2}-C_{1}, C_{3}\rightarrow C_{3}-C_{1}$$ we have:

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

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

Expanding along $$C_{3}$$ we have:

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

Hence, the given result is proved.

(ii) $$\begin{vmatrix} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{vmatrix}$$

Applying $$R_{1}\rightarrow R_{1}+R_{2}+R_{3}$$ we have:

$$\\\Delta = \begin{vmatrix} 3y+k & 3y+k & 3y+k \\ y & y+k & y \\ y & y & y+k \end{vmatrix}$$ $$= (3y+k) \begin{vmatrix} 1 & 1 & 1 \\ y & y+k & y \\ y & y & y+k \end{vmatrix}\\$$

Applying $$C_{2}\rightarrow C_{2}-C_{1}, C_{3}\rightarrow C_{3}-C_{1}$$ we have:

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

Expanding along $$C_{3}$$ we have:

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

Hence, the given result is proved.

Q.11: By using properties of determinants, show that:

(i) $$\begin {vmatrix} a – b – c & 2a & 2a \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end {vmatrix} = \left ( a + b + c \right )^{3}\\$$

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

Sol:

$$\Delta = \begin {vmatrix} a – b – c & 2a & 2a \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end {vmatrix}$$

(i)

Applying $$R_{1} \rightarrow R_{1} + R_{2} + R_{3}$$, we have:

$$\Delta = \begin {vmatrix} a + b + c & a + b + c & a + b + c \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end {vmatrix} \\$$

= $$\left ( a + b + c \right ) \begin {vmatrix} 1 & 1 & 1 \\ 2b & b – c – a & 2b\\ 2c & 2c & c – a – b \end {vmatrix}$$

Applying $$C_{2} \rightarrow C_{2} – C_{1} , C_{3} \rightarrow C_{3} – C_{1}$$, we have:

$$\Delta = \left ( a + b + c \right ) \begin {vmatrix} 1 & 0 & 0 \\ 2b & – \left ( a + b + c \right ) & 0 \\ 2c & 0 & – \left ( a + b + c \right ) \end {vmatrix} \\$$

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

Expanding along C3, we have:

∆ = (a + b + c)3 (-1) (-1) = (a + b + c)3

Hence, the given result is proved.

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

(ii) Applying C1 → C1 + C2 + C3, we have:

$$\\\Delta = \begin{vmatrix} 2 \left ( x + y + z \right ) & x & y \\ 2 \left ( x + y + z \right ) & y + z + 2x & y \\ 2 \left ( x + y + z \right ) & x & z + x + 2y \end {vmatrix}$$

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

Applying R2 → R2 – R1 and R3 → R3 – R1, we have:

$$\Delta = 2 \left ( x + y + z \right ) \begin {vmatrix} 1 & x & y \\ 0 & x + y + z & 0 \\ 0 & 0 & x + y + z \end {vmatrix}\\$$

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

Expanding along R3, we have:

∆ = 2(x + y + z)3 (1) (1 – 0) = 2(x + y + z)3

Hence, the given results are proved.

Q.12: By using properties of determinants, show that:

$$\begin {vmatrix} 1 & x & x ^ {2} \\ x ^ {2} & 1 & x \\ x & x ^ {2} & 1 \end {vmatrix} = \left ( 1 – x ^{3} \right ) ^ {2}$$

Sol:

$$\Delta = \begin {vmatrix} 1 & x & x ^ {2} \\ x ^ {2} & 1 & x \\ x & x ^ {2} & 1 \end {vmatrix} \\$$

Applying R1 → R1 + R2 + R3, we have:

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

= $$\\\left ( 1 + x + x ^ {2} \right ) \begin {vmatrix} 1 & 1 & 1 \\ x ^ {2} & 1 & x \\ x & x ^ {2} & 1 \end {vmatrix}$$

Applying C2 → C2 – C1 and C3 → C3 – C1, we have:

$$\\ \Delta = \left ( 1 + x + x ^ {2} \right ) \begin {vmatrix} 1 & 0 & 0 \\ x ^ {2} & 1 – x ^ {2} & x – x ^ {2} \\ x & x ^ {2} – x & 1 – x \end {vmatrix}$$

= $$\\\left ( 1 + x + x ^ {2} \right ) \left ( 1 – x \right ) \left ( 1 – x \right ) \begin {vmatrix} 1 & 0 & 0 \\ x ^ {2} & 1 + x & x \\ x & -x &aamp; 1 \end {vmatrix}\\$$

= $$\\\left ( 1 – x ^ {3} \right ) \left ( 1 – x \right ) \begin {vmatrix} 1 & 0 & 0 \\ x ^ {2} & 1 + x & x \\ x & -x & 1 \end {vmatrix}\\$$

Expanding along R1, we have:

$$\\\Delta = \left ( 1 – x ^ {3} \right ) \left ( 1 – x \right ) \left ( x \right ) \begin {vmatrix} 1 + x & x \\ -x & 1 \end {vmatrix}\\$$

= (1 – x3) (1 – x) (1 + x + x2)

= (1 – x3) (1 – x3)

= (1 – x3)2

Hence, the given result is proved.

Q.13: By using properties of determinants, show that:

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

Sol:

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

Applying R1 → R1 + bR3 and R2 → R2 – aR3, we have:

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

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

Expanding along R1, we have:

$$\\ \Delta = \left ( 1 + a ^ {2} + b ^ {2} \right ) ^ {2} \begin {bmatrix} (1) \begin {vmatrix} 1 & a \\ -2a & 1 – a^ {2} – b^ {2} \end {vmatrix} \\ -b \begin {vmatrix} 0 & 1 \\ 2b & -2a \end {vmatrix} \end {bmatrix} \\$$

= (1 + a2 + b2)2 [1 – a2 – b2 + 2a2 –b (-2b)]

= (1 + a2 + b2)2 (1 + a2 + b2)

= (1 + a2 + b2)3

Q.14: By using properties of determinants, show that:

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

Sol:

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

Taking out common factors a, b, c from R1, R2 and R3 respectively, we have:

$$\Delta = abc \begin {vmatrix} a + \frac {1} {b} & b & c \\ a & b + \frac {1} {b} & c \\ a & b & c + \frac {1} {c} \end {vmatrix}\\$$

Applying R2 → R2 – R1 and R3 → R3 – R1, we have:

$$\\ \Delta = abc \begin {vmatrix} a + \frac {1} {b} & b & c \\ – \frac {1} {a} & \frac {1} {b} & 0 \\ – \frac {1} {a} & 0 & \frac {1} {c} \end {vmatrix}\\$$

Applying C1 → aC1, C2 → bC2, and C3 → cC3, we have:

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

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

Expanding along R3, we have:

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

= -1(-c2) + (a2 + 1 + b2) = 1 + a2 + b2 + c2

Hence, the given result is proved.

Let A be a square matrix of order 3 × 3, KA is equal to

(1). k|A|

(2). k2|A|

(3). k3|A|

(4). 3k|A|

(3)

A is a square matrix of order 3 × 3

Let A = $$\begin {vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end {vmatrix} \\$$

Then, kA = $$\begin{vmatrix} ka_{1} & kb_{1} & kc_{1} \\ ka_{2} & kb_{2} & kc_{2} \\ ka_{3} & kb_{3} & kc_{3} \end {vmatrix}$$

Therefore, $$\\|kA| = \begin {vmatrix} ka_{1} & kb_{1} & kc_{1} \\ ka_{2} & kb_{2} & kc_{2} \\ ka_{3} & kb_{3} & kc_{3} \end {vmatrix}\\$$

$$\\ k ^ {3} = \begin {vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end {vmatrix} \; \; \; \; \; \; \; \; \; \; \left ( Taking \; out \; common \; factors \; k \; from \; each \; row \right )$$

= k3|A|

Therefore, |kA| = k3|A|

Hence, the correct answer is C.

Q.16: Which of the following is correct?

(1) Determinant is a square matrix.

(2) Determinant is a number associated to a matrix.

(3) Determinant is a number associated to a square matrix.

Sol:

(3)

We know that to every square matrix, A = [aij] of order n. we can associate a number called the determinant of a square matrix A, where aij = (i, j)th element of A.

Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is C.