Q: Using the property of determinants and without expanding, prove that: $$ \left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|=0 $$

$\Delta=\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$ $=\left|\begin{array}{ccc}a-c & b-a & c-b \\ b-c & c-a & a-b \\ -(a-c) & -(b-a) & -(c-b)\end{array}\right| \quad\left[R_{1} \rightarrow R_{1}+R_{2}\right]$ $$ =\left|\begin{array}{lll} a-c & b-a & c-b \\ b-c & c-a & a-b \\ a-c & b-a & c-b \end{array}\right| $$ Here, the two rows $R_{1}$ and $R_{3}$ are identical. Hence, $\Delta=0$

Q: Using the property of determinants and without expanding, prove that: $\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$

$\Delta=\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|$ $=\frac{1}{c}\left|\begin{array}{ccc}0 & a c & -b c \\ -a & 0 & -c \\ b & c & 0\end{array}\right| \quad\left[R_{1} \rightarrow c R_{1}\right]$ $=\frac{1}{c}\left|\begin{array}{ccc}a b & a c & 0 \\ -a & 0 & -c \\ b & c & 0\end{array}\right| \quad\left[R_{1} \rightarrow R_{1}-b R_{2}\right]$ $=\frac{a}{c}\left|\begin{array}{ccc}b & c & 0 \\ -a & 0 & -c \\ b & c & 0\end{array}\right|$ Here, the two rows $R_{1}$ and $R_{3}$ are identical. Hence, $\Delta=0$

Q: Using the property of determinants and without expanding, prove that: $\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

$\Delta=\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|$ $=a b c\left|\begin{array}{ccc}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right|$ [Taking out factors $a, b, c$ from $R_{1}, R_{2}, R_{3}$ ] $=a^{2} b^{2} c^{2}\left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right|$ [Taking out factors $a, b, c$ from $C_{1}, C_{2}, C_{3}$ ] $=a^{2} b^{2} c^{2}\left|\begin{array}{ccc}-1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0\end{array}\right|$ $=a^{2} b^{2} c^{2}(-1)\left|\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right| \left[R_{2} \rightarrow R_{2}+R_{1}\right.$ and $\left.R_{3} \rightarrow R_{3}+R_{1}\right]$ $=-a^{2} b^{2} c^{2}(0-4)$ $=4 a^{2} b^{2} c^{2}$

Q: By using properties of determinants show that: $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

Let $\Delta=\left|\begin{array}{ccc}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|$ $$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} x & x^{2} & y z \\ y-x & y^{2}-x^{2} & z x-y z \\ z-x & z^{2}-x^{2} & x y-y z \end{array}\right| \quad\left[R_{2} \rightarrow R_{2}-R_{1} \text { and } R_{3} \rightarrow R_{3}-R_{1}\right] \\ & =\left|\begin{array}{ccc} x & x^{2} & y z \\ -(x-y) & -(x-y)(x+y) & z(x-y) \\ (z-x) & (z-x)(z+x) & -y(z-x) \end{array}\right| \\ & =(x-y)(z-x)\left|\begin{array}{ccc} x & x^{2} & y z \\ -1 & -x-y & z \\ 1 & (z+x) & -y \end{array}\right| \\ \Delta & =(x-y)(z-x)\left|\begin{array}{ccc} x & x^{2} & y z \\ -1 & -x-y & z \\ 0 & z-y & z-y \end{array}\right| \\ & =(x-y)(z-x)(z-y)\left|\begin{array}{ccc} x & x^{2} & y z \\ -1 & -x-y & z \\ 0 & 1 & 1 \end{array}\right| \\ & =[(x-y)(z-x)(z-y)]\left[\begin{array}{cc} (-1) & x \\ -1 & z \end{array}|+1| \begin{array}{cc} x & x^{2} \\ -1 & -x-y \end{array}\right] \\ & =(x-y)(z-x)(z-y)\left[(-x z-y z)+\left(-x^{2}-x y+x^{2}\right)\right] \\ & =-(x-y)(z-x)(z-y)(x y+y z+z x) \\ & =(x-y)(y-z)(z-x)(x y+y z+z x) \end{aligned} $$ Hence, $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

Q: By using properties of determinants show that: (i) $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$ (ii) $\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$

(i) $$ \Delta=\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| $$ $$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \\ & =(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \\ & =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c) \end{array}\right| \\ & =(a+b+c)^{3}\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -1 & 0 \\ 2 c & 0 & -1 \end{array}\right| \\ & =(a+b+c)^{3}(-1)(-1) \\ & =(a+b+c)^{3} \end{aligned} $$ Hence, proved. (ii) $$ \begin{array}{rlr} \Delta & =\left|\begin{array}{ccc} x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y \end{array}\right| & \\ \Delta & =\left|\begin{array}{ccc} 2(x+y+z) & x & y \\ 2(x+y+z) & y+z+2 x & y \\ 2(x+y+z) & x & z+x+2 y \end{array}\right| & {\left[C_{1} \rightarrow C_{1}+C_{2}+C_{3}\right]} \\ & =2(x+y+z)\left|\begin{array}{ccc} 1 & x & y \\ 1 & y+z+2 x & y \\ 1 & x & z+x+2 y \end{array}\right| & \\ & =2(x+y+z)\left|\begin{array}{ccc} 1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z \end{array}\right| & {\left[R_{2} \rightarrow R_{2}-R_{1} \text { and } R_{3} \rightarrow R_{3}-R_{1}\right]} \\ & =2(x+y+z)^{3}\left|\begin{array}{ccc} 1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right| & \\ & =2(x+y+z)^{3}(1)(1-0) & \\ & =2(x+y+z)^{3} & \end{array} $$ Hence, proved.

Question 1

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

$\left|\begin{array}{lll}x & a & x+a \ y & b & y+b \ z & c & z+c\end{array}ight|=0$

Accepted Answer

$$

\begin{aligned}

\Delta & =\left|\begin{array}{lll}

x & a & x+a \

y & b & y+b \

z & c & z+c

\end{array}ight| \

& =\left|\begin{array}{lll}

x & a & x \

y & b & y \

z & c & z

\end{array}ight|+\left|\begin{array}{lll}

x & a & a \

y & b & b \

z & c & c

\end{array}ight|

\end{aligned}

$$

Here, two columns of each determinant are identical.

Hence,

$$

\begin{aligned}

\Delta & =0+0 \

& =0

\end{aligned}

$$

Question 2

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

$$

\left|\begin{array}{lll}

a-b & b-c & c-a \

b-c & c-a & a-b \

c-a & a-b & b-c

\end{array}ight|=0

$$

Accepted Answer

$\Delta=\left|\begin{array}{lll}a-b & b-c & c-a \ b-c & c-a & a-b \ c-a & a-b & b-c\end{array}ight|$

$=\left|\begin{array}{ccc}a-c & b-a & c-b \ b-c & c-a & a-b \ -(a-c) & -(b-a) & -(c-b)\end{array}ight| \quad\left[R_{1} ightarrow R_{1}+R_{2}ight]$

$$

=\left|\begin{array}{lll}

a-c & b-a & c-b \

b-c & c-a & a-b \

a-c & b-a & c-b

\end{array}ight|

$$

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

Hence, $\Delta=0$

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}ight|=0

$$

Accepted Answer

$$

\begin{aligned}

\Delta & =\left|\begin{array}{lll}

2 & 7 & 65 \

3 & 8 & 75 \

5 & 9 & 86

\end{array}ight| \

& =\left|\begin{array}{lll}

2 & 7 & 63+2 \

3 & 8 & 72+3 \

5 & 9 & 81+5

\end{array}ight| \

& =\left|\begin{array}{lll}

2 & 7 & 63 \

3 & 8 & 72 \

5 & 9 & 81

\end{array}ight|+\left|\begin{array}{lll}

2 & 7 & 2 \

3 & 8 & 3 \

5 & 9 & 5

\end{array}ight| \

& =\left|\begin{array}{ccc}

2 & 7 & 9(7) \

3 & 8 & 9(8) \

5 & 9 & 9(9)

\end{array}ight|+0 \

& =9\left|\begin{array}{ccc}

2 & 7 & 7 \

3 & 8 & 8 \

5 & 9 & 9

\end{array}ight| \

& =0

\end{aligned}

$$

Question 4

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

$$

\left|\begin{array}{lll}

1 & b c & a(b+c) \

1 & c a & b(c+a) \

1 & a b & c(a+b)

\end{array}ight|=0

$$

Accepted Answer

$$

\begin{array}{rlr}

\Delta & =\left|\begin{array}{lll}

1 & b c & a(b+c) \

1 & c a & b(c+a) \

1 & a b & c(a+b)

\end{array}ight| & \

& =\left|\begin{array}{lll}

1 & b c & a b+b c+c a \

1 & c a & a b+b c+c a \

1 & a b & a b+b c+c a

\end{array}ight| & {\left[C_{3} ightarrow C_{3}+C_{2}ight]}

\end{array}

$$

Here, the two columns $C_{1}$ and $C_{3}$ are proportional.

Hence, $\Delta=0$

Question 5

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

$$

\left|\begin{array}{lll}

b+c & q+r & y+z \

c+a & r+p & z+x \

a+b & p+q & x+y

\end{array}ight|=2\left|\begin{array}{lll}

a & p & x \

b & q & y \

c & r & z

\end{array}ight|

$$

Accepted Answer

$$

\begin{align*}

\Delta & =\left|\begin{array}{ccc}

b+c & q+r & y+z \

c+a & r+p & z+x \

a+b & p+q & x+y

\end{array}ight| \

& =\left|\begin{array}{ccc}

b+c & q+r & y+z \

c+a & r+p & z+x \

a & p & x

\end{array}ight|+\left|\begin{array}{ccc}

b+c & q+r & y+z \

c+a & r+p & z+x \

b & q & y

\end{array}ight| \

& =\Delta_{1}+\Delta_{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots(1) 	ag{1}

\end{align*}

$$

Now,

$$

\begin{array}{rlr}

\Delta_{1} & =\left|\begin{array}{ccc}

b+c & q+r & y+z \

c+a & r+p & z+x \

a & p & x

\end{array}ight| & \

& =\left|\begin{array}{ccc}

b+c & q+r & y+z \

c & r & z \

a & p & x

\end{array}ight| & {\left[R_{2} ightarrow R_{2}-R_{3}ight]} \

& =\left|\begin{array}{ccc}

b & q & y \

c & r & z \

a & p & x

\end{array}ight| & {\left[R_{1} ightarrow R_{1}-R_{2}ight]} \

& =(-1)^{2}\left|\begin{array}{ccc}

a & p & x \

b & q & y \

c & r & z

\end{array}ight| & \

\Delta_{1} & =\left|\begin{array}{ccc}

a & p & x \

b & q & y \

c & r & z

\end{array}ight| & {\left[R_{1} \leftrightarrow R_{3} 	ext { and } R_{2} \leftrightarrow R_{3}ight]} \

\Delta_{2} & =\left|\begin{array}{ccc}

b+c & q+r & y+z \

c+a & r+p & z+x \

b & q & y

\end{array}ight| & {\left[R_{1} ightarrow R_{1}-R_{3}ight]} \

\Delta_{2} & =\left|\begin{array}{ccc}

c & r & z \

c+a & r+p & z+x \

b & q & y

\end{array}ight| & {\left[R_{2} ightarrow R_{2}-R_{1}ight]} \

\Delta_{2} & =\left|\begin{array}{ccc}

c & r & z \

a & p & x \

b & q & y

\end{array}ight| & \

\Delta_{2} & =(-1)^{2}\left|\begin{array}{ccc}

a & p & x \

b & q & y \

c & r & z

\end{array}ight| & {\left[R_{1} \leftrightarrow R_{2} 	ext { and } R_{2} \leftrightarrow R_{3}ight]}

\end{array}

$$

From (1),(2) and (3), we have

$$

\begin{aligned}

\Delta & =\left|\begin{array}{lll}

a & p & x \

b & q & y \

c & r & z

\end{array}ight|+\left|\begin{array}{lll}

a & p & x \

b & q & y \

c & r & z

\end{array}ight| \

& =2\left|\begin{array}{lll}

a & p & x \

b & q & y \

c & r & z

\end{array}ight|

\end{aligned}

$$

Hence, $\left|\begin{array}{lll}b+c & q+r & y+z \ c+a & r+p & z+x \ a+b & p+q & x+y\end{array}ight|=2\left|\begin{array}{lll}a & p & x \ b & q & y \ c & r & z\end{array}ight|$ proved.

Question 6

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

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

Accepted Answer

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

$=\frac{1}{c}\left|\begin{array}{ccc}0 & a c & -b c \ -a & 0 & -c \ b & c & 0\end{array}ight| \quad\left[R_{1} ightarrow c R_{1}ight]$

$=\frac{1}{c}\left|\begin{array}{ccc}a b & a c & 0 \ -a & 0 & -c \ b & c & 0\end{array}ight| \quad\left[R_{1} ightarrow R_{1}-b R_{2}ight]$

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

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

Hence, $\Delta=0$

Question 7

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

$\left|\begin{array}{ccc}-a^{2} & a b & a c \ b a & -b^{2} & b c \ c a & c b & -c^{2}\end{array}ight|=4 a^{2} b^{2} c^{2}$

Accepted Answer

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

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

[Taking out factors $a, b, c$ from $R_{1}, R_{2}, R_{3}$ ]

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

[Taking out factors $a, b, c$ from $C_{1}, C_{2}, C_{3}$ ]

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

$=a^{2} b^{2} c^{2}(-1)\left|\begin{array}{ll}0 & 2 \ 2 & 0\end{array}ight| \left[R_{2} ightarrow R_{2}+R_{1}ight.$ and $\left.R_{3} ightarrow R_{3}+R_{1}ight]$

$=-a^{2} b^{2} c^{2}(0-4)$

$=4 a^{2} b^{2} c^{2}$

Question 8

By using properties of determinants show that:

(i) $\left|\begin{array}{lll}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}\end{array}ight|=(a-b)(b-c)(c-a)$

(ii) $\left|\begin{array}{ccc}1 & 1 & 1 \ a & b & c \ a^{3} & b^{3} & c^{3}\end{array}ight|=(a-b)(b-c)(c-a)(a+b+c)$

Accepted Answer

(i) Let $\Delta=\left|\begin{array}{lll}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}\end{array}ight|$

$$

\begin{array}{rlr}

\Delta & =\left|\begin{array}{ccc}

0 & a-c & a^{2}-c^{2} \

0 & b-c & b^{2}-c^{2} \

1 & c & c^{2}

\end{array}ight| & \

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

0 & -1 & -a-c \

0 & 1 & b+c \

1 & c & c^{2}

\end{array}ight| & \

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

0 & 0 & -a+b \

0 & 1 & b+c \

1 & c & c^{2}

\end{array}ight| & {\left[R_{1} ightarrow R_{1}+R_{2}ight]} \

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

0 & 0 & -1 \

0 & 1 & b+c \

1 & c & c^{2}

\end{array}ight| & \

& =(a-b)(b-c)(c-a)\left|\begin{array}{cc}

0 & -1 \

1 & b+c

\end{array}ight| & \

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

\end{array}

$$

Hence, $\left|\begin{array}{lll}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}\end{array}ight|=(a-b)(b-c)(c-a) \quad$ proved.

(ii) Let $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \ a & b & c \ a^{3} & b^{3} & c^{3}\end{array}ight|$

$\Delta=\left|\begin{array}{ccc}0 & 0 & 1 \ a-c & b-c & c \ a^{3}-c^{3} & b^{3}-c^{3} & c^{3}\end{array}ight| \quad\left[C_{1} ightarrow C_{1}-C_{3}ight.$ and $\left.C_{2} ightarrow C_{2}-C_{3}ight]$

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

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

Applying $C_{1} ightarrow C_{1}+C_{2}$,

$$

\begin{aligned}

\Delta & =(c-a)(b-c)\left|\begin{array}{ccc}

0 & 0 & 1 \

0 & 1 & c \

\left(b^{2}-a^{2}ight)+(b c-a c) & \left(b^{2}+b c+c^{2}ight) & c^{3}

\end{array}ight| \

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

0 & 0 & 1 \

0 & 1 & c \

-(a+b+c) & \left(b^{2}+b c+c^{2}ight) & c^{3}

\end{array}ight| \

& =(a-b)(b-c)(c-a)(a+b+c)\left|\begin{array}{ccc}

0 & 0 & 1 \

0 & 1 & c \

-1 & \left(b^{2}+b c+c^{2}ight) & c^{3}

\end{array}ight| \

& =(a-b)(b-c)(c-a)(a+b+c)(-1)\left|\begin{array}{ll}

0 & 1 \

1 & c

\end{array}ight| \

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

\end{aligned}

$$

Hence, $\left|\begin{array}{ccc}1 & 1 & 1 \ a & b & c \ a^{3} & b^{3} & c^{3}\end{array}ight|=(a-b)(b-c)(c-a)(a+b+c) \quad$ proved.

Question 9

By using properties of determinants show that:

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

Accepted Answer

Let $\Delta=\left|\begin{array}{ccc}x & x^{2} & y z \ y & y^{2} & z x \ z & z^{2} & x y\end{array}ight|$

$$

\begin{aligned}

\Delta & =\left|\begin{array}{ccc}

x & x^{2} & y z \

y-x & y^{2}-x^{2} & z x-y z \

z-x & z^{2}-x^{2} & x y-y z

\end{array}ight| \quad\left[R_{2} ightarrow R_{2}-R_{1} 	ext { and } R_{3} ightarrow R_{3}-R_{1}ight] \

& =\left|\begin{array}{ccc}

x & x^{2} & y z \

-(x-y) & -(x-y)(x+y) & z(x-y) \

(z-x) & (z-x)(z+x) & -y(z-x)

\end{array}ight| \

& =(x-y)(z-x)\left|\begin{array}{ccc}

x & x^{2} & y z \

-1 & -x-y & z \

1 & (z+x) & -y

\end{array}ight| \

\Delta & =(x-y)(z-x)\left|\begin{array}{ccc}

x & x^{2} & y z \

-1 & -x-y & z \

0 & z-y & z-y

\end{array}ight| \

& =(x-y)(z-x)(z-y)\left|\begin{array}{ccc}

x & x^{2} & y z \

-1 & -x-y & z \

0 & 1 & 1

\end{array}ight| \

& =[(x-y)(z-x)(z-y)]\left[\begin{array}{cc}

(-1) & x \

-1 & z

\end{array}|+1| \begin{array}{cc}

x & x^{2} \

-1 & -x-y

\end{array}ight] \

& =(x-y)(z-x)(z-y)\left[(-x z-y z)+\left(-x^{2}-x y+x^{2}ight)ight] \

& =-(x-y)(z-x)(z-y)(x y+y z+z x) \

& =(x-y)(y-z)(z-x)(x y+y z+z x)

\end{aligned}

$$

Hence, $\left|\begin{array}{lll}x & x^{2} & y z \ y & y^{2} & z x \ z & z^{2} & x y\end{array}ight|=(x-y)(y-z)(z-x)(x y+y z+z x)$

Question 10

By using properties of determinants show that:

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

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

Accepted Answer

(i)

$$

\begin{aligned}

\Delta & =\left|\begin{array}{ccc}

x+4 & 2 x & 2 x \

2 x & x+4 & 2 x \

2 x & 2 x & x+4

\end{array}ight| \

\Delta & =\left|\begin{array}{ccc}

5 x+4 & 5 x+4 & 5 x+4 \

2 x & x+4 & 2 x \

2 x & 2 x & x+4

\end{array}ight| \

& =(5 x+4)\left|\begin{array}{ccc}

1 & 1 & 1 \

2 x & x+4 & 2 x \

2 x & 2 x & x+4

\end{array}ight| \

& =(5 x+4)\left|\begin{array}{ccc}

1 & 0 & 0 \

2 x & -x+4 & 0 \

2 x & 0 & -x+4

\end{array}ight| \

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

1 & 0 & 0 \

2 x & 1 & 0 \

2 x & 0 & 1

\end{array}ight| \

& =(5 x+4)(4-x)^{2}\left|\begin{array}{cc}

1 & 0 \

2 x & 1

\end{array}ight| \

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

\end{aligned}

$$

Hence, $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \ 2 x & x+4 & 2 x \ 2 x & 2 x & x+4\end{array}ight|=(5 x+4)(4-x)^{2} \quad$ proved.

(ii) $\Delta=\left|\begin{array}{ccc}y+k & y & y \ y & y+k & y \ y & y & y+k\end{array}ight|$

$$

\begin{array}{rlr}

\Delta & =\left|\begin{array}{ccc}

3 y+k & 3 y+k & 3 y+k \

y & y+k & y \

y & y & y+k

\end{array}ight| & {\left[R_{1} ightarrow R_{1}+R_{2}+R_{3}ight]} \

& =(3 y+k)\left|\begin{array}{ccc}

1 & 1 & 1 \

y & y+k & y \

y & y & y+k

\end{array}ight| & \

& =(3 y+k)\left|\begin{array}{ccc}

1 & 0 & 0 \

y & k & 0 \

y & 0 & k

\end{array}ight| & {\left[C_{2} ightarrow C_{2}-C_{1} 	ext { and } C_{3} ightarrow C_{3}-C_{1}ight]} \

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

1 & 0 & 0 \

y & 1 & 0 \

y & 0 & 1

\end{array}ight| &

\end{array}

$$

Expanding along $C_{3}$

$$

\begin{aligned}

\Delta & =k^{2}(3 y+k)\left|\begin{array}{ll}

1 & 0 \

y & 1

\end{array}ight| \

& =k^{2}(3 y+k)

\end{aligned}

$$

Hence, $\left|\begin{array}{ccc}y+k & y & y \ y & y+k & y \ y & y & y+k\end{array}ight|=k^{2}(3 y+k) \quad$ proved.

Question 11

By using properties of determinants show that:

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

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

Accepted Answer

(i)

$$

\Delta=\left|\begin{array}{ccc}

a-b-c & 2 a & 2 a \

2 b & b-c-a & 2 b \

2 c & 2 c & c-a-b

\end{array}ight|

$$

\begin{aligned}

\Delta & =\left|\begin{array}{ccc}

a+b+c & a+b+c & a+b+c \

2 b & b-c-a & 2 b \

2 c & 2 c & c-a-b

\end{array}ight| \

& =(a+b+c)\left|\begin{array}{ccc}

1 & 1 & 1 \

2 b & b-c-a & 2 b \

2 c & 2 c & c-a-b

\end{array}ight| \

& =(a+b+c)\left|\begin{array}{ccc}

1 & 0 & 0 \

2 b & -(a+b+c) & 0 \

2 c & 0 & -(a+b+c)

\end{array}ight| \

& =(a+b+c)^{3}\left|\begin{array}{ccc}

1 & 0 & 0 \

2 b & -1 & 0 \

2 c & 0 & -1

\end{array}ight| \

& =(a+b+c)^{3}(-1)(-1) \

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

\end{aligned}

$$

Hence, proved.

(ii)

$$

\begin{array}{rlr}

\Delta & =\left|\begin{array}{ccc}

x+y+2 z & x & y \

z & y+z+2 x & y \

z & x & z+x+2 y

\end{array}ight| & \

\Delta & =\left|\begin{array}{ccc}

2(x+y+z) & x & y \

2(x+y+z) & y+z+2 x & y \

2(x+y+z) & x & z+x+2 y

\end{array}ight| & {\left[C_{1} ightarrow C_{1}+C_{2}+C_{3}ight]} \

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

1 & x & y \

1 & y+z+2 x & y \

1 & x & z+x+2 y

\end{array}ight| & \

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

1 & x & y \

0 & x+y+z & 0 \

0 & 0 & x+y+z

\end{array}ight| & {\left[R_{2} ightarrow R_{2}-R_{1} 	ext { and } R_{3} ightarrow R_{3}-R_{1}ight]} \

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

1 & x & y \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight| & \

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

& =2(x+y+z)^{3} &

\end{array}

$$

Hence, proved.

Question 12

By using properties of determinants show that:

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

Accepted Answer

$$

\begin{aligned}

\Delta & =\left|\begin{array}{ccc}

1 & x & x^{2} \

x^{2} & 1 & x \

x & x^{2} & 1

\end{array}ight| \

& =\left|\begin{array}{ccc}

1+x+x^{2} & 1+x+x^{2} & 1+x+x^{2} \

x^{2} & 1 & x \

x & x^{2} & 1

\end{array}ight| \quad\left[R_{1} ightarrow R_{1}+R_{2}+R_{3}ight] \

& =\left(1+x+x^{2}ight)\left|\begin{array}{ccc}

1 & 1 & 1 \

x^{2} & 1 & x \

x & x^{2} & 1

\end{array}ight| \

\Delta & =\left(1+x+x^{2}ight)\left|\begin{array}{ccc}

1 & 0 & 0 \

x^{2} & 1-x^{2} & x-x^{2} \

x & x^{2}-x & 1-x

\end{array}ight| \quad\left[C_{2} ightarrow C_{2}-C_{1} 	ext { and } C_{3} ightarrow C_{3}-C_{1}ight] \

\Delta & =\left(1+x+x^{2}ight)(1-x)(1-x)\left|\begin{array}{ccc}

1 & 0 & 0 \

x^{2} & 1+x & x \

x & -x & 1

\end{array}ight| \

& =\left(1-x^{3}ight)(1-x)\left|\begin{array}{ccc}

1 & 0 & 0 \

x^{2} & 1+x & x \

x & -x & 1

\end{array}ight|

\end{aligned}

$$

Expanding along $R_{1}$

$$

\begin{aligned}

\Delta & =\left(1-x^{3}ight)(1-x)(1)\left|\begin{array}{cc}

1+x & x \

-x & 1

\end{array}ight| \

& =\left(1-x^{3}ight)(1-x)\left(1+x+x^{2}ight) \

& =\left(1-x^{3}ight)\left(1-x^{3}ight) \

& =\left(1-x^{3}ight)^{2}

\end{aligned}

$$

Hence, proved.

Question 13

By using properties of determinants show that:

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

Accepted Answer

$$

\begin{aligned}

\Delta & =\left|\begin{array}{ccc}

1+a^{2}-b^{2} & 2 a b & -2 b \

2 a b & 1-a^{2}+b^{2} & 2 a \

2 b & -2 a & 1-a^{2}-b^{2}

\end{array}ight| \

& =\left|\begin{array}{ccc}

1+a^{2}+b^{2} & 0 & -b\left(1+a^{2}+b^{2}ight) \

0 & 1+a^{2}+b^{2} & a\left(1+a^{2}+b^{2}ight) \

2 b & -2 a & 1-a^{2}-b^{2}

\end{array}ight| \quad\left[R_{1} ightarrow R_{1}+b R_{3} 	ext { and } R_{2} ightarrow R_{2}-a R_{3}ight] \

& =\left(1+a^{2}+b^{2}ight)^{2}\left|\begin{array}{ccc}

1 & 0 & -b \

0 & 1 & a \

2 b & -2 a & 1-a^{2}-b^{2}

\end{array}ight| \

& =\left(1+a^{2}+b^{2}ight)^{2}\left[(1)\left|\begin{array}{cc}

1 & a \

-2 a & 1-a^{2}-b^{2}

\end{array}ight|-b\left|\begin{array}{cc}

0 & 1 \

2 b & -2 a

\end{array}ight|ight] \

& =\left(1+a^{2}+b^{2}ight)^{2}\left[1-a^{2}-b^{2}+2 a^{2}-b(-2 b)ight] \

& =\left(1+a^{2}+b^{2}ight)^{2}\left(1+a^{2}+b^{2}ight) \

& =\left(1+a^{2}+b^{2}ight)^{3}

\end{aligned}

$$

Hence, proved.

Question 14

By using properties of determinants show that:

$$

\left|\begin{array}{ccc}

a^{2}+1 & a b & a c \

a b & b^{2}+1 & b c \

c a & c b & c^{2}+1

\end{array}ight|=1+a^{2}+b^{2}+c^{2}

$$

Accepted Answer

$\Delta=\left|\begin{array}{ccc}a^{2}+1 & a b & a c \ a b & b^{2}+1 & b c \ c a & c b & c^{2}+1\end{array}ight|$

Taking out common factors $a, b, c$ from $R_{1}, R_{2}, R_{3}$ respectively,

$\Delta=a b c\left|\begin{array}{ccc}a+\frac{1}{a} & b & c \ a & b+\frac{1}{b} & c \ a & b & c+\frac{1}{c}\end{array}ight|$

$=a b c\left|\begin{array}{ccc}a+\frac{1}{a} & b & c \ -\frac{1}{a} & \frac{1}{b} & 0 \ -\frac{1}{a} & 0 & \frac{1}{c}\end{array}ight| \quad\left[R_{2} ightarrow R_{2}-R_{1}ight.$ and $\left.R_{3} ightarrow R_{3}-R_{1}ight]$

$=a b c 	imes \frac{1}{a b c}\left|\begin{array}{ccc}a^{2}+1 & b^{2} & c^{2} \ -1 & 1 & 0 \ -1 & 0 & 1\end{array}ight| \quad\left[C_{1} ightarrow a C_{1}, C_{2} ightarrow b C_{2}ight.$ and $\left.C_{3} ightarrow c C_{3}ight]$

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

$=-1\left|\begin{array}{cc}b^{2} & c^{2} \ 1 & 0\end{array}ight|+1\left|\begin{array}{cc}a^{2}+1 & b^{2} \ -1 & 1\end{array}ight|$

$=-1\left(-c^{2}ight)+\left(a^{2}+1+b^{2}ight)$

$=1+a^{2}+b^{2}+c^{2}$

Hence, proved.

Question 15

Let $A$ be a square matrix of order $3 	imes 3$, then $|k A|$ is equal to:

(A) $k|A|$

(B) $k^{2}|A|$

(C) $k^{3}|A|$

(D) $3^{k}|A|$

Accepted Answer

Let $A=\left(\begin{array}{lll}a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3}\end{array}ight)$

Then,

$$

\begin{aligned}

k A & =\left(\begin{array}{lll}

k a_{1} & k b_{1} & k c_{1} \

k a_{2} & k b_{2} & k c_{2} \

k a_{3} & k b_{3} & k c_{3}

\end{array}ight) \

|k A| & =\left|\begin{array}{lll}

k a_{1} & k b_{1} & k c_{1} \

k a_{2} & k b_{2} & k c_{2} \

k a_{3} & k b_{3} & k c_{3}

\end{array}ight|

\end{aligned}

$$

Taking out common factors $k$ from each row

$$

\begin{aligned}

|k A| & =k^{3}\left|\begin{array}{lll}

a_{1} & b_{1} & c_{1} \

a_{2} & b_{2} & c_{2} \

a_{3} & b_{3} & c_{3}

\end{array}ight| \

& =k^{3}|A|

\end{aligned}

$$

The correct option is C.

Question 16

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 the above.

Accepted Answer

We know that to every square matrix, $A=\left[a_{i j}ight]$ of order $n$, we can associate a number called the determinant of square matrix $A$, where $a_{i j}=(i, j)^{	ext {th }}$ element of $A$.

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

Hence, the correct option is C.

\section*{EXERCISE 4.3}