QBManager

Matrix and Determinant

79004 If $A = [\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}], B = [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}]$
and $A X = B$ , then $z$ is equal to

1 1

2 -1

3 -3

4 3

Explanation:

(D) : It is given that, $AX = B$
$\Rightarrow [\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
This will have a solution if $A^{- 1}$ exists, i.e., $| A | \neq 0$
$| A | = 1 (8 - 9) - 1 (4 - 3) + 1 (3 - 2)$
$\Rightarrow - 1 - 1 + 1 = - 1 \neq 0$
Hence, $A^{- 1}$ exists.
Now, $X = A^{- 1} B$
$X = {[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}]}^{- 1} [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
Co-factor matrix of $A = [\begin{array}{ccc} - 1 & - 1 & - 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 1 \end{array}]$
$∴ adj A = [\begin{array}{ccc} - 1 & - 1 & 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 1 \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |} = [\begin{array}{ccc} 1 & 1 & - 1 \\ 1 & - 3 & 2 \\ - 1 & 2 & - 1 \end{array}]$
$X = A^{- 1} B = [\begin{array}{ccc} 1 & 1 & - 1 \\ 1 & - 3 & 2 \\ - 1 & 2 & - 1 \end{array}] [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
$\Rightarrow [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 7 + 16 - 22 \\ 7 - 48 + 44 \\ - 7 + 32 - 22 \end{matrix}] = [\begin{matrix} 1 \\ 3 \\ 3 \end{matrix}]$
We get, $z = 3$

COMEDK-2012

Matrix and Determinant

79005 If $[\begin{array}{cc} - 2 & 5 \\ 3 & - 1 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}] [\begin{matrix} 3 \\ - 1 \end{matrix}]$ , then $(x, y)$ is

1

(1, 2)

2

(- 1, 2)

3

(1, - 2)

4

(2, 1)

Explanation:

(D) : We have,
$[\begin{array}{cc} - 2 & 5 \\ 3 & - 1 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}] [\begin{matrix} 3 \\ - 1 \end{matrix}]$
$[\begin{matrix} - 2 x + 5 y \\ 3 x - y \end{matrix}] = [\begin{array}{l} 1 \\ 5 \end{array}]$
Comparing these value, we get -
$- 2 x + 5 y = 1, 3 x - y = 5 y = 3 x - 5$
$- 2 x + 5 (3 x - 5) = 1 13 x = 26 \Rightarrow x = 2$
Substituting $x = 2$ , we get- $y = 6 - 5 = 1$
Hence, $(x, y) = (2, 1)$

COMEDK-2014

Matrix and Determinant

79006 If $f (x) = | \begin{array}{ccc} a & - 1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a \end{array} |$ , then $f (2 x) - f (x)$ equals

1

a (2 a + 3 x)

2 ax

(2 x + 3 a)

3

a x (2 a + 3 x)

4

x (2 a + 3 x)

Explanation:

(C) : We have,
$f (x) = | \begin{array}{ccc} a & - 1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a \end{array} | \Rightarrow f (x) = a | \begin{array}{ccc} 1 & - 1 & 0 \\ x & a & - 1 \\ x^{2} & a x & a \end{array} |$
$f (x) = a [1 (a^{2} + a x) + 1 (a x + x^{2}) + 0]$
$= a (a^{2} + 2 ax + x^{2})$
$f (x) = a (a + x)^{2}$
$∴ f (2 x) = a (a + 2 x)^{2}$
So, $f (2 x) - f (x) = a [(a + 2 x)^{2} - (a + x)^{2}]$
$= a [a^{2} + 4 x^{2} + 4 a x - a^{2} - x^{2} - 2 a x]$
$= a (3 x^{2} + 2 a x) = a x (2 a + 3 x)$

SRM JEE-2008

Matrix and Determinant

79007 If $p + q + r = 0 = a + b + c$ , then the value of the
determinant $| \begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array} |$ is

1 0

2

pa + qb + rc

3 1

4 -1

Explanation:

(A) : Here,
Let, $Δ = | \begin{array}{ccc} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{array} |$
$Δ = p a (a^{2} r q - p^{2} c b) - q b (q^{2} c a - b^{2} p r) + r c (q p c^{2} - r^{2} a b)$
$Δ =$ prqa $^{3} - p^{3} abc - q^{3} abc + {pqrb}^{3} + {pqrc}^{3} - r^{3} abc$
$= pqr (a^{3} + b^{3} + c^{3}) - abc (p^{3} + q^{3} + r^{3})$
$∵ a + b + c = 0 = p + q + r$ (given)
$∴ a^{3} + b^{3} + c^{3} = 3 a b c$
and $p^{3} + q^{3} + r^{3} = 3 pqr$
So, $Δ = pqr (3 abc) - abc (3 pqr)$
$Δ = 0$

SRM JEE-2009

Matrix and Determinant

79004 If $A = [\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}], B = [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}]$
and $A X = B$ , then $z$ is equal to

1 1

2 -1

3 -3

4 3

Explanation:

(D) : It is given that, $AX = B$
$\Rightarrow [\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
This will have a solution if $A^{- 1}$ exists, i.e., $| A | \neq 0$
$| A | = 1 (8 - 9) - 1 (4 - 3) + 1 (3 - 2)$
$\Rightarrow - 1 - 1 + 1 = - 1 \neq 0$
Hence, $A^{- 1}$ exists.
Now, $X = A^{- 1} B$
$X = {[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}]}^{- 1} [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
Co-factor matrix of $A = [\begin{array}{ccc} - 1 & - 1 & - 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 1 \end{array}]$
$∴ adj A = [\begin{array}{ccc} - 1 & - 1 & 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 1 \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |} = [\begin{array}{ccc} 1 & 1 & - 1 \\ 1 & - 3 & 2 \\ - 1 & 2 & - 1 \end{array}]$
$X = A^{- 1} B = [\begin{array}{ccc} 1 & 1 & - 1 \\ 1 & - 3 & 2 \\ - 1 & 2 & - 1 \end{array}] [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
$\Rightarrow [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 7 + 16 - 22 \\ 7 - 48 + 44 \\ - 7 + 32 - 22 \end{matrix}] = [\begin{matrix} 1 \\ 3 \\ 3 \end{matrix}]$
We get, $z = 3$

COMEDK-2012

Matrix and Determinant

79005 If $[\begin{array}{cc} - 2 & 5 \\ 3 & - 1 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}] [\begin{matrix} 3 \\ - 1 \end{matrix}]$ , then $(x, y)$ is

1

(1, 2)

2

(- 1, 2)

3

(1, - 2)

4

(2, 1)

Explanation:

(D) : We have,
$[\begin{array}{cc} - 2 & 5 \\ 3 & - 1 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}] [\begin{matrix} 3 \\ - 1 \end{matrix}]$
$[\begin{matrix} - 2 x + 5 y \\ 3 x - y \end{matrix}] = [\begin{array}{l} 1 \\ 5 \end{array}]$
Comparing these value, we get -
$- 2 x + 5 y = 1, 3 x - y = 5 y = 3 x - 5$
$- 2 x + 5 (3 x - 5) = 1 13 x = 26 \Rightarrow x = 2$
Substituting $x = 2$ , we get- $y = 6 - 5 = 1$
Hence, $(x, y) = (2, 1)$

COMEDK-2014

Matrix and Determinant

79006 If $f (x) = | \begin{array}{ccc} a & - 1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a \end{array} |$ , then $f (2 x) - f (x)$ equals

1

a (2 a + 3 x)

2 ax

(2 x + 3 a)

3

a x (2 a + 3 x)

4

x (2 a + 3 x)

Explanation:

(C) : We have,
$f (x) = | \begin{array}{ccc} a & - 1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a \end{array} | \Rightarrow f (x) = a | \begin{array}{ccc} 1 & - 1 & 0 \\ x & a & - 1 \\ x^{2} & a x & a \end{array} |$
$f (x) = a [1 (a^{2} + a x) + 1 (a x + x^{2}) + 0]$
$= a (a^{2} + 2 ax + x^{2})$
$f (x) = a (a + x)^{2}$
$∴ f (2 x) = a (a + 2 x)^{2}$
So, $f (2 x) - f (x) = a [(a + 2 x)^{2} - (a + x)^{2}]$
$= a [a^{2} + 4 x^{2} + 4 a x - a^{2} - x^{2} - 2 a x]$
$= a (3 x^{2} + 2 a x) = a x (2 a + 3 x)$

SRM JEE-2008

Matrix and Determinant

79007 If $p + q + r = 0 = a + b + c$ , then the value of the
determinant $| \begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array} |$ is

1 0

2

pa + qb + rc

3 1

4 -1

Explanation:

(A) : Here,
Let, $Δ = | \begin{array}{ccc} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{array} |$
$Δ = p a (a^{2} r q - p^{2} c b) - q b (q^{2} c a - b^{2} p r) + r c (q p c^{2} - r^{2} a b)$
$Δ =$ prqa $^{3} - p^{3} abc - q^{3} abc + {pqrb}^{3} + {pqrc}^{3} - r^{3} abc$
$= pqr (a^{3} + b^{3} + c^{3}) - abc (p^{3} + q^{3} + r^{3})$
$∵ a + b + c = 0 = p + q + r$ (given)
$∴ a^{3} + b^{3} + c^{3} = 3 a b c$
and $p^{3} + q^{3} + r^{3} = 3 pqr$
So, $Δ = pqr (3 abc) - abc (3 pqr)$
$Δ = 0$

SRM JEE-2009

Matrix and Determinant

79004 If $A = [\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}], B = [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}]$
and $A X = B$ , then $z$ is equal to

1 1

2 -1

3 -3

4 3

Explanation:

(D) : It is given that, $AX = B$
$\Rightarrow [\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
This will have a solution if $A^{- 1}$ exists, i.e., $| A | \neq 0$
$| A | = 1 (8 - 9) - 1 (4 - 3) + 1 (3 - 2)$
$\Rightarrow - 1 - 1 + 1 = - 1 \neq 0$
Hence, $A^{- 1}$ exists.
Now, $X = A^{- 1} B$
$X = {[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}]}^{- 1} [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
Co-factor matrix of $A = [\begin{array}{ccc} - 1 & - 1 & - 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 1 \end{array}]$
$∴ adj A = [\begin{array}{ccc} - 1 & - 1 & 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 1 \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |} = [\begin{array}{ccc} 1 & 1 & - 1 \\ 1 & - 3 & 2 \\ - 1 & 2 & - 1 \end{array}]$
$X = A^{- 1} B = [\begin{array}{ccc} 1 & 1 & - 1 \\ 1 & - 3 & 2 \\ - 1 & 2 & - 1 \end{array}] [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
$\Rightarrow [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 7 + 16 - 22 \\ 7 - 48 + 44 \\ - 7 + 32 - 22 \end{matrix}] = [\begin{matrix} 1 \\ 3 \\ 3 \end{matrix}]$
We get, $z = 3$

COMEDK-2012

Matrix and Determinant

79005 If $[\begin{array}{cc} - 2 & 5 \\ 3 & - 1 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}] [\begin{matrix} 3 \\ - 1 \end{matrix}]$ , then $(x, y)$ is

1

(1, 2)

2

(- 1, 2)

3

(1, - 2)

4

(2, 1)

Explanation:

(D) : We have,
$[\begin{array}{cc} - 2 & 5 \\ 3 & - 1 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}] [\begin{matrix} 3 \\ - 1 \end{matrix}]$
$[\begin{matrix} - 2 x + 5 y \\ 3 x - y \end{matrix}] = [\begin{array}{l} 1 \\ 5 \end{array}]$
Comparing these value, we get -
$- 2 x + 5 y = 1, 3 x - y = 5 y = 3 x - 5$
$- 2 x + 5 (3 x - 5) = 1 13 x = 26 \Rightarrow x = 2$
Substituting $x = 2$ , we get- $y = 6 - 5 = 1$
Hence, $(x, y) = (2, 1)$

COMEDK-2014

Matrix and Determinant

79006 If $f (x) = | \begin{array}{ccc} a & - 1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a \end{array} |$ , then $f (2 x) - f (x)$ equals

1

a (2 a + 3 x)

2 ax

(2 x + 3 a)

3

a x (2 a + 3 x)

4

x (2 a + 3 x)

Explanation:

(C) : We have,
$f (x) = | \begin{array}{ccc} a & - 1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a \end{array} | \Rightarrow f (x) = a | \begin{array}{ccc} 1 & - 1 & 0 \\ x & a & - 1 \\ x^{2} & a x & a \end{array} |$
$f (x) = a [1 (a^{2} + a x) + 1 (a x + x^{2}) + 0]$
$= a (a^{2} + 2 ax + x^{2})$
$f (x) = a (a + x)^{2}$
$∴ f (2 x) = a (a + 2 x)^{2}$
So, $f (2 x) - f (x) = a [(a + 2 x)^{2} - (a + x)^{2}]$
$= a [a^{2} + 4 x^{2} + 4 a x - a^{2} - x^{2} - 2 a x]$
$= a (3 x^{2} + 2 a x) = a x (2 a + 3 x)$

SRM JEE-2008

Matrix and Determinant

79007 If $p + q + r = 0 = a + b + c$ , then the value of the
determinant $| \begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array} |$ is

1 0

2

pa + qb + rc

3 1

4 -1

Explanation:

(A) : Here,
Let, $Δ = | \begin{array}{ccc} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{array} |$
$Δ = p a (a^{2} r q - p^{2} c b) - q b (q^{2} c a - b^{2} p r) + r c (q p c^{2} - r^{2} a b)$
$Δ =$ prqa $^{3} - p^{3} abc - q^{3} abc + {pqrb}^{3} + {pqrc}^{3} - r^{3} abc$
$= pqr (a^{3} + b^{3} + c^{3}) - abc (p^{3} + q^{3} + r^{3})$
$∵ a + b + c = 0 = p + q + r$ (given)
$∴ a^{3} + b^{3} + c^{3} = 3 a b c$
and $p^{3} + q^{3} + r^{3} = 3 pqr$
So, $Δ = pqr (3 abc) - abc (3 pqr)$
$Δ = 0$

SRM JEE-2009

Matrix and Determinant

79004 If $A = [\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}], B = [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}]$
and $A X = B$ , then $z$ is equal to

1 1

2 -1

3 -3

4 3

Explanation:

(D) : It is given that, $AX = B$
$\Rightarrow [\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
This will have a solution if $A^{- 1}$ exists, i.e., $| A | \neq 0$
$| A | = 1 (8 - 9) - 1 (4 - 3) + 1 (3 - 2)$
$\Rightarrow - 1 - 1 + 1 = - 1 \neq 0$
Hence, $A^{- 1}$ exists.
Now, $X = A^{- 1} B$
$X = {[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{array}]}^{- 1} [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
Co-factor matrix of $A = [\begin{array}{ccc} - 1 & - 1 & - 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 1 \end{array}]$
$∴ adj A = [\begin{array}{ccc} - 1 & - 1 & 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 1 \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |} = [\begin{array}{ccc} 1 & 1 & - 1 \\ 1 & - 3 & 2 \\ - 1 & 2 & - 1 \end{array}]$
$X = A^{- 1} B = [\begin{array}{ccc} 1 & 1 & - 1 \\ 1 & - 3 & 2 \\ - 1 & 2 & - 1 \end{array}] [\begin{matrix} 7 \\ 16 \\ 22 \end{matrix}]$
$\Rightarrow [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 7 + 16 - 22 \\ 7 - 48 + 44 \\ - 7 + 32 - 22 \end{matrix}] = [\begin{matrix} 1 \\ 3 \\ 3 \end{matrix}]$
We get, $z = 3$

COMEDK-2012

Matrix and Determinant

79005 If $[\begin{array}{cc} - 2 & 5 \\ 3 & - 1 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}] [\begin{matrix} 3 \\ - 1 \end{matrix}]$ , then $(x, y)$ is

1

(1, 2)

2

(- 1, 2)

3

(1, - 2)

4

(2, 1)

Explanation:

(D) : We have,
$[\begin{array}{cc} - 2 & 5 \\ 3 & - 1 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}] [\begin{matrix} 3 \\ - 1 \end{matrix}]$
$[\begin{matrix} - 2 x + 5 y \\ 3 x - y \end{matrix}] = [\begin{array}{l} 1 \\ 5 \end{array}]$
Comparing these value, we get -
$- 2 x + 5 y = 1, 3 x - y = 5 y = 3 x - 5$
$- 2 x + 5 (3 x - 5) = 1 13 x = 26 \Rightarrow x = 2$
Substituting $x = 2$ , we get- $y = 6 - 5 = 1$
Hence, $(x, y) = (2, 1)$

COMEDK-2014

Matrix and Determinant

79006 If $f (x) = | \begin{array}{ccc} a & - 1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a \end{array} |$ , then $f (2 x) - f (x)$ equals

1

a (2 a + 3 x)

2 ax

(2 x + 3 a)

3

a x (2 a + 3 x)

4

x (2 a + 3 x)

Explanation:

(C) : We have,
$f (x) = | \begin{array}{ccc} a & - 1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a \end{array} | \Rightarrow f (x) = a | \begin{array}{ccc} 1 & - 1 & 0 \\ x & a & - 1 \\ x^{2} & a x & a \end{array} |$
$f (x) = a [1 (a^{2} + a x) + 1 (a x + x^{2}) + 0]$
$= a (a^{2} + 2 ax + x^{2})$
$f (x) = a (a + x)^{2}$
$∴ f (2 x) = a (a + 2 x)^{2}$
So, $f (2 x) - f (x) = a [(a + 2 x)^{2} - (a + x)^{2}]$
$= a [a^{2} + 4 x^{2} + 4 a x - a^{2} - x^{2} - 2 a x]$
$= a (3 x^{2} + 2 a x) = a x (2 a + 3 x)$

SRM JEE-2008

Matrix and Determinant

79007 If $p + q + r = 0 = a + b + c$ , then the value of the
determinant $| \begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array} |$ is

1 0

2

pa + qb + rc

3 1

4 -1

Explanation:

(A) : Here,
Let, $Δ = | \begin{array}{ccc} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{array} |$
$Δ = p a (a^{2} r q - p^{2} c b) - q b (q^{2} c a - b^{2} p r) + r c (q p c^{2} - r^{2} a b)$
$Δ =$ prqa $^{3} - p^{3} abc - q^{3} abc + {pqrb}^{3} + {pqrc}^{3} - r^{3} abc$
$= pqr (a^{3} + b^{3} + c^{3}) - abc (p^{3} + q^{3} + r^{3})$
$∵ a + b + c = 0 = p + q + r$ (given)
$∴ a^{3} + b^{3} + c^{3} = 3 a b c$
and $p^{3} + q^{3} + r^{3} = 3 pqr$
So, $Δ = pqr (3 abc) - abc (3 pqr)$
$Δ = 0$

SRM JEE-2009

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Question Bank Series

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