Using properties of determinants prove that: 1 1 1 a b c b c c a a b = ( a - b ) ( b - c ) ( c - a ) .

1 1 1 a b c b c c a a b = 0 0 1 a - b b - c c b c - c a c a - a b a b [By applying C 1 → C 1 - C 2 , C 2 → C 2 - C 3 ] = 0 0 1 a - b b - c c - c ( a - b ) - a ( b - c ) a b = a - b b - c 0 0 1 1 1 c - c - a a b [By taking ( a - b ) and ( b - c ) common from C 1 and C 2 respectively.] Expanding along R 1 , we get = ( a - b ) ( b - c ) [ 0 + 0 + 1 { - a - ( - c ) } ] = ( a - b ) ( b - c ) ( c - a ) Hence, 1 1 1 a b c b c c a a b = a - b b - c c - a

Using properties of determinants prove that: 1 b + c b 2 + c 2 1 c + a c 2 + a 2 1 a + b a 2 + b 2 = a - b b - c c - a .

Given, 1 b + c b 2 + c 2 1 c + a c 2 + a 2 1 a + b a 2 + b 2 = 0 b - a b 2 - a 2 0 c - b c 2 - b 2 1 a + b a 2 + b 2 [By applying R 1 → R 1 - R 2 , R 2 → R 2 - R 3 ] = 0 b - a ( b - a ) ( b + a ) 0 c - b ( c - b ) ( c + b ) 1 a + b a 2 + b 2 = b - a c - b 0 1 b + a 0 1 c + b 1 a + b a 2 + b 2 [By taking ( b - a ) and c - b common from R 1 and R 2 respectively] By expanding along C 1 , we get = ( b - a ) ( c - b ) [ 0 + 0 + 1 { ( c + b ) - ( b + a ) } ] = ( a - b ) ( b - c ) ( c - a )

Using properties of determinants prove that: 1 1 + p 1 + p + q 2 3 + 2 p 1 + 3 p + 2 q 3 6 + 3 p 1 + 6 p + 3 q = 1 .

Given, 1 1 + p 1 + p + q 2 3 + 2 p 1 + 3 p + 2 q 3 6 + 3 p 1 + 6 p + 3 q = 1 1 1 + p 2 3 1 + 3 p 3 6 1 + 6 p [By applying C 2 → C 2 - p C 1 and C 3 → C 3 - q C 1 ] = 1 1 1 2 3 1 3 6 1 [By applying C 3 → C 3 - p C 2 ] = 0 0 1 1 2 1 2 5 1 [By applying C 1 → C 1 - C 3 and C 2 → C 2 - C 3 ] Expanding along R 1 , we get = 0 - 0 + 1 1 ( 5 ) - ( 2 ) ( 2 ) = 5 - 4 = 1 Hence, 1 1 + p 1 + p + q 2 3 + 2 p 1 + 3 p + 2 q 3 6 + 3 p 1 + 6 p + 3 q = 1

Using properties of determinants prove that: a + x y z x a + y z x y a + z = a 2 a + x + y + z

Given, a + x y z x a + y z x y a + z = a + x + y + z y z a + x + y + z a + y z a + x + y + z y a + z [Applying C 1 → C 1 + C 2 + C 3 ] = ( a + x + y + z ) 1 y z 1 a + y z 1 y a + z [Taking ( a + x + y + z ) common from C 1 ] = ( a + x + y + z ) 1 y z 0 a 0 0 0 a [Applying R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ] By expanding along C 1 , we get = ( a + x + y + z ) 1 a 2 - 0 - 0 + 0 = a 2 ( a + x + y + z )

Using properties of determinants prove that: x a a a x a a a x = x + 2 a x - a 2 .

Given, x a a a x a a a x = x + 2 a x + 2 a x + 2 a a x a a a x [By applying R 1 → R 1 + R 2 + R 3 ] = x + 2 a 1 1 1 a x a a a x [Taking out x + 2 a common from R 1 ] = x + 2 a 1 1 1 0 x - a - ( x - a ) a a x [By applying R 2 → R 2 - R 3 ] = x + 2 a x - a 1 1 1 0 1 - 1 a a x [By taking x - a common from R 2 ] Expand along C 1 , we get = ( x + 2 a ) ( x - a ) 1 x + a - 0 + a - 1 - 1 = ( x + 2 a ) ( x - a ) 2

Using properties of determinants prove that: x + 4 2 x 2 x 2 x x + 4 2 x 2 x 2 x x + 4 = 5 x + 4 x - 4 2

Given, x + 4 2 x 2 x 2 x x + 4 2 x 2 x 2 x x + 4 = 5 x + 4 5 x + 4 5 x + 4 2 x x + 4 2 x 2 x 2 x x + 4 [Applying R 1 → R 1 + R 2 + R 3 ] = 5 x + 4 1 1 1 2 x x + 4 2 x 2 x 2 x x + 4 [Taking 5 x + 4 common from R 1 ] = ( 5 x + 4 ) 1 0 0 2 x 4 - x 0 2 x 0 4 - x [Applying C 2 → C 2 - C 1 , C 3 → C 3 - C 1 ] = ( 5 x + 4 ) 1 ( 4 - x ) 2 - 0 + 0 = ( 5 x + 4 ) ( x - 4 ) 2 Hence proved

E M B I B E

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 6 - Determinants>EXERCISE>Q 2

HARD

12th CBSE

IMPORTANT

Earn 100

Evaluate : $|\begin{array}{l} 29 & 26 & 22 \\ 25 & 31 & 27 \\ 63 & 54 & 46 \end{array}|$

50% studentsanswered this correctly

Important Questions on Determinants

EASY

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 6 - Determinants>EXERCISE>Q 3

Evaluate : $|\begin{matrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{matrix}|$

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 6 - Determinants>EXERCISE>Q 4

Evaluate : $|\begin{array}{l} 1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2} \end{array}|$ .

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 6 - Determinants>EXERCISE>Q 5

Using properties of determinants prove that: $|\begin{array}{l} 1 & 1 & 1 \\ a & b & c \\ b c & c a & a b \end{array}| = (a - b) (b - c) (c - a)$ .

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 6 - Determinants>EXERCISE>Q 6

Using properties of determinants prove that: $|\begin{matrix} 1 & b + c & b^{2} + c^{2} \\ 1 & c + a & c^{2} + a^{2} \\ 1 & a + b & a^{2} + b^{2} \end{matrix}| = (a - b) (b - c) (c - a)$ .

HARD

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 6 - Determinants>EXERCISE>Q 7

Using properties of determinants prove that: $|\begin{matrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2 p & 1 + 3 p + 2 q \\ 3 & 6 + 3 p & 1 + 6 p + 3 q \end{matrix}| = 1$ .

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 6 - Determinants>EXERCISE>Q 8

Using properties of determinants prove that: $|\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}| = a^{2} (a + x + y + z)$

HARD

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 6 - Determinants>EXERCISE>Q 9

Using properties of determinants prove that: $|\begin{array}{lll} x & a & a \\ a & x & a \\ a & a & x \end{array}| = (x + 2 a) {(x - a)}^{2}$ .

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 6 - Determinants>EXERCISE>Q 10

Using properties of determinants prove that: $|\begin{matrix} x + 4 & 2 x & 2 x \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix}| = (5 x + 4) {(x - 4)}^{2}$