Evaluate 2 3 - 5 7 1 - 2 - 3 4 1 by two methods:

Let ∆ = 2 3 - 5 7 1 - 2 - 3 4 1 First method ∆ = - 1 1 + 1 2 1 + 8 + - 1 1 + 2 3 7 - 6 + - 1 1 + 3 - 5 28 + 3 = 2 1 + 8 - 3 7 - 6 - 5 28 + 3 = 18 - 3 - 155 = - 140 Second method is the Sarus Method, where we adjoin the first two columns to the right to get 2 3 - 5 2 3 7 1 - 2 7 1 - 3 4 1 - 3 4 = ( 2 × 1 × 1 + 3 × - 2 × - 3 - 5 × 7 × 4 ) - ( - 5 × 1 × - 3 + 2 × - 2 × 4 + 3 × 7 × 1 ) = ( 2 + 18 - 140 ) - ( 15 - 16 + 21 ) = - 120 - 20 = - 140

Evaluate 2 3 - 5 7 1 - 2 - 3 4 1 by two methods:

Let ∆ = 2 3 - 5 7 1 - 2 - 3 4 1 First method ∆ = - 1 1 + 1 2 1 + 8 + - 1 1 + 2 3 7 - 6 + - 1 1 + 3 - 5 28 + 3 = 2 1 + 8 - 3 7 - 6 - 5 28 + 3 = 18 - 3 - 155 = - 140 Second method is the Sarus Method, where we adjoin the first two columns to the right to get 2 3 - 5 2 3 7 1 - 2 7 1 - 3 4 1 - 3 4 = ( 2 × 1 × 1 + 3 × - 2 × - 3 - 5 × 7 × 4 ) - ( - 5 × 1 × - 3 + 2 × - 2 × 4 + 3 × 7 × 1 ) = ( 2 + 18 - 140 ) - ( 15 - 16 + 21 ) = - 120 - 20 = - 140

If A = 2 5 2 1 and B = 4 - 3 2 5 , verify that | A B | = | A | | B | .

Given, A = 2 5 2 1 and B = 4 - 3 2 5 Consider LHS A B = 2 5 2 1 4 - 3 2 5 = 8 + 10 - 6 + 25 8 + 2 - 6 + 5 = 18 19 10 - 1 A B = - 18 - 190 = - 208 Consider RHS A = 2 - 10 = - 8 B = 20 - - 6 = 26 A B = - 8 × 26 = - 208 ∴ LHS = RHS

If A = 1 0 1 0 1 2 0 0 4 , then show that | 3 A | = 27 | A | .

A = 1 0 1 0 1 2 0 0 4 ⇒ 3 A = 3 0 3 0 3 6 0 0 12 ⇒ 3 A = - 1 1 + 1 3 36 - 0 + - 1 1 + 2 0 0 - 0 + - 1 1 + 3 3 0 - 0 = 3 36 - 0 - 0 0 - 0 + 3 0 - 0 = 3 × 36 = 108 . . . 1 ⇒ A = - 1 1 + 1 1 4 - 0 + - 1 1 + 2 0 0 - 0 + - 1 1 + 3 1 0 - 0 = 1 4 - 0 - 0 0 - 0 + 1 0 - 0 = 4 ⇒ 27 A = 27 × 4 = 108 . . . 2 ∴ 3 A = 27 A

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 5

HARD

12th CBSE

IMPORTANT

Earn 100

Evaluate $|\begin{matrix} 2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1 \end{matrix}|$ by two methods:

Important Questions on Determinants

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 6

Evaluate:

∆ = |\begin{matrix} 0 & \sin α & - \cos α \\ - \sin α & 0 & \sin β \\ \cos α & - \sin β & 0 \end{matrix}|

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 7

Evaluate:

$∆ = |\begin{matrix} \cos α \cos β & \cos α \sin β & - \sin α \\ - \sin β & \cos β & 0 \\ \sin α \cos β & \sin α \sin β & \cos α \end{matrix}|$

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 8

If

A = [\begin{array}{l} 2 & 5 \\ 2 & 1 \end{array}]

and

B = [\begin{array}{r} 4 & - 3 \\ 2 & 5 \end{array}]

, verify that

| A B | = | A | | B | .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 9

If

A =

[\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix}]

, then show that

| 3 A | = 27 | A | .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 10

Find the value of $x$ , if $|\begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix}| = |\begin{matrix} 2 x & 4 \\ 6 & x \end{matrix}|$ is of the form $\pm \sqrt{a}$ , then $a =$

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 10

Find the value of $x$ , if $|\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}| = |\begin{matrix} x & 3 \\ 2 x & 5 \end{matrix}|$

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 10

Find the value of $x$ , if $|\begin{matrix} 3 & x \\ x & 1 \end{matrix}| = |\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}|$ , $\pm a \sqrt{a}$ , then $a =$