If A = 5 5 α α 0 α 5 α 0 0 5 and A 2 = 2 find the value(s) of α .

Given that, A = 5 5 α α 0 α 5 α 0 0 5 and A 2 = 2 To find: value(s) of α . Now A 2 = 5 5 α α 0 α 5 α 0 0 5 5 5 α α 0 α 5 α 0 0 5 = 25 25 α + 5 α 2 5 α + 25 α 2 + 5 α 0 α 2 5 α 2 + 25 α 0 0 25 ∴ A 2 = 25 25 α + 5 α 2 5 α + 25 α 2 + 5 α 0 α 2 5 α 2 + 25 α 0 0 25 Expanding along C 1 , = 25 α 2 5 α 2 + 25 α 0 25 - 0 + 0 = 25 ( 25 α 2 - 0 ) = 625 α 2 Also given A 2 = 25 ⇒ 625 α 2 = 25 ⇒ α 2 = 25 625 = 1 25 ⇒ α = ± 1 25 = ± 1 5 Hence, values of α = 1 5 , - 1 5 .

Show that one root of the equation x + a b c b x + c a c a x + b = 0 is - ( a + b + c ) .

We have to prove that [ - ( a + b + c ) ] is a root of the equation x + a b c b x + c a c a x + b = 0 . Applying C 1 → C 1 + C 2 + C 3 , ⇒ x + a + b + c b c b + x + c + a x + c a c + a + x + b a x + b = 0 Taking out ( x + a + b + c ) common from C 1 , we get ⇒ ( x + a + b + c ) 1 b c 1 x + c a 1 a x + b = 0 Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 [to create two zeros in the first column] ⇒ ( x + a + b + c ) 1 b c 0 x + c - b a - c 0 a - b x + b - c = 0 Expanding by first column, we get ⇒ ( x + a + b + c ) [ 1 x + c - b a - c a - b x + b - c - 0 b c a - b x + b - c + 0 b c x + c - b a - c ] = 0 ⇒ ( x + a + b + c ) [ ( x + c - b ) ( x + b - c ) - ( a - c ) ( a - b ) ] = 0 ⇒ ( x + a + b + c ) [ { x 2 - ( c - b ) 2 } - ( a 2 - a b - c a + b c ) ] = 0 ⇒ ( x + a + b + c ) [ { x 2 - ( c 2 + b 2 - 2 b c ) } - a 2 + a b + c a - b c ) ] = 0 ⇒ ( x + a + b + c ) [ x 2 - c 2 - b 2 - a 2 + a b + c a + b c ] = 0 ∴ Either ⇒ ( x + a + b + c ) = 0 ⇒ x = - ( a + b + c ) or ( x 2 - c 2 - b 2 - a 2 + a b + c a + b c ) = 0 ⇒ x = ± ( a 2 + b 2 + c 2 - a b - b c - c a ) 1 2 ∴ One root is - ( a + b + c ) and other are ± ( a 2 + b 2 + c 2 - a b - b c - c a ) 1 2 .

Without expanding, show that a b c x y z p q r = x y z p q r a b c = y b q x a p z c r .

We have to show that a b c x y z p q r = x y z p q r a b c = y b q x a p z c r L . H . S . = a b c x y z p q r operating R 1 ↔ R 2 ,we have = ( - 1 ) x y z a b c p q r operating R 2 ↔ R 3 , we get = ( - 1 ) 2 x y z p q r a b c = x y z p q r a b c Hence, proved. Interchanging row and columns, we have = x p a y q b z r c [ ∵ A = A ' ] Operating C 2 ↔ C 3 , = ( - 1 ) x a p y b q z c r Operating R 1 ↔ R 2 , we have = ( - 1 ) 2 y b q x a p z c r = y b q x a p z c r = R . H . S . Hence, the given result is proved.

E M B I B E

Understanding ISC Mathematics Class 12 Volume 1>Chapter 4 - Determinants>EXERCISE 4.2

EASY

12th ICSE

IMPORTANT

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If $A$ is a square matrix of order $3$ and $|A| = 5$ . Find the value of $|3 A|$ .

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Important Questions on Determinants

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 4 - Determinants>EXERCISE 4.2

If

A

is a square matrix of order

3

and

|A| = - 2

. Find the value of

|- 5 A|

.

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 4 - Determinants>EXERCISE 4.2

If

A

is a matrix of order

3 \times 3

and its determinant is

4

. Find the value of

|3 A|

.

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 4 - Determinants>EXERCISE 4.2

If

A

is a matrix of order

3 \times 3

and

|3 A| = k |A|

, then write the value of

k

.

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 4 - Determinants>EXERCISE 4.2

If $A = [\begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix}]$ and $B = [\begin{matrix} - 1 & 1 \\ 3 & 6 \end{matrix}]$ , then find the value of $|AB|$ .

HARD

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 4 - Determinants>EXERCISE 4.2

If

A = [\begin{matrix} 5 & 5 α & α \\ 0 & α & 5 α \\ 0 & 0 & 5 \end{matrix}]

and

|A^{2}| = 2

find the value(s) of

α

.

HARD

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 4 - Determinants>EXERCISE 4.2

Using the properties of determinants, solve the following equation for $x .$

$|\begin{matrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{matrix}| = 0 .$

HARD

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 4 - Determinants>EXERCISE 4.2

Show that one root of the equation $|\begin{matrix} x + a & b & c \\ b & x + c & a \\ c & a & x + b \end{matrix}| = 0$ is $- (a + b + c)$ .

HARD

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 4 - Determinants>EXERCISE 4.2

Without expanding, show that

$|\begin{matrix} a & b & c \\ x & y & z \\ p & q & r \end{matrix}| = |\begin{matrix} x & y & z \\ p & q & r \\ a & b & c \end{matrix}| = |\begin{matrix} y & b & q \\ x & a & p \\ z & c & r \end{matrix}|$ .