Question 1: Evaluate the determinants in Exercises 1 and 2.

Answer

= 2(−1) − 4(−5) = − 2 + 20 = 18

Question 2: Evaluate the determinants in Exercises 1 and 2.

Answer

= (cos θ)(cos θ) − (−sin θ)(sin θ) = cos² θ+ sin² θ = 1

= (x2 − x + 1)(x + 1) − (x − 1)(x + 1)

= x3 − x2 + x + x2 − x + 1 − (x2 − 1)

= x3 + 1 − x2 + 1

= x3 − x2 + 2

 

Question 3:

If , then show that |2A| = 4|A|

Answer

The given matrix is

∴ R.H.S. = 4|A| = 4 x (-6) = -24 

∴ L.H.S. = R.H.S.

 

Question 4:

If  , then show that |3A| = 27 |A|.

Answer

The given matrix is

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C₁) for easier calculation.

∴ 27|A| = 27(4) = 108                   ……(i)

= 3 (36 - 0) = 3(36) = 108              ……(ii)

From equations (i) and (ii), we have:

|3A| = 27|A|

Hence, the given result is proved.

 

Question 5: Evaluate the determinants 

Answer

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

By expanding along the first row, we have:

= 3(1 + 6) +4 (1 + 4) +5 (3 - 2)

= 3(7) + 4(5) + 5(1)

= 21 + 20 + 5 = 46

By expanding along the first row, we have:

= 0 – 1(0 - 6) + 2 (-3 - 0)

= -1 (-6) + 2 (-3)

= 6 – 6 = 0

By expanding along the first column, we have:

= 2 (0 -5) -0 + 3 (1 + 4)

= -10 + 15 = 5

Question 6:  

Answer

 By expanding along the first row, we have:

= 1(-9 + 12) -1 (-18 + 15) -2 (8 - 5)

= 1(3) -1 (-3) -2 (3)

= 3 + 3 – 6

= 0

 

Question 7: Find values of x, if

Answer

⇒ 2 x 1 – 5X4 = 2x X x – 6 X 4

⇒ 2 – 20 = 2x² – 24

⇒ x² = 3

⇒ 2 X 5 – 3 X 4 = x X 5 – 3 X 2x

⇒ 10 – 12 = 5x – 6x

 ⇒ -2 = -x

⇒ x = 2

 

Question 8:

If , then x is equal to (A) 6 (B) ±6 (C) −6 (D) 0

Answer

Answer: B

⇒ x² – 36 = 36 - 36

⇒ x² – 36 = 0

⇒ x² – 36

⇒ x = ±6

Hence, the correct answer is B.