Reeya asked in Science & MathematicsMathematics · 4 years ago

# Help me with this determinants question?

Prove that the value of the following determinant is ZERO:

|sin^2(23)° sin^2(67)° cos180°|

|-sin^2(67) -sin^2(23)° cos^2(180)°|

|cos(180)° sin^2(23)° sin^2(67)°|

I have been trying for hours but keep getting the wrong answer?

Relevance
• 4 years ago

If lAl = l a b c l

. . . . . .l d e f l

. . . . . .l g h i l . . . then lAl = aei + bfg +cdh - ceg - bdi - afh

Then the determinant in this question equals :-

(sin²23.-sin²23.sin²67) + (sin²67.cos²180.cos180) + (cos180.-sin²67.sin²23) - (cos180.-sin²23.cos180) - (sin²67.-sin²67.sin²67) - (sin²23.cos²180.sin²23)

Replacing cos180 = -1 and cos²180 = 1

(sin²23.-sin²23.sin²67) - sin²67 + (sin²67.sin²23) + sin²23 - (sin²67.-sin²67.sin²67) - (sin²23.sin²23)

Tidying up signs.

(sin²67.sin²67.sin²67) + (sin²67.sin²23) + sin²23 - [(sin²23.sin²23.sin²67) + (sin²23.sin²23) + sin²67]

As, sin²θ + sin²(90 - θ) = 1 then replace sin²67 with (1 - sin²23)

(1 - sin²23)(1 - sin²23)(1 - sin²23) + (1 - sin²23)sin²23 + sin²23 - [sin²23.sin²23 (1 - sin²23) + (sin²23.sin²23) + (1 - sin²23)]

After expanding the bracketed terms

1 - 3sin²23 + 3sin⁴23 - sin⁶23 + sin²23 - sin⁴23 + sin²23 - sin⁴23 + sin⁶23 - sin⁴23 - 1 + sin²23

And all terms cancel confirming that the determinant is ZERO.

• ?
Lv 7
4 years agoReport

See my answer. The easy way.

• ?
Lv 7
4 years ago

Do this the easy way!

value of the determinant?

|sin^2(23)° sin^2(67)° cos(180°)|

|-sin^2(67°) -sin^2(23°) cos^2(180°)|

|cos(180)° sin^2(23)° sin^2(67)°|

substitute

cos(180°)=-1, cos^2(180°) = 1

sin(23°) = cos(67°) = a

sin(67°) = cos(23°) = b

| a^2 b^2 -1 |

| -b^2 -a^2 1 |

| -1 a^2 b^2 |

Instead of evaluating this we can use row and column operations.

add columns 2 and 3 to column 1

giving for col 1

a^2+b^2 - 1

-(a^2+b^2) + 1

a^2+b^2 - 1

Note that

a^2+b^2= sin^2(23°)+cos^2(23°) = 1

All entries in column 1 are now zero.

Therefore the determinant is zero!

• Reeya4 years agoReport

Thanks for your answer.It is simpler and better than the one above.