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secondary 4 | E Maths
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Date Posted: 3 years ago
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1 β+2 1
2 5 -3
3 4 1
Finding the determinant of a 3×3 matrix isn't in the O Level syllabus.
Anyway,
For a general 3×3 matrix of the format :
a1 b1 c1
a2 b2 c2
a3 b3 c3
The determinant
= (a1b2c3 + b1c2a3 + c1a2b3) - (a3b2c1 + b3c2a1 + c3a2b1)
So, for this matrix,
Determinant
= (1×5×1 + (β+2)×(-3)×(3) + 1×2×4) - (3×5×1 + 4×(-3)×1 + 1×2×(β+2))
= (5 - 9β - 18 + 8) - (15 - 12 + 2β + 4)
= (-5 - 9β) - (7 + 2β)
= -12 - 11β
So -12 - 11β = 0
11β = -12
β = -12/11 = -1 1/11
2 5 -3
3 4 1
Finding the determinant of a 3×3 matrix isn't in the O Level syllabus.
Anyway,
For a general 3×3 matrix of the format :
a1 b1 c1
a2 b2 c2
a3 b3 c3
The determinant
= (a1b2c3 + b1c2a3 + c1a2b3) - (a3b2c1 + b3c2a1 + c3a2b1)
So, for this matrix,
Determinant
= (1×5×1 + (β+2)×(-3)×(3) + 1×2×4) - (3×5×1 + 4×(-3)×1 + 1×2×(β+2))
= (5 - 9β - 18 + 8) - (15 - 12 + 2β + 4)
= (-5 - 9β) - (7 + 2β)
= -12 - 11β
So -12 - 11β = 0
11β = -12
β = -12/11 = -1 1/11
See 1 Answer
1 β+2 1
2 5 -3
3 4 1
Finding the determinant of a 3×3 matrix isn't in the O Level syllabus.
Anyway,
For a general 3×3 matrix of the format :
a1 b1 c1
a2 b2 c2
a3 b3 c3
The determinant
= (a1b2c3 + b1c2a3 + c1a2b3) - (a3b2c1 + b3c2a1 + c3a2b1)
So, for this matrix,
Determinant
= (1×5×1 + (β+2)×(-3)×(3) + 1×2×4) - (3×5×1 + 4×(-3)×1 + 1×2×(β+2))
= (5 - 9β - 18 + 8) - (15 - 12 + 2β + 4)
= (-5 - 9β) - (7 + 2β)
= -12 - 11β
So -12 - 11β = 0
11β = -12
β = -12/11 = -1 1/11
2 5 -3
3 4 1
Finding the determinant of a 3×3 matrix isn't in the O Level syllabus.
Anyway,
For a general 3×3 matrix of the format :
a1 b1 c1
a2 b2 c2
a3 b3 c3
The determinant
= (a1b2c3 + b1c2a3 + c1a2b3) - (a3b2c1 + b3c2a1 + c3a2b1)
So, for this matrix,
Determinant
= (1×5×1 + (β+2)×(-3)×(3) + 1×2×4) - (3×5×1 + 4×(-3)×1 + 1×2×(β+2))
= (5 - 9β - 18 + 8) - (15 - 12 + 2β + 4)
= (-5 - 9β) - (7 + 2β)
= -12 - 11β
So -12 - 11β = 0
11β = -12
β = -12/11 = -1 1/11
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