Determinants

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New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

A |x+2x+3x+2ax+3x+4x+2bx+4x+5x+2c|=0

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a year ago

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Vishal Baghel

Contributor-Level 10

Given equations are :-

2x+3y+10z=4

4x−6y+5z=1

6x+9y−20z=2

This system of equation can be written, in matrix form, as AX= B, Where

⇒ x = 2, y = 3, z = 5.

New answer posted

a year ago

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Vishal Baghel

Contributor-Level 10

L . H . S . = | s i n α c o s α c o s ( α + δ ) s i n β c o s β c o s ( β + δ ) s i n γ c o s γ c o s ( γ + δ ) | = 1 s i n δ c o s δ | s i n α s i n δ c o s α c o s δ c o s ( α + δ ) s i n β s i n δ c o s β c o s δ c o s ( β + δ ) s i n γ s i n δ c o s γ c o s δ c o s ( γ + δ ) |

Applying cos(A+B)=cosAcosB−sinAsinB in c3

c1→c1+c3?=1sinδcosδ|cosαcosδcosαcosδcosαcosδ−sinαsinδcosβcosδcosβcosδcosβcosδ−sinβsinδcosγcosδcosγcosδcosγcosδ−sinγsinδ|c1=c2∴?=0=R.H.S.

New answer posted

a year ago

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Vishal Baghel

Contributor-Level 10

L . H . S . = | 1 1 + p 1 + p + q 2 3 + 2 p 4 + 3 p + 2 q 3 6 + 3 p 1 0 + 6 p + 3 q | R 2 → R 2 − 2 R 1 R 3 → R 3 − 3 R 1 = | 1 1 + p 1 + p + q 0 1 2 + p 0 3 7 + 3 p | R 3 → R 3 − 3 R 2 = | 1 1 + p 1 + p + q 0 1 2 + p 0 0 1 |

Expanding along c1

?=|12+p01|=1=R.H.S.

New answer posted

a year ago

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Vishal Baghel

Contributor-Level 10

L . H . S . = | 3 a − a + b − a + c − b + a 3 b − b + c − c + a − c + b 3 c | c 1 → c 1 + c 2 + c 3 = | a + b + c − a + b − a + c a + b + c 3 b − b + c a + b + c − c + b 3 c | ( a + b + c ) | 1 − a + b − a + c 1 3 b − b + c 1 − c + b 3 c |

R2→R2−R1and R3→R3−R1

=(a+b+c)|1−a+b−a+c03b+a−b−b+a0−a+b+a−b3c+a−c|

Expanding along c1

=(a+b+c)|2b+aa−ba−c2c+a|=(a+b+c)[4bc+2ab+2ac+a2+ac+ab−bc]=(a+b+c)(3ab+3bc+3ac)=3(a+b+c)(ab+bc+ac)=R.H.S.

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

L . H . S . = | x x 2 1 + p x 3 y y 2 1 + p y 3 z z 2 1 + p z 3 | = | x x 2 1 y y 2 1 z z 2 1 | + | x x 2 p x 3 y y 2 p y 3 z z 2 p z 3 | = ? 1 + ? 2 − − − − − ( 1 ) N o w ? 2 = | x x 2 p x 3 y y 2 p y 3 z z 2 p z 3 | = p x y z | 1 x x 2 1 y y 2 1 z z 2 | c 1 ↔ c 3 = − p x y z | x 2 x 1 y 2 y 1 z 2 z 1 | c 1 ↔ c 2 = p x y z | x x 2 1 y y 2 1 z z 2 1 | = p x y z ? 1

Putting value in (1)

⇒?1+pxyz?1=(1+pxyz)?1−−−−−(2)Now?1=|xx21yy21zz21|

R2→R2−R1 and R3→R3−R1

=|xx21y−xy2−x20z−xz2−x20|

Expanding along c3

?1=|(y−x)(y−x)(y+x)(z−x)(z−x)(z+x)|=(y−x)(z−x)|1y+x1z+x|=(y−x)(z−x)(z+x−y−x)=(x−y)(y−z)(z−x)

Putting ?1 in (2)

L.H.S.=(1+pxyz)(x−y)(y−z)(z−x)=R.H.S

New answer posted

a year ago

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Vishal Baghel

Contributor-Level 10

  L . H . S . = | α α 2 β + γ β β 2 γ + α γ γ 2 α + β | c 3 → c 3 + c 1 = | α α 2 α + β + γ β β 2 α + β + γ γ γ 2 α + β + γ | = ( α + β + γ ) | α α 2 1 β β 2 1 γ γ 2 1 |

R 2 → R 2 − R 1 and R 3 → R 3 − R 1

= ( α + β + γ ) | α α 2 1 β − α β 2 − α 2 0 γ − α γ 2 − α 2 0 |

Expanding along c 3

= ( α + β + γ ) | β − α β 2 − α 2 γ − α γ 2 − α 2 | = ( α + β + γ ) | β − α ( β − α ) ( β + α ) γ − α ( γ − α ) ( γ + α ) | ⇒ ( α + β + γ ) ( β − α ) ( γ − α ) | 1 β + γ 1 γ + α | ⇒ ( α + β + γ ) ( β − α ) ( γ − α ) { γ + 2 − ( β − γ ) } ⇒ ( α + β + γ ) ( β − α ) ( γ − α ) ( γ − β ) ⇒ ( α + β + γ ) ( α − β ) ( β − γ ) ( γ − α ) = R . H . S .

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Expanding along C1

=2 (x+y) { (−x) (x−y)−y.y}

=2 (x+y) (−x2+xy−y2)

=−2 (x+y) (x2−xy+y2)

=−2 {x3−x2y+xy2+x2y−xy2+y3}

= −2 (x3+y3)

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

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