Determinants

Theorem: If A be k × k matrix then |pA|=p^k|A|.

Given, p=4,k=2 and |A|=5.

|4A|=4² × 5

=16 × 5

=80

Theorem: If Let A be k × k matrix then |pA|=p^k|A|.

Given: k=3 and p=3.

|3A|=3³ × |A|

=27|A|.

Comparing above with k|A| gives k=27.

Theorem: If A be k × k matrix then |pA|=p^k|A|.

Given: p=2, k=3 and|A|=4

|2A|=2³ × |A|

=8 × 4

= 32

Theorem: A_ij is found by deleting i^th rowand j^th column, the determinant of left matrix is called cofactor with multiplied by (-1)^(i+j).

Given: i=3 and j=2.

A₃₂=(-1)⁽³⁺²⁾(2 × 4-6 × 5)

=-1 × (-22)

=22

a₃₂=5

a₃₂A₃₂= 5 × 22

=110

Theorem: This evaluation can be done in two different ways either by taking out the common things and then calculating the determinants or simply take determinant.

I will prefer first method because with that chances of silly mistakes reduces.

Take out x+1 from second row.

(x+1) ×

⇒(x+1) × (x²-x+1-(x-1))

⇒ (x+1) × (x²-2x+2)

⇒ x³-2x²+2x+x²-2x+2

⇒ x³-x²+2 .

This we can very simply go through directly.

((a+ib)(a-ib))-((-c+id)(c+id)).

⇒ (a²+b²)-(-c² -d²).

⇒ a²+b² + c²+d²

∵ i × i=-1

Here the determinant is compared so we need to take determinant both sides then find x.

12x+14=32-42

⇒ 12x=-10-14

⇒ 12x=-24

⇒ x=-2

this question is having the same logic as above.

2x²-40=18+14

⇒ 2x²=72

⇒ x²=36

⇒ x=.

Simply by equating both sides we can get the value of x.

2x²+2x-2(x²+4x+3)=-12

⇒ -6x-6=-12

⇒ -6x=-6

⇒

Find the determinant of A and then multiply it by 3

|A|=2

3|A|=3 × 2

=6

It is determinant multiplied by a scalar number 2, just find determinant of matrix and multiply it by 2.

2 × (35-20)

2 × 15= 30

Find determinant

× √ 24-√ 20 × √ 5

√ 144-√ 100.

=12-10

=2.

After finding determinant we will get a trigonometric identity.

2cos²θ +2sin²θ

=2

∵ sin²θ + cos²θ = 1

After finding determinant we will get a trigonometric identity.

cos²α +sin²α

=1

∵ sin²θ + cos²θ = 1

After finding determinant we will get,

Sin60° = = cos30°

Cos60° = = sin30°

sin 60° × cos30° + sin30° × cos60°

=

= 1.

By directly opening this determinant

cos65° × cos25° -sin25° × sin65°

= cos(65°+25°) ∵ cosAcosB-sinAsinB=cos(A+B)

= cos90°

= 0

∵ cosAcosB-sinAsinB=cos(A+B)

cos15°cos75° - sin75°sin15°

= cos(15°+75°) ∵ cosAcosB-sinAsinB=cos(A+B)

= cos90°

= 0

∵ cosAcosB-sinAsinB=cos(A+B)

We know that expansion of determinant with respect to first row is a₁₁A₁₁+a₁₂A₁₂+a₁₃A₁₃.

0(3 × 6-5 × 4)-2(2 × 6-4 × 4)+0(2 × 5-4 × 3)

= 8.

We know that C₁⇒ C₁-C₂, would not change anything for the determinant.

Applying the same in above determinant, we get

Now it can clearly be seen that C₁=8 × C₃

Applying above equation we get,

We know that if a row or column of a determinant is 0. Then it is singular determinant.

For A to be singular matrix its determinant should be equal to 0.

0= (3-2x) × 4-(x+1) × 2

0= 12-8x-2x-2

0=10-10x

X=1.

=14 × (-7)-9 × (-8)

= -26

= 3√3 × √3 – (-√5 × √5)

= 14.

[R₂’ = (1/2)R₂]

[R₂’ = R₂ - R₃]

[R₁’ = R₁ - R₃]

[R₃’ = 2R₃]

= ( - 14){(2 × 13) - (2 × 12)} - 5{(2 × )– ( - 3) × 13} - 5{( - 3) × 12 - 2 × }

[expanding by the first row]

= - 14 × (26 - 24) - 5(81 + 39) - 5( - 36 - 81)

= - 14 × 2 - 5 × 120 - 5 × ( - 117) = - 28 - 600 + 585 = - 43

[R₁’ = R₁ - R₂]

[R₂’ = 2R₂]

[R₂’ = R₂ - R₃]

[R₃’ = 2R₃]

= 4(8 × 23 - 8 × 27) - 5{8 × - ( - 13) × 23} - 5{( - 13) × 27 - 8 × }

[expansion by first row]

= 132

[R₁’ = R₁/6]

Now, for any determinant, if at least two rows are identical, then the value of the determinant becomes zero.

Here, the first and third rows are identical.

So, the value of the above determinant evaluated = 0

Expanding by first row, we get,

1(9 × 25 - 16 × 16) + 4(16 × 9 - 4 × 25) + 9(4 × 16 - 9 × 9) = - 31 + 176 - 153 = - 8

[C₁’ = C₁ - C₂ & C₂’ = C₂ - C₃]

[C₁’ = C₁/(a - b) & C₂’ = C₂/(b - c)]

= (a - b)(b - c)[0 + 0 + 1{ - a - ( - c)}] [expansion by first row]

= (a - b)(b - c)(c - a)

[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

[R₁’ = R₁/(b - a) & R₂’ = R₂/(c - b)]

= (b - a)(c - b)[0 + 0 + 1{(c + b) - (b + a)}] [expansion by first column]

= (a - b)(b - c)(c - a)

[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

[R₁’ = R₁ - R₂]

[R₂’ = R₂*2]

[R₂’ = R₂ + R₃]

= (1/2)[0 + 3(1 + q) - (1 + 6p + 3q) + p(6 + 3p - 3p)] [expansion by first row]

= (1/2)(3 + 3q - 1 - 6p - 3q + 6p) = 1

[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

[R₁’ = R₁/a & R₂’ = R₂/a]

= a²[a + z - ( - y) - ( - x)] [expansion by first row]

= a²(a + x + y + z)

[R₁’ = R₁ + R₂ + R₃]

[R₁’ = R₁/(x + 2a)]

[R₂’ = R₂ - R₃]

[R₂’ = R₂/(x - a)]

= (x + 2a)(x - a)[x - ( - a) + ( - a - 0) + ( - a)] [expansion by first row]

= (x + 2a)(x - a)(x + a - a - a) = (x + 2a)(x - a)²

[R₁’ = R₁ + R₂ + R₃]

[R₁’ = R₁/(5x + 4)]

[R₂’ = R₂ - R₃]

[R₂’ = R₂/(x - 4)]

= (5x + 4)(x - 4)[ - (x + 4) - 2x + 2x - 0 + 0 - ( - 2x)] [expansion by first row]

= (5x + 4)(x - 4)( - x - 4 + 2x) = (5x + 4)(x - 4)²

[R₁’ = R₁ + R₂ + R₃]

[R₁’ = R₁/(5x + )]

[R₂’ = R₂ - R₃]

[R₂’ = R₂/(x - )]

= (5x + )(x - )[ - (x + ) - 2x + 2x - 0 + 0 - ( - 2x)] [expansion by first row]

= (5x + )(x - )( - x - + 2x) = (5x +)(x -)²

[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

[R₁’ = R₁/(a - 1) & R₂’ = R₂/(a - 1)]

= (a - 1)²[a + 1 - 0 - 2] [expansion by first row]

= (a - 1)³

[R₁’ = R₁ + R₂ + R₃]

[R₁’ = R₁/3(x + y)]

[R₂’ = R₂ - R₃]

[R₂’ = R₂/y]

[R₁’ = R₁ - R₂]

= 3y(x + y)[0 + 3(x + y) - x + 0] [expansion by first row]

= 3y(x + y)(3y) = 9y²(x + y)

[C₁’ = C₁ + C₂ + C₃]

[C₁’ = C₁/(x + y + z)]

[transforming row and column]

[C₁’ = C₁ - C₂ & C₂’ = C₂ - C₃]

= (x + y + z)[0 + 0 + ( - x - 2y)( - y - 2z) - ( - x + y)(2y + z)] [expansion by first row]

= (x + y + z)(xy + 2y² + 2xz + 4yz + 2xy - 2y² + xz - yz)

= (x + y + z)(3xy + 3yz + 3xz)

= 3(x + y + z)(xy + yz + zx)

[C₁’ = C₁/x, C₂’ = C₂/y & C₃’ = C₃/z]

[C₁’ = C₁ - C₂ & C₂’ = C₂ - C₃]

[C₁’ = C₁/(x - y)& C₂’ = C₂/(y - z)]

= xyz(x - y)(y - z)(0 + 0 + y + z - x - y) [expansion by first row]

= xyz(x - y)(y - z)( z - x)

[R₁’ = R₁ + R₂ + R₃]

[R₁’ = R₁/(a + b + c)]

= (a + b + c)[2(b - c)c - b(c - a) + (c + a)(c - a) - (a + b)(b - c)] [expansion by first row]

= (a + b + c)(2bc - 2c² - bc + ab + c² - a² - ab - b² + ac + bc

= (a + b + c)(ab + bc + ac - a² - b² - c²)

= 3abc - a³ - b³ - c³

[R₁’ = R₁ + R₂ + R₃]

[R₁’ = R₁/2]

[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

= 2[c{a(a + b) - ( - ac)} + 0 + a{c(b - c) - ac}] [expansion by first row]

= 2(a²c + abc + ac² + abc - ac² - a²c)

= 4abc

[R₁’ = 3R₁]

[R₁’ = R₁ - R₂]

[R₂’ = 2R₂]

[R₂’ = R₂ - R₃]

= (1/6)[0 + 0 + 6a{a(a + b) - a(2a + b) [expansion by first column]

= - a³

[R₁’ = R₁ + R₂]

[R₁’ = R₁/(a + b)]

[R₂’ = R₂ + R₃]

[R₂’ = R₁/(b + c)]

[R₁’ = R₁ + R₂]

= (a + b)(b + c){0 + 2( - b + a + b + c) + 0} [expansion by first row]

= 2(a + b)(b + c)(c + a)

[R₁’ = xR₁ & R₂’ = yR₂]

[R₁’ = R₁ + R₂ - R₃]

= (1/xy)[0 + 0 + (ax² + 2bxy + cy²){by(bx + cy) - cy(ax + by)}[expansion by first row] .

= (1/xy)( ax² + 2bxy + cy²)(b²xy + bcy² - acxy - bcy²)

= (b² - ac)(ax² + 2bxy + cy²)

[R₂’ = R₂ - R₃]

[R₂’ = R₂/4]

[transforming row and column]

[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

[R₁’ = R₁/(a - b) & R₂’ = R₂/(b - c)]

[R₁’ = R₁ - R₂]

[R₁’ = R₁/(a - c)]

= 4(a - b)(b - c)(a - c)(c² - 2c + 1 - bc - c² + 2c + 0 + bc + c² - c²) [expansion by first row]

= 4(a - b)(b - c)(c - a)

[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

[R₁’ = R₁ - R₂]

[R₁’ = R₁/2]

[transforming row and column]

[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

[R₁’ = R₁ - R₂]

= 2{0 + 0 + 2(0 - 2)} [expansion by first row]

. = - 8

[C₃’ = 2C₃]

[C₁’ = C₁ - C₃]

[C₁’ = C₁ + C₂]

[C₁’ = C₁/(l² + m² + n²)]

[transforming row and column]

[C₁’ = C₁ - C₂ & C₂’ = C₂ - C₃]

[C₁’ = C₁/(l - m) & R₂’ = C₂/(l - m)]

= (l² + m² + n²)(l - m)(m - n){0 + 0 - l(l + m) + n(m + n)} [expansion by first row]

= (l² + m² + n²)(l - m)(m - n){0 + 0 - l(l + m) + n(m + n)}

= (l² + m² + n²)(l - m)(m - n)( - l² - ml + mn + n²)

= (l² + m² + n²)(l - m)(m - n){(n² - l²) + m(n - l)}

= (l² + m² + n²)(l - m)(m - n)(n - l)(l + m + n)

[C₃’ = 2C₃]

[C₁’ = C₁ - C₃]

[C₁’ = C₁ + C₂]

[C₁’ = C₁/(a² + b² + c²)]

[transforming row and column]

[C₁’ = C₁ - C₂ & C₂’ = C₂ - C₃]

[C₁’ = C₁/(a - b) & C₂’ = C₂/(b - c)]

= (a² + b² + c²)(a - b)(b - c){0 + 0 - a(a + b) + c(b + c)} [expansion by first row]

= (a² + b² + c²)(a - b)(b - c){0 + 0 - a(a + b) + c(b + c)}

= (a² + b² + c²)(a - b)(b - c)( - a² - ba + bc + c²)

= (a² + b² + c²)(a - b)(b - c){(c² - a²) + b(c - a)}

= (a² + b² + c²)(a - b)(b - c)(c - a)(a + b + c)

[R₁’ = R₁ + R₂ + R₃]

[R₁’ = R₁/2]

[R₁’ = R₁ - R₂]

= 2[c²{(c² + a²)(a² + b²) - b²c²} + 0 + a²{b²c² - c²(c² + a²)}] [expansion by first row]

= 2[c²(c²a² + a⁴ + b²c² + a²b² - b²c²) + a²(b²c² - c⁴ - a²c²)]

= 2[a²c⁴ + a⁴c² + a²b²c² + a²b²c² - a²c⁴ - a⁴c²]

= 4a²b²c²

Operating R₁→R₁+ bR �₃, R₂→R₂- aR₃

Taking (1+ a²+b²) from R₁ and R₂

Operating R₃→ R₃ - 2bR₁ + 2aR₂

Taking (1+a²+b²) from R₃

Expanding with respect to C₁

= (1+a²+b²)³ 1×[ 1-0]

= (1+a²+b²)³

Hence proved

Operating C₁→aC₁

Operating C₁→C₁+bC₂+cC₃

Taking (a²+b²+c²) common from C₁

Operating R₁→R₁-R₃, R₂→R₂-R₃

Operating C₂→C₂-C₃

Taking (a+b+c) common from C₂

Expanding with respect to C₁

= (a²+b²+c²)(a+b+c)

_{Expanding with R1}

_=b²c²(a²c+abc-abc-a²b)-bc(a³c²+a²bc²-a²b²c-a³b²)+(b+c)(a³bc²-a³b²c)

_=a²b³c²-a²b³c²-a³bc²-a²b³c²+a²b³c²+a³b³c+a³b²c²-a³b³c+a³bc³-a³b²c²

₌₀

Operating R₁→aR₁, R₂→bR₂, R₃→cR₃

Taking a, b, c common from C₁, C₂, C₃ respectively

Operating R₁→R₁- R₃, R₂→R₂- R₃

Taking (a+b+c) common from R₁, R₂

Operating R₃→ R₃- R₁- R₂

Operating C₁→aC₁, C₂→bC₂

Operating C₁→C₁+C₃, C₂→C₂+C₃

Taking a, b, 2ab from R₁, R₂, R₃

Expanding with R₃

Taking (b-a) common from C₁, C₃

Operating R₂→R₂-R₁+R₃

[Properties of determinants say that if 1 row or column has only 0 as its elements, the value of the determinant is 0]

= 0

Hence Proved

Taking a, b, c from C₁, C₂, C₃

Operating R₁→R₁-R₃, R₂→R₂-R₃

Taking (a²+b²+c²) common from R₁, R₂

Operating R₃→R₃+R₁+R₂

Taking (a²+b²+c²) common from C₃ �

Expanding with C₃

= abc(a²+b²+c²)³×1×(1-0)

= abc (a²+b²+c²)³

Hence proved

Given that α, β, γ are in an AP, which means 2β=α+γ

Operating R₃→R₃-2R₂+R₁

[we know that 2β=α+γ]

Operating R₁→R₁-R₃, R₂→R �₂-R₃

[By the properties of determinants, we know that if all the elements of a row or column is 0, then the value of the determinant is also 0]

=0

Hence proved

Operating R₁→R₁-R₂, R₂→R₂-R₃

Expanding with C₃

= (2(a+2) – 2(a+3))

= (2a+4-2a-6)

= -2

By properties of determinants, we can split the given determinant into 2 parts

→

Taking x, y, z common from R₁, R₂, R₃ respectively

→

Operating R₁→R₁-R₃, R₂→R₂- R₃

→

→

Taking (x-z) and (y-z) common from R₁, R₂

→

Expanding with R₃

→y²+yz+z²-x²-xz-z² = xyz(xy²+xyz+xz²+zy²+yz²+z³-x²y-xyz-yz²-x²z-xz²-z³)

→ (y-x)(y+x) +z(y-x) =xyz(xy²+zy² -x²y -x²z)

→(y-x)(x+y+z)=xyz(xy(y-x)+z(y²-x²))

→(y-x)(x+y+z)= xyz(xy(y-x)+z(x+y)(y-x))

→(y-x)(x+y+z) = xyz(xy(y-x)+(xz+yz)(y-x))

→(y-x)(x+y+z)= xyz(y-x)(xy+xz+yz)

→x+y+z = xyz(xy+xz+yz)

Hence Proved

Operating R₁→R₁-R₂, R₂→R₂-R₃

Taking (a-b), (b-c) common from R₁, R₂ respectively

Operating R₁→R₁- R₂

Taking (a-c) common from R₁

Expanding with C₁

= (a-c)(a-b)(b-c)×(2b²+2bc+2c²-ab-b²-bc-ac-bc-c²+a²+ab+ac)

=-(c-a)(b-c)(a-b)(a²+b²+c²)

Hence Proved

Operating R₁→R₁-R₂, R₂→R₂-R₃

→

→

Taking (a-b) and (b-c) from R₁, R₂

→

Method 1:

For the two determinants to be equal, their difference must be 0.

Since 2 columns have only 0 as their elements, by properties of determinants

=0

Method 2:

Expanding both with C₁

LHS

=(a-b)(b-c)(-a+c)

RHS

=(a-b)(b-c)(b+c-a-b)

=(a-b)(b-c)(-a+c)

∴ LHS = RHS

_{Operating R1→R1}-R₃, R_2→R2-R3

→

Taking (a-c) and (b-c) common from R₁, R₂

→

Method 1:

If the determinants are equal, their difference must also be equal.

(a-c) and (b-c) get cancelled.

Since all elements of C₁ are 0, by properties of determinants,

=0

∴ The 2 determinants are equal.

Method 2:

Expanding with C₁

→(a-c)(b-c)(b+c-a-c) = (a-c)(b-c)(b-a)

→(a-c)(b-c)(b-a)=(a-c)(b-c)(b-a)

∴ RHS and LHS are equal

Operating R₁→R₁-R₂

Taking (x-2) common from R₁

Here, we can see that x-2 is a factor of the determinant.

We can say that when x-2 is put in the equation, we get 0.

x-2=0

→x=2

_{Operating R1→R1}-R_2, R_2→R2-R₃

Expanding with C₁

0=(x-c)(b-c)(b²-2bc+c²+3bc-x²+2xb-b²-3xb)

0= (x-c)(b-c)(bc+c²-x²-xb)

0=(x-c)(b-c)(-b(-c+x)-(c-x)(-c-x))

0=(x-c)²(b-c)(-b-c-x)

Either x-c=0 or b-c=0 or (-b-c-x)=0

∴x=c or b=c or x=-(b+c)

If b=c, x=b

∴ x=c or x=b or x=-(b+c)

_{Operating C1→C1}+C₂+C₃

Taking (x+a+b+c) common from C₁

Operating R₁→R₁-R₃, R₂→R₂-R₃

Expanding with C₁

0=(x+a+b+c)(0+x²)

0=x²(x+a+b+c)

Either x²=0 or (x+a+b+c)=0

∴ x=0 or x=-(a+b+c)

_{Operating C1→C1}+C₂+C₃

Taking (3x-2) common from C₁

Operating R₁→R₁-R₃, R₂→R₂-R₃

Expanding with C₁

0= (3x-2)(0+(3x-11)²)

0=(3x-2)(3x-11)²

Either 3x-2=0 or 3x-11=0

∴

Operating C₁→C₁+C₂+C₃

Taking (x+9) common from C₁

Operating R₁→R₁-R₃, R₂→R₂-R₃

0= (x+9)(0-x+x²+1-x)

0=(x+9)(x²-2x+1)

0=(x+9)(x-1)²

∴ Either x+9=0 or x-1=0

x=-9, x=1

_{Operating R1→R1}+R₂+R₃

Taking (x+9) common from R₁

Operating C₁→C₁-C₃, C₂→C₂-C₃

Expanding with R₁

0=(x+9)(0-(x-2)(7-x))

0=(x+9)(7-x)(2-x)

Either x+9=0 or 7-x=0 or 2-x=0

∴ x=-9 or x=7 or x=2

_{Expanding with R1}

_{0= x(-3x}²-6x-2x²+6x)+6(2x+4+3x-9)-1(4x-9x)

_0=x(-5x²)+6(5x-5)-1(-5x)

_{0= -5x}³+30x-30+5x

_0=-5x³+35x-30

_x³-7x+6=0

_x³-x-6x+6=0

_x(x²-1)-6(x-1)=0

_{x(x-1)(x+1)-6(x-1)=0}

_(x-1)(x²+x-6)=0

_(x-1)(x²+3x-2x-6)=0

_{(x-1)(x(x+3)-2(x+3)(=0}

_{(x-1)(x+3)(x-2)=0}

_{Either x-1=0 or x+3=0 or x-2=0}

_{∴x=1 or x=-3 or x=2}

Operating C₁→aC₁

Operating C₁→C₁+bC₂+cC₃

Taking (a²+b²+c²)

Operating C₂→C₂-bC₁, C₃→C₃-cC₃

Expanding with R₃

a+b+c)

= (a²+b²+c²)(a+b+c)

Hence Proved

Area of a triangle =

Expanding with C₃

= 37.5 sq. units

Expanding with C₃

= 34 sq. units

Expanding with R₃

=28 sq. units

Expanding with R₁

= 9 sq. units

Operating R₁→ R₁-R₃, R₂→ R₂-R₃

Expanding with C₃

= -23.5 sq. units = 23.5 sq units

Expanding with C₃

=0

Since the area between the 3 points is 0, the three points lie in a straight line, i.e. they are collinear.

Expanding with C₃

=0

Since the area between the 3 points is 0, the three points lie in a straight line, i.e. they are collinear.

Expanding with C₃

=0

Since the area between the 3 points is 0, the three points lie in a straight line, i.e. they are collinear.

Since they are collinear, the area will be 0

→

Expanding with C₃

→

→

→ 10k – 50=0

→ 10k=50

∴ k=5

Since they are collinear, the area will be 0

→

Expanding with C₃

→ 0 = (7k-11)-(35-55)+(5-5k)

→ 0= 2k-14

→ 2k=14

∴ k=7

Since they are collinear, the area will be 0

→

Expanding with C₃

→ 0 = (10-4k)-(5+4)+(k+2)

→ 0=-3k+3

→ 3k=3

∴ k= 1

Expanding with C₃

→ 70 = (20-4k)-(8+6k)+(8+30)

→ 70= -10k+50

→ 20=-2k

→ k=-2

Expanding with C₁

→ 8=-2(4-k)

→ -4=4-k

→ k=8

Since the points are collinear, the area they enclose is 0

Expanding with C₁

→ 0 = a(b-1)+(-b)

→ 0= ab-a-b

→ a+b=ab

→

→

Hence proved

To find: Value of

Formula used: (i)

We have,

On expanding the above,

⇒ {cos 70°} {cos 20°} – {sin 70°} {sin 20°}

On applying formula

⇒ {sin (90 – 70)} {sin (90 – 20)} - {sin 70°} {sin 20°}

⇒ {sin 20°} {sin 70°} - {sin 70°} {sin 20°}

= 0

To find: Value of

Formula used: (i)

We have,

On expanding the above,

⇒ {cos 15°} {cos 15°} – {sin 15°} {sin 15°}

On applying formula

= cos (15 + 15)

= cos (30°)

To find: Value of

Formula used: (i)

We have,

On expanding the above,

⇒ (sin 23°) (cos 7°) – (cos 23°) (-sin 7°)

⇒ (sin 23°) (cos 7°) + (cos 23°) (sin 7°)

On applying formula

= sin (23 + 7)

= sin (30°)

To find: Value of

Formula used: i² = -1

We have,

On expanding the above,

⇒ (a + ib) (a – ib) – (-c + id) (c + id)

⇒ (a² – iab + iba – i²b²) - (-c² – icd + icd + i²d²)

⇒ {a² – iab + iba – (-1)b²} – {-c² – icd + icd + (-1)d²}

⇒ {a² – iab + iba + 1b²} – {-c² – icd + icd - 1d²}

⇒ a² + b² + c² + d²

To find: Value of

Formula used: = 1

We have,

On expanding the above along 1^st column

⇒ 1 - +

⇒

⇒ … (i)

As = 1,

⇒

⇒

Using the above obtained value of in eqn. (i)

⇒

⇒

⇒

⇒ 1 – 1 = 0

To find: Value of

Formula used: (i) = 1

(ii)

We have,

On expanding the above along 1^st column

⇒ 1 - +

⇒

⇒ … (i)

As = 1,

⇒

⇒

Using the above obtained value of in eqn. (i)

⇒

⇒

⇒

⇒ 1 – 1 = 0

To find: Value of

We have,

⇒

Applying

⇒

Applying

⇒

Taking 4 common from R₁

⇒

Applying

⇒

Taking -2 common from R₁

⇒

Applying

⇒

Applying

⇒

Taking 9 common from R₁

⇒

Expanding along R₁

⇒ -8 [1[(3)(43)-(16)(0)] – 0 [(4)(43)-(0)(0)] - 2 [(4)(16)-(3)(0)]]

⇒ -8 [[(129)-(0)] - 2 [(64)-(0)]]

⇒ -8 [129 – 128]

⇒ -8

To find: Value of

We have,

⇒

Taking 2 common from R₂

⇒

Taking 6 common from R₃

⇒

Applying

⇒

Applying

⇒

Expanding column 1

⇒ 12 [1{(1)(14)-(6)(2)}]

⇒ 12 [1{(14)-(12)}]

⇒ 12[2]

⇒ 24

To find: Value of

We have,

Applying

⇒

⇒

Applying

⇒

⇒

If every element of a row is 0 then the value of the determinant will be 0

To find: Value of

We have,

Applying

⇒

Applying

⇒

Expanding along C₁

⇒ [1{(1)(3p-2)-(3)(p-2)}]

⇒ 1

To find: Value of

We have,

Applying

⇒

Applying

⇒

We know, x³ – y³ = ( x – y) (x²+xy+y²)

⇒

Taking (b-a) common from C₂

⇒

Taking (c-a) common from C₂

⇒

Expanding along C₁

⇒ (b - a) (c – a)[1{(1)(c² + ca + a²) – (b² + ab + a²)(1)}]

⇒ (b - a) (c – a)[c² + ca + a² – b² - ab - a²]

⇒ (b - a) (c – a)[c² – b² + ca - ab]

⇒ (b - a) (c – a)[(c – b) (c + b) + a(c – b)]

⇒ (b - a) (c – a)[(a + b + c)(c – b)]

⇒ (a – b) (b – c) (c – a) (a + b + c)

To find: Value of

Formula Used: sin(A+B) = sinAcosB+cosAsinB

We have,

Applying

⇒

Applying

⇒

We know, sin(A+B) = sinAcosB+cosAsinB

⇒

Applying

⇒

⇒

= 0

When two columns are identical then the value of determinant is 0

To find: Nature of

We have,

Applying

⇒

Taking (a+b+c) common from R₁

⇒

Expanding along R₁

⇒ (a+b+c)[1{(b)(c)-(a)(a)} – 1{(b)(b)-(c)(a)} + 1{(a)(b)-(c)(c)}]

⇒ (a+b+c)[1{bc-a²} – 1{b²-ca} + 1{ba - c²}]

⇒ (a+b+c)[bc - a² –b² + ca + ab - c²]

⇒ -(a+b+c)[ c² + a² + b² - ca - bc – ba ]

⇒

⇒

⇒

⇒

Clearly, we can see that the answer is negative

To find: Value of

We have,

Applying

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Expanding along R₂

⇒ [x{(2x)(8x) - (8x+8y)(0)}]

⇒ [x{16x²}]

⇒ x³

To find: Value of

We have,

Applying

⇒

Applying

⇒

Expanding along C₃

⇒ [1{(a²-1)(a-1) – (a-1)(2a – 2)}]

⇒ [1{(a-1)(a+1)(a-1) – (a-1)2(a – 1)}]

⇒ [{(a+1)(a-1)² – 2(a-1)²}]

⇒ [{(a-1)² (a+1-2)}]

⇒ [{(a-1)² (a-1)}]

⇒ (a-1)³

To find: Value of

We have,

Applying

⇒

Applying

⇒

Expanding along C₁

⇒ [a{(a) (a+b) – (a)(2a+b)}]

⇒ [a{(a² + ab) – (2a²+ab)}]

⇒ [a{a² + ab – 2a² - ab}]

⇒ [a{–a²}]

⇒ -a³

To find: Value of

We have,

Applying +

⇒

⇒

⇒ (a+b+c)

Expanding along R₁

⇒ (a+b+c)[2{(c) (c) – (b) (a)} -1{(c+a)(c)-(a+b)(a)} + 1{(c+a)(b)-(a+b)(c)}]

⇒ (a+b+c)[2{c² – ab} -1{c²+ac-a²-ab} + 1{bc+ba-ac-bc}]

⇒ (a+b+c)[2c² – 2ab - c² – ac + a² + ab + ba - ac]

⇒ (a+b+c)[c² + a² - 2ac]

⇒ (a+b+c)(c – a)²

To find: Value of

We have,

Applying

⇒

Expanding along R₁

⇒ [x{(1)(1+y)-(1)(1)}]

⇒ [x{1+y-1}]

⇒ xy

To find: Value of

We have,

Applying

⇒

⇒

Taking (b – a) common

⇒

Applying

⇒

⇒

Taking (c - b) common

⇒

Expanding along C₃

⇒ (b – a) (c – b) [1{(c) (1) – (a) (1)}]

⇒ (b – a) (c – b) (c – a)

⇒ (a – b) (b – c) (c – a)

To find: Value of

We have,

Applying

⇒

Taking 2 common

⇒

Applying

⇒

Expanding along R₁

⇒ 2 [c{(c + a) (a + b) – (b) (c)} + a{(b)(c) – (c) (c + a)}]

⇒ 2 [c{(ac + cb +a² + ab – bc} + a{(bc – c² - ac)}]

⇒ 2 [c{(ac + a² + ab)} + a{(bc – c² - ac)}]

⇒ 2 [ac² + ca² + abc + abc – ac² – a²c]

⇒ 2 [2abc]

⇒ 4abc

To find: Value of

We have,

Applying

⇒

Taking (a - b) common

⇒

Applying

⇒

Taking (c-a) common

⇒

Expanding along R₁

= (b – a)(c – a)[0 – 1(1 – (a – b)) + (b + c)(0)]

= (b – a)(c – a)(-1 + a – b)

= (b – a)(c – a)(a – b – 1)

= (b – a)(ac – bc – c – a² + ab + a)

= (abc – b²c – bc – a²b + ab² + ab – a²c + abc + ac + a³ + a²b + a²)

= 4abc

To find: Value of

We have,

Applying

⇒

Applying

⇒

Expanding along R₁

⇒ [2{(5)(x+14) – (6)(x+10)} – 3{(4)(x+14) – (6)(x+7)} + 4 {(4)(x+10) – (5)(x+7)}]

⇒ [2{5x + 70 – 6x – 60} – 3{4x + 56 -6x - 42} + 4 {4x + 40 – 5x - 35}]

⇒ [2{10 - x} – 3{14 -2x} + 4 {5 - x}]

⇒ [20 – 2x – 42 + 6x + 20 - 4x]

⇒ -2

To find: Value of x

We have,

Applying

⇒

Applying

⇒

Expanding along R₁

⇒ [1{(x)(-2) – (6)(2)}] = 0

⇒ [1{-2x – 12}] = 0

⇒ -2x-12 = 0

⇒ -2x = 12

⇒ x = -6

To find: Value of x

We have,

Applying

⇒

Applying

⇒

Expanding along R₁

⇒ [(2x-7){(x)(x) – (6)(2)} + (14-x){(2)(6) – (x)(7)] = 0

⇒ [(2x-7){x²– 12} + (14-x){12 – 7x}] = 0

⇒ [2x³ – 24x – 7x² + 84 + 168 – 98x -12x + 7x²] = 0

⇒ [2x³ – 134x + 252] = 0

⇒ [x³ – 67x + 126] = 0

By Hit and trial x = -2, 3, -7

To find: Value of x

We have,

Applying

⇒

Applying

⇒

Expanding along R₁

⇒ [x-2{(-1)(-40) – (-4)(-11)} -1 {(x-4)(-40) – (-4)(x-8)} + 2 {(x-4)(-11) – (-1)(x-8)] = 0

⇒ [(x-2){40-44} -1 {(-40x + 160 + 4x – 32} + 2 {-11x + 44 + x - 8}] = 0

⇒ [(x-2){-4} -1 {(- 36x + 128} + 2 {-10x+36}] = 0

⇒ [-4x + 8 + 36x - 128 - 20x + 72] = 0

⇒ 12x – 48 = 0

⇒ x = 4

To find: Value of x

We have,

Applying

⇒

Applying

⇒

Taking 2 common from R₁

⇒

Taking 2 common from R₂

⇒

Applying

⇒

Expanding along R₁

⇒ 4[x{(x)(a+x) – (-x)(a-2x)}] – (-x){(0)(a+x) – (-x)(a)}] = 0

⇒ 4[x{ax + x² +ax -2x²}] – (-x){ax}] = 0

⇒ 4[x{2ax - x²}] + ax²] = 0

⇒ 4[2ax² – x³ + ax²] = 0

⇒ – x² + 3ax = 0

⇒ -x(x – 3a) = 0

⇒ x = 0 , or x = 3a

To find: Value of x

We have,

Applying

⇒

Applying

⇒

Expanding along R₁

⇒ (3x-11){(3x-11)(3x-8) – (3)(11-3x)} – (11-3x){(0)((3x-8) – (11-3x)(3)} = 0

⇒ (3x-11){(3x-11)(3x-8+3)} – (11-3x){–(11-3x)(3)} = 0

⇒ (3x-11)²(3x-5)} + (3x-11){(3x-11)(3)} = 0

⇒ (3x-11)²(3x-5)} + (3x-11)²(3)} = 0

⇒ (3x-11)²(3x-5+3) = 0

⇒ (3x-11)²(3x-2) = 0

⇒

To find: Area of ABC

Given: A(-2,4), B(2,-6) and C(5,4)

Formula used:

We have, A(-2,4), B(2,6) and C(5,4)

Expanding along R₁

⇒

⇒

⇒

⇒

⇒

⇒ 35 sq. units

To find: Area of ABC

Given: A(3,-2), B(k,2) and C(8,8)

The formula used:

We have, A(3,-2), B(k,2) and C(8,8)

Expanding along R₁

⇒

⇒

⇒ -18 +2k - 16 + 8k -16 = 0

⇒ 10k -50 = 0

⇒ k = 5