Question 1: Write minors and co-factors of each element of first column of the following matrices and hence evaluate determinant.
Solution:
i) Let Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth column.
Here, a11 = 5
Minor of a11 = M11 = -1
Note: In 2×2 matrix, minor is obtained for a particular element, by deleting that row and column were element is present.
Minor of a12 = M12 = 0
Minor of a21 = M21 = 20
Minor of a22 = M22 = 0
As M12 and M22 are zero so we don’t consider them. Hence we have got only two minors for this determinant.
M11 = -1 & M21 = 20
Now, co-factors for the determinants are
C11 = (-1)1+1 x M11 {∵Cij =(-1)1+1 x Mij}
= (+1)x(-1)
= -1
C21 = (-1)2+1 x M21
= (-1)3 x 20
= -20
Evaluating the determinant,
|A| = a11 x C11 + a21 x C21
=5 x (-1) + 0 x (-20)
= -5
Solution:
Let Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth column.
Minor of a11 = M11 = 3
Note: In 2×2 matrix, minor is obtained for a particular element, by deleting that row and column were element is present.
Minor of a21 = M21 = 4
Now, co-factors for the determinants are
C11 = (-1)1+1 x M11 {∵Cij =(-1)i+j x Mij}
= (+1) x 3
= 3
C21 = (-1)2+1 x M21
= (-1)3 x 4
= -4
Evaluating the determinant,
|A| = a11 x C11 + a21 x C21
=-1 x 3 + 2 x (-4)
=-11
Solution:
Let Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth column.
Cij = (-1)i+j x Mij
Given,
We have,
M11 = -1×2 – 5×2
M11 = -12
M21 = -3×2 – 5×2
M21 = -16
M31 = -3×2 – (-1) x 2
M31 = -4
Co-factors of the determinant are as follows,
C11 = (-1)1+1 x M11
= 1x-12
= -12
C21 = (-1)2+1 x M21
= (-1)3 x -16
= 16
C31 = (-1)3+1 x M31
= (1)4 x (-4)
= -4
To evaluate the determinant expand along first column,
|A| = a11 x C11 + a21 x C21+ a31 x C31
=1x(-12) + 4×16 + 3x(-4)
= -12 + 64 – 12
= 40
Solution:
Let Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth column.
Also, Cij = (-1)i+j x Mij
Given,
We have,
M11 = b x ab – c x ca
M11 = ab2 – ac2
M21 = a x ab – c x bc
M21 = a2b – c2b
M31 = a x ca – b x bc
M31 = a2c – b2c
Co-factors of the determinant are as follows,
C11 = (-1)1+1 x M11
= 1 x (ab2 – ac2)
= ab2 – ac2
C21 = (-1)2+1 x M21
= (-1)3 x (a2b – c2b)
= c2b – a2b
C31 = (-1)3+1 x M31
= (1)4 x (a2c – b2c)
= a2c – b2c
To evaluate the determinant expand along first column,
|A| = a11 x C11 + a21 x C21+ a31 x C31
=1 x (ab2 – ac2) + 1 x (c2b – a2b) + 1 x (a2c – b2c)
= ab2 – ac2 + c2b – a2b + a2c – b2c
Solution:
Let Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth column.
Cij = (-1)i+j x Mij
Given,
We have,
M11 = 5×1 – 7×0
M11 = 5
M21 = 2×1 – 7×6
M21 = -40
M31 = 2×0 – 5×6
M31 = -30
Co-factors of the determinant are as follows,
C11 = (-1)1+1 x M11
= 1×5
= 5
C21 = (-1)2+1 x M21
= (-1)3 x -40
= 40
C31 = (-1)3+1 x M31
= (1)4 x (-30)
= -30
To evaluate the determinant expand along first column,
|A| = a11 x C11 + a21 x C21+ a31 x C31
=0x5 + 1×40 + 3x(-20)
= 0 + 40 – 90
= 50
Solution:
Let Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth column.
Cij = (-1)i+j x Mij
Given,
We have,
M11 = b x c – f x f
M11 = bc – f2
M21 = h x c – f x g
M21 = hc – fg
M31 = h x f – b x g
M31 = hf – bg
Co-factors of the determinant are as follows,
C11 = (-1)1+1 x M11
= 1x (bc – f2)
= bc – f2
C21 = (-1)2+1 x M21
= (-1)3 x (hc – fg)
= fg – hc
C31 = (-1)3+1 x M31
= (1)4 x (hf – bg)
= hf – bg
To evaluate the determinant expand along first column,
|A| = a11 x C11 + a21 x C21+ a31 x C31
=a x (bc – f2) + h x (fg – hc) + g x (hf – bg)
= abc – af2 + hgf – h2c + ghf –bg2
Solution:
Let Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth column
Also, Cij = (-1)i+j x Mij
Given,
From the matrix we have,
M11 = 0(-1 x 0 – 5 x 1) – 1(1 x 0 – (-1) x 1) + (-2)(1 x 5 – (-1) x (-1))
M11 = -9
M21 = -1(-1 x 0 – 5 x 1) – 0(1 x 0 – (-1) x 1) + (1 x 5 – (-1) x (-1))
M21 = 9
M31 = -1(1 x 0 – 5 x (-2)) – 0(0 x 0 – (-1) x (-2)) + 1(0 x 5 – (-1) x 1)
M31 = -9
M41 = -1(1 x 1 – (-1) x (-2)) – 0(0 x 1 – 1 x (-2)) + 1(0 x (-1) – 1 x 1)
M41 = 0
Co-factors of the determinant are as follows,
C11 = (-1)1+1 x M11
= 1x (-9)
= -9
C21 = (-1)2+1 x M21
= (-1)3 x 9
= -9
C31 = (-1)3+1 x M31
= (-1)4 x -9
= -9
C41 = (-1)4+1 x M41
= (-1)5 x 0
= 0
To evaluate the determinant expand along first column,
|A| = a11 x C11 + a21 x C21+ a31 x C31+ a41 x C41
=2 x (-9) + (-3) x (-9) + 1 x (-9) + 2 x 0
= -18 + 27 – 9
= 0
Question 2: Evaluate following determinants
Solution:
Given,
Cross multiplying the values inside the determinant,
|A| = (5x + 1) – (-7)x
|A| = 5x2 = 8x
Solution:
Given,
{
Solution:
Given,
∣A∣ = cos15°×cos75°+sin15°×sin75°
As per formula
cos(A−B)=cosAcosB+sinAsinB
Substitute this in |A| so we get,
∣A∣ = cos(75−15)°
∣A∣ = cos60°
∣A∣ = 0.5
Solution:
∣A∣ = (a+ib)(a−ib)−(c+id)(−c+id)
Expanding the brackets we get,
∣A∣=(a+ib)(a−ib)+(c+id)(c−id)
|A| = a2-i2b2+c2-i2d2
We know i2 = -1
|A| = a2-1b2+c2-(-1)d2
|A| = a2+b2+c2+d2
Question 3: Evaluate the following:
Solution:
In the given formula, ∣AB∣=∣A∣∣B∣
Cross multiplying the terms in |A|
∣A∣ = 2(17×12−5×20)−3(13×12−5×15)+7(13×20−15×17)
= 2(204−100)−3(156−75)+7(260−255)
= 2×104−3×81+7×5
= 208−243+45
= 0
Now ∣A∣2=∣A∣×∣A∣
∣A∣2=0
Question 4: Show that,
Solution:
Method 1:
Given,
Let the given determinant as A,
Using sin(a+B) = sinA×cosB+cosA×sinB
∣A∣ = sin10°×cos80°+cos10°×sin80°
∣A∣ = sin(10+80)°
∣A∣ = sin90°
∣A∣ = 1
Method 2:
∣A∣ = sin10°×cos80°+cos10°×sin80°
[∴cosθ = sin(90−θ)]
∣A∣ = sin10°cos(90°−10°)+cos10°sin(90°−10°)
∣A∣ = sin10°sin10°+cos10°cos10°
∣A∣ = sin210°+cos210°
[∴sin2θ+cos2θ = 1]
∣A∣ = 1
Question 5: Evaluate the following determinant by two methods.
Solution:
Method 1
Expanding along the first row
∣A∣ = 2(1×1−4×−2)−3(7×1−(−2)×−3)−5(7×4−1×(−3))
∣A∣ = 2(1+8)−3(7−6)−5(28+3)
∣A∣ = 2×9−3×1−5×31
∣A∣ = 18−3−155
∣A∣ = −140
Method 2
Here it is Sarus Method, we adjoin the first two columns.
Expanding along second column,∣A∣ = 2(1×1−4×(−2))−7(3×1−4×(−5))−3(3×(−2)−1×(−5)) ∣A∣ = 2(1+8)−7(3+20)−3(−6+5) ∣A∣ = 2×9−7×23−3×(−1) ∣A∣ = 18−161+3 ∣A∣ = −140
Question 6: Evaluate the following:
Solution:
∣A∣ = 0(0−sinβ(−sinβ))−sinα(−sinα×0−sinβcosα)−cosα((−sinα)(−sinβ)−0×cosα) ∣A∣ = 0+sinαsinβcosα−cosαsinαsinβ ∣A∣ = 0
Question 7:
Solution:
Expand C3, we have ∣A∣ = sinα(−sinαsin2β − cos2βsinα) + cosα(cosαcos2β + cosαsin2β) ∣A∣ = sin2α(sin2β + cos2β) + cos2α(cos2β + sin2β) ∣A∣ = sin2α(1) + cos2α(1) ∣A∣ = 1
Question 8: If
Solution:
Let’s take LHS,
∣AB∣ = −18−190 ∣AB∣ = −208 Now taking RHS and calculating,
∣A∣ = 2−10 ∣A∣ = −8 ∣B∣ = 20−(−6) ∣B∣ = 26 ∣A∣∣B∣ = −8×26 ∣A∣∣B∣ = −208 ∴LHS = RHS Hence, it is proved.
Question 9: If
Solution:
Evaluate along the first column,
Now every element with 3, = 3(36−0) − 0 + 0 = 108 Now, according to the question, ∣3A∣ = 27∣A∣ Substituting the values we get, 108 = 27(4) 108 = 108 Hence, proved.
Question 10: Find the values of x, if:
Solution:
2−20 = 2x2−24 −18 = 2x2−24 2x2 = 6 Taking the square root, x2 = 3 x = ±√3
Solution:
2 × 5 − 3 × 4 = 5 × x − 3 × 2x 10 − 12 = 5x − 6x −2 = −x x = 2
Solution:
3(1)−x(x) = 3(1)−2(4) 3−x2 = 3−8 −x2 = −8 x2 = 8 x = ±2√2
Solution:
3x(4)−7(2) = 10 12x−14 = 10 12x = 24 x = 24/12 x = 2
Solution:
Cross multiplying elements from LHS, (x+1)(x+2)−(x−3)(x−1) = 12+1 x2 + 3x + 2 − x2+4x − 3 = 13 7x−1 = 13 7x = 14 x = 2
Solution:
2x(x)−5(8) = 6(3)−5(8) 2x2−40 = 18−40 2x2 = 18 x2 = 9 x = ±3
Question 11: Find integral value of x, if
Solution:
Here we have to take the determinant of the 3×3 matrix x2(8−1)−x(0−3)+1(0−6) 8x2−x2+3x−6 = 28 7x2+3x−6 = 28 7x2+3x−34 = 0 Factorization of the above equation we get, (7x+17)(x−2) = 0 x = 2 Integral value of x is 2. Thus, x = −17/7 is not an integer.
Question 12: For what value of x the matrix A is singular?
Solution:
Matrix A is singular if, ∣A∣ = 0
Cross−multiply the elements in the determinant, 8 + 8x − 21 + 7x = 0 15x − 13 = 0 15x = 13 x = 13/15
Solution:
Matrix A is singular if ∣A∣=0 Expanding along first row,
∣A∣ = (x−1)[(x−1)2−1] − 1[x−1−1] + 1[1−x+1] ∣A∣ = (x−1)(x2+1−2x−1) − 1(x−2) + 1(2−x)
Expanding the brackets to factorize |A| = (x−1)(x2−2x) − x + 2 + 2 − x |A| = (x-1) × x × (x-2) + (4-2x) |A| = (x−1)× x ×(x−2) + 2(2−x) |A| = (x−1)× x ×(x−2) − 2(x−2) [∴ Take (x−2) as common] |A| = (x−2)[x(x−1)−2] Since A is a singular matrix, so ∣A∣ = 0 (x−2)(x2−x−2) = 0 There are two cases, Case1: (x−2) = 0 x = 2 Case2: x2−x−2 = 0 x2−2x + x−2 = 0 x(x−2) + 1(x−2) = 0 (x−2)(x+1) = 0 x = 2,−1 ∴ x = 2 or −1