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In this article, we will be going to solve the entire exercise 6.1 of our RD Sharma textbook.
In linear algebra, the determinant is a scalar value that can be computed from the elements of a square matrix. It provides important information about the matrix and the linear transformations it represents. Determinants have applications in various fields, including solving systems of linear equations, analyzing matrix invertibility, and finding eigenvalues.
The determinant of a square matrix A is denoted as or |A|. For a 2x2 matrix, the determinant is calculated as:
Read More: Linear Algebra
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 2x2 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 2x2 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 = -1x2 – 5x2
M11 = -12
M21 = -3x2 – 5x2
M21 = -16
M31 = -3x2 – (-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) + 4x16 + 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 = 5x1 – 7x0
M11 = 5
M21 = 2x1 – 7x6
M21 = -40
M31 = 2x0 – 5x6
M31 = -30
Co-factors of the determinant are as follows,
C11 = (-1)1+1 x M11
= 1x5
= 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 + 1x40 + 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
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
Solution:
Inthegivenformula, ∣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
Solution:
Method 1:
Given,
LetthegivendeterminantasA,
Usingsin(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
Solution:
Method 1
Expandingalongthefirstrow
∣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
Solution:
∣A∣ = 0(0−sinβ(−sinβ))−sinα(−sinα×0−sinβcosα)−cosα((−sinα)(−sinβ)−0×cosα) ∣A∣ = 0+sinαsinβcosα−cosαsinαsinβ ∣A∣ = 0
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
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.
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.
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
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.
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
1).Evaluate the determinant:
|3 -1|
|2 4|
2).If |a b| = 6, find the value of |2a 2b|
|c d| |2c 2d|
3).Solve for x:
|x+1 3| = 10
| 2 5|
4).Calculate the determinant:
| 1 2 -1|
|-2 0 3|
| 4 1 2|
5).If |a b| = 3 and |c d| = 4, find |ac bd|
|c d| |a b| |ca db|
6).Prove that |a+p b+p| = |a b|
|c+p d+p| |c d|
7).Evaluate:
|cos θ -sin θ|
|sin θ cos θ|
8).Find the value of k for which the following determinant is zero:
|k 2|
|3 k-1|
9).If |a b| = 5, find the value of |a² ab|
|c d| |ac bd|
10).Solve the system of equations using determinants:
2x + 3y = 7
4x - y = 5
Chapter 6 of RD Sharma Class 12 introduces determinants, which are numerical values associated with square matrices. Key points include:
