Free BCS-012 Solved Assignment | July-2024 & January-2025 | Basic Mathematics | IGNOU

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Aspect

Details

Programme Title

BACHELOR OF COMPUTER APPLICATIONS

Course Code

BCS-012

Course Title

Basic Mathematics

Assignment Code

BCS-012

University

Indira Gandhi National Open University (IGNOU)

Type

Free IGNOU Solved Assignment 

Language

English

Session

July 2024 – January 2025

Submission Date

31st March for July session, 30th September for January session

BCS-012 Solved Assignment

Q1: For what value of ‘ k k kkk ‘ the points ( k + 1 , 2 k ) , ( k , 2 2 k ) ( k + 1 , 2 k ) , ( k , 2 2 k ) (-k+1,2k),(k,2-2k)(-k+1,2 k),(k, 2-2 k)(k+1,2k),(k,22k) and ( 4 k , 6 2 k ) ( 4 k , 6 2 k ) (-4-k,6-2k)(-4-k, 6-2 k)(4k,62k) are collinear.
Q2: Solve the following system of equations by using Matrix Inverse Method.
3 x + 4 y + 7 z = 14 2 x y + 3 z = 4 2 x + 2 y 3 z = 0 3 x + 4 y + 7 z = 14 2 x y + 3 z = 4 2 x + 2 y 3 z = 0 {:[3x+4y+7z=14],[2x-y+3z=4],[2x+2y-3z=0]:}\begin{aligned} & 3 x+4 y+7 z=14 \\ & 2 x-y+3 z=4 \\ & 2 x+2 y-3 z=0 \end{aligned}3x+4y+7z=142xy+3z=42x+2y3z=0
Q3: Use principle of Mathematical Induction to prove that:
1 1 2 + 1 2 3 + + 1 n ( n + 1 ) = n n + 1 1 1 2 + 1 2 3 + + 1 n ( n + 1 ) = n n + 1 (1)/(1**2)+(1)/(2**3)+dots dots dots dots+(1)/(n(n+1))=(n)/(n+1)\frac{1}{1 * 2}+\frac{1}{2 * 3}+\ldots \ldots \ldots \ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}112+123++1n(n+1)=nn+1
Q4: How many terms of the sequence 3 , 3 , 3 3 , 3 , 3 , 3 3 , sqrt3,3,3sqrt3,dots\sqrt{3}, 3,3 \sqrt{3}, \ldots3,3,33, must be taken to get the sum 39 + 13 3 39 + 13 3 39+13sqrt339+13 \sqrt{3}39+133 ?
Q5: If y = a e m x + b e m x y = a e m x + b e m x y=ae^(mx)+be^(-mx)y=a e^{m x}+b e^{-m x}y=aemx+bemx, Prove that d 2 y / d x 2 = m 2 y d 2 y / d x 2 = m 2 y d^(2)y//dx^(2)=m^(2)yd^2 y / d x^2=m^2 yd2y/dx2=m2y
Q6: Integrate function f ( x ) = x / [ ( x + 1 ) ( 2 x 1 ) ] f ( x ) = x / [ ( x + 1 ) ( 2 x 1 ) ] f(x)=x//[(x+1)(2x-1)]f(x)=x /[(x+1)(2 x-1)]f(x)=x/[(x+1)(2x1)] w.r.t x x xxx
Q7: If 1, w , w 2 w , w 2 w,w^(2)w, w^2w,w2 are Cube Roots of unity show that ( 1 + w ) 2 ( 1 + w ) 3 + w 2 = 0 ( 1 + w ) 2 ( 1 + w ) 3 + w 2 = 0 (1+w)^(2)-(1+w)^(3)+w^(2)=0(1+w)^2-(1+w)^3+w^2=0(1+w)2(1+w)3+w2=0.
Q8: If α , β α , β alpha,beta\alpha, \betaα,β are roots of equation 2 x 2 3 x 5 = 0 2 x 2 3 x 5 = 0 2x^(2)-3x-5=02 x^2-3 x-5=02x23x5=0, them find a Quadratic equation whose roots are α 2 , β 2 α 2 , β 2 alpha^(2),beta^(2)\alpha^2, \beta^2α2,β2
Q9: Solve the inequality 3 5 ( x 2 ) 5 3 ( 2 x ) 3 5 ( x 2 ) 5 3 ( 2 x ) (3)/(5)(x-2) <= (5)/(3)(2-x)\frac{3}{5}(x-2) \leq \frac{5}{3}(2-x)35(x2)53(2x) and graph the solution set.
Q10: If a positive number exceeds its positive square root by 12 , then find the number.
Q11: Find the area bounded by the curves x 2 = y x 2 = y x^(2)=y\mathrm{x}^2=\mathrm{y}x2=y and y = x y = x y=xy=xy=x. Q12: Find the inverse of the matrix A = ( 1 6 4 2 4 1 1 2 5 ) A = 1 6 4 2 4 1 1 2 5 A=([1,6,4],[2,4,-1],[-1,2,5])A=\left(\begin{array}{ccc}1 & 6 & 4 \\ 2 & 4 & -1 \\ -1 & 2 & 5\end{array}\right)A=(164241125), if it exists,
Q13: If m m mmm times the m th m th m^(“th “)m^{\text {th }}mth term of an A.P. is n n nnn times its n th n th n^(“th “)n^{\text {th }}nth term, show that ( m + n ) th ( m + n ) th (m+n)^(“th “)(m+n)^{\text {th }}(m+n)th term of the A.P. is zero.
Q14: Show that
i) lim n 0 | x | x lim n 0 | x | x lim_(n rarr0)(|x|)/(x)\lim _{n \rightarrow 0} \frac{|x|}{x}limn0|x|x does not exist
ii) f ( x ) = | x | f ( x ) = | x | f(x)=|x|\mathrm{f}(x)=|x|f(x)=|x| is continuous at x = 0 x = 0 x=0x=0x=0.
Q15: Suriti wants to Invest at most 12000 in saving certificates and National Saving Bonds. She has to invest at least 2000 in Saving certificates and at least 4000 in National Saving Bonds. If Rate of
Interest on saving certificates is 8 % 8 % 8%8 \%8% per annum and rate of interest on national saving bond is 10 % 10 % 10%10 \%10% per annum. How much money should she invest to earn maximum yearly income? Find also the maximum yearly income.
Q16: A spherical balloon is being Inflated at the rate of 900 cm 3 / sec 900 cm 3 / sec 900cm^(3)//sec900 \mathrm{~cm}^3 / \mathrm{sec}900 cm3/sec. How fast is the Radius of the balloon Increasing when the Radius is 15 cm .

Expert Answer

BCS-012 Solved Assignment

Question:-01

For what value of ‘ k k kkk ‘ the points ( k + 1 , 2 k ) , ( k , 2 2 k ) ( k + 1 , 2 k ) , ( k , 2 2 k ) (-k+1,2k),(k,2-2k)(-k+1,2 k),(k, 2-2 k)(k+1,2k),(k,22k) and ( 4 k , 6 2 k ) ( 4 k , 6 2 k ) (-4-k,6-2k)(-4-k, 6-2 k)(4k,62k) are collinear.

Answer:

To determine the value of k k kkk such that the points ( k + 1 , 2 k ) ( k + 1 , 2 k ) (-k+1,2k)(-k+1, 2k)(k+1,2k), ( k , 2 2 k ) ( k , 2 2 k ) (k,2-2k)(k, 2 – 2k)(k,22k), and ( 4 k , 6 2 k ) ( 4 k , 6 2 k ) (-4-k,6-2k)(-4 – k, 6 – 2k)(4k,62k) are collinear, we need to use the condition that the area of the triangle formed by three collinear points is zero.
The formula for the area of a triangle formed by three points ( x 1 , y 1 ) ( x 1 , y 1 ) (x_(1),y_(1))(x_1, y_1)(x1,y1), ( x 2 , y 2 ) ( x 2 , y 2 ) (x_(2),y_(2))(x_2, y_2)(x2,y2), and ( x 3 , y 3 ) ( x 3 , y 3 ) (x_(3),y_(3))(x_3, y_3)(x3,y3) is:
Area = 1 2 | x 1 ( y 2 y 3 ) + x 2 ( y 3 y 1 ) + x 3 ( y 1 y 2 ) | Area = 1 2 x 1 ( y 2 y 3 ) + x 2 ( y 3 y 1 ) + x 3 ( y 1 y 2 ) “Area”=(1)/(2)|x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))|\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|Area=12|x1(y2y3)+x2(y3y1)+x3(y1y2)|
For the given points ( k + 1 , 2 k ) ( k + 1 , 2 k ) (-k+1,2k)(-k + 1, 2k)(k+1,2k), ( k , 2 2 k ) ( k , 2 2 k ) (k,2-2k)(k, 2 – 2k)(k,22k), and ( 4 k , 6 2 k ) ( 4 k , 6 2 k ) (-4-k,6-2k)(-4 – k, 6 – 2k)(4k,62k), we can substitute their coordinates into this formula and set the area equal to zero.

Step 1: Write the coordinates

Let the three points be:
  • ( x 1 , y 1 ) = ( k + 1 , 2 k ) ( x 1 , y 1 ) = ( k + 1 , 2 k ) (x_(1),y_(1))=(-k+1,2k)(x_1, y_1) = (-k + 1, 2k)(x1,y1)=(k+1,2k)
  • ( x 2 , y 2 ) = ( k , 2 2 k ) ( x 2 , y 2 ) = ( k , 2 2 k ) (x_(2),y_(2))=(k,2-2k)(x_2, y_2) = (k, 2 – 2k)(x2,y2)=(k,22k)
  • ( x 3 , y 3 ) = ( 4 k , 6 2 k ) ( x 3 , y 3 ) = ( 4 k , 6 2 k ) (x_(3),y_(3))=(-4-k,6-2k)(x_3, y_3) = (-4 – k, 6 – 2k)(x3,y3)=(4k,62k)

Step 2: Substitute into the area formula

The area formula becomes:
Area = 1 2 | ( k + 1 ) ( ( 2 2 k ) ( 6 2 k ) ) + k ( ( 6 2 k ) 2 k ) + ( 4 k ) ( 2 k ( 2 2 k ) ) | = 0 Area = 1 2 ( k + 1 ) ( ( 2 2 k ) ( 6 2 k ) ) + k ( ( 6 2 k ) 2 k ) + ( 4 k ) ( 2 k ( 2 2 k ) ) = 0 “Area”=(1)/(2)|(-k+1)((2-2k)-(6-2k))+k((6-2k)-2k)+(-4-k)(2k-(2-2k))|=0\text{Area} = \frac{1}{2} \left| (-k+1)((2-2k) – (6-2k)) + k((6-2k) – 2k) + (-4-k)(2k – (2 – 2k)) \right| = 0Area=12|(k+1)((22k)(62k))+k((62k)2k)+(4k)(2k(22k))|=0
Simplify the expression step by step.

Step 3: Simplify each term

  1. ( 2 2 k ) ( 6 2 k ) = 2 2 k 6 + 2 k = 4 ( 2 2 k ) ( 6 2 k ) = 2 2 k 6 + 2 k = 4 (2-2k)-(6-2k)=2-2k-6+2k=-4(2 – 2k) – (6 – 2k) = 2 – 2k – 6 + 2k = -4(22k)(62k)=22k6+2k=4
    So, the first term becomes ( k + 1 ) ( 4 ) = 4 ( k 1 ) ( k + 1 ) ( 4 ) = 4 ( k 1 ) (-k+1)(-4)=4(k-1)(-k + 1)(-4) = 4(k – 1)(k+1)(4)=4(k1).
  2. ( 6 2 k ) 2 k = 6 2 k 2 k = 6 4 k ( 6 2 k ) 2 k = 6 2 k 2 k = 6 4 k (6-2k)-2k=6-2k-2k=6-4k(6 – 2k) – 2k = 6 – 2k – 2k = 6 – 4k(62k)2k=62k2k=64k
    So, the second term becomes k ( 6 4 k ) = 6 k 4 k 2 k ( 6 4 k ) = 6 k 4 k 2 k(6-4k)=6k-4k^(2)k(6 – 4k) = 6k – 4k^2k(64k)=6k4k2.
  3. 2 k ( 2 2 k ) = 2 k 2 + 2 k = 4 k 2 2 k ( 2 2 k ) = 2 k 2 + 2 k = 4 k 2 2k-(2-2k)=2k-2+2k=4k-22k – (2 – 2k) = 2k – 2 + 2k = 4k – 22k(22k)=2k2+2k=4k2
    So, the third term becomes ( 4 k ) ( 4 k 2 ) = ( 4 ) ( 4 k 2 ) + ( k ) ( 4 k 2 ) = 16 k + 8 4 k 2 + 2 k ( 4 k ) ( 4 k 2 ) = ( 4 ) ( 4 k 2 ) + ( k ) ( 4 k 2 ) = 16 k + 8 4 k 2 + 2 k (-4-k)(4k-2)=(-4)(4k-2)+(-k)(4k-2)=-16 k+8-4k^(2)+2k(-4 – k)(4k – 2) = (-4)(4k – 2) + (-k)(4k – 2) = -16k + 8 – 4k^2 + 2k(4k)(4k2)=(4)(4k2)+(k)(4k2)=16k+84k2+2k.
    Simplifying the third term gives 14 k + 8 4 k 2 14 k + 8 4 k 2 -14 k+8-4k^(2)-14k + 8 – 4k^214k+84k2.

Step 4: Combine all terms

4 ( k 1 ) + ( 6 k 4 k 2 ) + ( 14 k + 8 4 k 2 ) = 0 4 ( k 1 ) + ( 6 k 4 k 2 ) + ( 14 k + 8 4 k 2 ) = 0 4(k-1)+(6k-4k^(2))+(-14 k+8-4k^(2))=04(k – 1) + (6k – 4k^2) + (-14k + 8 – 4k^2) = 04(k1)+(6k4k2)+(14k+84k2)=0
Simplifying:
4 k 4 + 6 k 4 k 2 14 k + 8 4 k 2 = 0 4 k 4 + 6 k 4 k 2 14 k + 8 4 k 2 = 0 4k-4+6k-4k^(2)-14 k+8-4k^(2)=04k – 4 + 6k – 4k^2 – 14k + 8 – 4k^2 = 04k4+6k4k214k+84k2=0
( 4 k + 6 k 14 k ) + ( 4 4 k 2 4 k 2 ) + 8 = 0 ( 4 k + 6 k 14 k ) + ( 4 4 k 2 4 k 2 ) + 8 = 0 (4k+6k-14 k)+(-4-4k^(2)-4k^(2))+8=0(4k + 6k – 14k) + (-4 – 4k^2 – 4k^2) + 8 = 0(4k+6k14k)+(44k24k2)+8=0
4 k 8 k 2 + 4 = 0 4 k 8 k 2 + 4 = 0 -4k-8k^(2)+4=0-4k – 8k^2 + 4 = 04k8k2+4=0

Step 5: Solve for k k kkk

This simplifies to:
8 k 2 4 k + 4 = 0 8 k 2 4 k + 4 = 0 -8k^(2)-4k+4=0-8k^2 – 4k + 4 = 08k24k+4=0
Dividing the entire equation by -4:
2 k 2 + k 1 = 0 2 k 2 + k 1 = 0 2k^(2)+k-1=02k^2 + k – 1 = 02k2+k1=0
This is a quadratic equation. Solve it using the quadratic formula:
k = 1 ± 1 2 4 ( 2 ) ( 1 ) 2 ( 2 ) = 1 ± 1 + 8 4 = 1 ± 9 4 k = 1 ± 1 2 4 ( 2 ) ( 1 ) 2 ( 2 ) = 1 ± 1 + 8 4 = 1 ± 9 4 k=(-1+-sqrt(1^(2)-4(2)(-1)))/(2(2))=(-1+-sqrt(1+8))/(4)=(-1+-sqrt9)/(4)k = \frac{-1 \pm \sqrt{1^2 – 4(2)(-1)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4}k=1±124(2)(1)2(2)=1±1+84=1±94
k = 1 ± 3 4 k = 1 ± 3 4 k=(-1+-3)/(4)k = \frac{-1 \pm 3}{4}k=1±34
Thus, k = 1 + 3 4 = 2 4 = 1 2 k = 1 + 3 4 = 2 4 = 1 2 k=(-1+3)/(4)=(2)/(4)=(1)/(2)k = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}k=1+34=24=12 or k = 1 3 4 = 4 4 = 1 k = 1 3 4 = 4 4 = 1 k=(-1-3)/(4)=(-4)/(4)=-1k = \frac{-1 – 3}{4} = \frac{-4}{4} = -1k=134=44=1.

Step 6: Conclusion

The value of k k kkk can be either 1 2 1 2 (1)/(2)\frac{1}{2}12 or 1 1 -1-11.

Question:-02

Solve the following system of equations by using Matrix Inverse Method.

3 x + 4 y + 7 z = 14 2 x y + 3 z = 4 2 x + 2 y 3 z = 0 3 x + 4 y + 7 z = 14 2 x y + 3 z = 4 2 x + 2 y 3 z = 0 {:[3x+4y+7z=14],[2x-y+3z=4],[2x+2y-3z=0]:}\begin{aligned} & 3x + 4y + 7z = 14 \\ & 2x – y + 3z = 4 \\ & 2x + 2y – 3z = 0 \end{aligned}3x+4y+7z=142xy+3z=42x+2y3z=0

Answer:

To solve the system of equations using the matrix inverse method, we represent the system in matrix form as A x = b A x = b Ax=bA \mathbf{x} = \mathbf{b}Ax=b, where:
A = [ 3 4 7 2 1 3 2 2 3 ] , x = [ x y z ] , b = [ 14 4 0 ] A = 3 4 7 2 1 3 2 2 3 , x = x y z , b = 14 4 0 A=[[3,4,7],[2,-1,3],[2,2,-3]],quadx=[[x],[y],[z]],quadb=[[14],[4],[0]]A = \begin{bmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 2 & 2 & -3 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 14 \\ 4 \\ 0 \end{bmatrix}A=[347213223],x=[xyz],b=[1440]

Step 1: Find the inverse of matrix A A AAA

To find the inverse of A A AAA, we first calculate its determinant.
det ( A ) = 3 ( ( 1 ) ( 3 ) 2 ( 3 ) ) 4 ( 2 ( 3 ) 2 ( 3 ) ) + 7 ( 2 ( 2 ) 2 ( 1 ) ) det ( A ) = 3 ( 1 ) ( 3 ) 2 ( 3 ) 4 2 ( 3 ) 2 ( 3 ) + 7 2 ( 2 ) 2 ( 1 ) “det”(A)=3((-1)(-3)-2(3))-4(2(-3)-2(3))+7(2(2)-2(-1))\text{det}(A) = 3 \left( (-1)(-3) – 2(3) \right) – 4 \left( 2(-3) – 2(3) \right) + 7 \left( 2(2) – 2(-1) \right)det(A)=3((1)(3)2(3))4(2(3)2(3))+7(2(2)2(1))
Simplifying:
det ( A ) = 3 ( 3 6 ) 4 ( 6 6 ) + 7 ( 4 + 2 ) det ( A ) = 3 3 6 4 6 6 + 7 4 + 2 “det”(A)=3(3-6)-4(-6-6)+7(4+2)\text{det}(A) = 3 \left( 3 – 6 \right) – 4 \left( -6 – 6 \right) + 7 \left( 4 + 2 \right)det(A)=3(36)4(66)+7(4+2)
det ( A ) = 3 ( 3 ) 4 ( 12 ) + 7 ( 6 ) det ( A ) = 3 ( 3 ) 4 ( 12 ) + 7 ( 6 ) “det”(A)=3(-3)-4(-12)+7(6)\text{det}(A) = 3(-3) – 4(-12) + 7(6)det(A)=3(3)4(12)+7(6)
det ( A ) = 9 + 48 + 42 = 81 det ( A ) = 9 + 48 + 42 = 81 “det”(A)=-9+48+42=81\text{det}(A) = -9 + 48 + 42 = 81det(A)=9+48+42=81
Since det ( A ) = 81 0 det ( A ) = 81 0 “det”(A)=81!=0\text{det}(A) = 81 \neq 0det(A)=810, matrix A A AAA is invertible.

Step 2: Find the adjoint of A A AAA

The adjoint of A A AAA, denoted as adj ( A ) adj ( A ) “adj”(A)\text{adj}(A)adj(A), is the transpose of the cofactor matrix of A A AAA. Let’s compute the cofactors of each element of A A AAA:
  • The cofactor of A 11 A 11 A_(11)A_{11}A11 (3) is:
    Cofactor ( A 11 ) = ( 1 ) 1 | 1 3 2 3 | = ( 1 ) ( ( 1 ) ( 3 ) 3 ( 2 ) ) = 3 6 = 9 Cofactor ( A 11 ) = ( 1 ) 1 1 3 2 3 = ( 1 ) ( 1 ) ( 3 ) 3 ( 2 ) = 3 6 = 9 “Cofactor”(A_(11))=(-1)^(1)*|[-1,3],[2,-3]|=(-1)*((-1)(-3)-3(2))=-3-6=-9\text{Cofactor}(A_{11}) = (-1)^1 \cdot \begin{vmatrix} -1 & 3 \\ 2 & -3 \end{vmatrix} = (-1) \cdot \left( (-1)(-3) – 3(2) \right) = -3 – 6 = -9Cofactor(A11)=(1)1|1323|=(1)((1)(3)3(2))=36=9
  • The cofactor of A 12 A 12 A_(12)A_{12}A12 (4) is:
    Cofactor ( A 12 ) = ( 1 ) 2 | 2 3 2 3 | = ( 2 ( 3 ) 3 ( 2 ) ) = 6 6 = 12 Cofactor ( A 12 ) = ( 1 ) 2 2 3 2 3 = 2 ( 3 ) 3 ( 2 ) = 6 6 = 12 “Cofactor”(A_(12))=(-1)^(2)*|[2,3],[2,-3]|=(2(-3)-3(2))=-6-6=-12\text{Cofactor}(A_{12}) = (-1)^2 \cdot \begin{vmatrix} 2 & 3 \\ 2 & -3 \end{vmatrix} = \left( 2(-3) – 3(2) \right) = -6 – 6 = -12Cofactor(A12)=(1)2|2323|=(2(3)3(2))=66=12
  • The cofactor of A 13 A 13 A_(13)A_{13}A13 (7) is:
    Cofactor ( A 13 ) = ( 1 ) 3 | 2 1 2 2 | = ( 1 ) ( 2 ( 2 ) ( 1 ) ( 2 ) ) = ( 4 + 2 ) = 6 Cofactor ( A 13 ) = ( 1 ) 3 2 1 2 2 = ( 1 ) 2 ( 2 ) ( 1 ) ( 2 ) = 4 + 2 = 6 “Cofactor”(A_(13))=(-1)^(3)*|[2,-1],[2,2]|=(-1)*(2(2)-(-1)(2))=-(4+2)=-6\text{Cofactor}(A_{13}) = (-1)^3 \cdot \begin{vmatrix} 2 & -1 \\ 2 & 2 \end{vmatrix} = (-1) \cdot \left( 2(2) – (-1)(2) \right) = – \left( 4 + 2 \right) = -6Cofactor(A13)=(1)3|2122|=(1)(2(2)(1)(2))=(4+2)=6
  • The cofactor of A 21 A 21 A_(21)A_{21}A21 (2) is:
    Cofactor ( A 21 ) = ( 1 ) 2 | 4 7 2 3 | = ( 4 ( 3 ) 7 ( 2 ) ) = 12 14 = 26 Cofactor ( A 21 ) = ( 1 ) 2 4 7 2 3 = 4 ( 3 ) 7 ( 2 ) = 12 14 = 26 “Cofactor”(A_(21))=(-1)^(2)*|[4,7],[2,-3]|=(4(-3)-7(2))=-12-14=-26\text{Cofactor}(A_{21}) = (-1)^2 \cdot \begin{vmatrix} 4 & 7 \\ 2 & -3 \end{vmatrix} = \left( 4(-3) – 7(2) \right) = -12 – 14 = -26Cofactor(A21)=(1)2|4723|=(4(3)7(2))=1214=26
  • The cofactor of A 22 A 22 A_(22)A_{22}A22 (-1) is:
    Cofactor ( A 22 ) = ( 1 ) 3 | 3 7 2 3 | = ( 1 ) ( 3 ( 3 ) 7 ( 2 ) ) = ( 9 14 ) = 23 Cofactor ( A 22 ) = ( 1 ) 3 3 7 2 3 = ( 1 ) 3 ( 3 ) 7 ( 2 ) = ( 9 14 ) = 23 “Cofactor”(A_(22))=(-1)^(3)*|[3,7],[2,-3]|=(-1)*(3(-3)-7(2))=-(-9-14)=23\text{Cofactor}(A_{22}) = (-1)^3 \cdot \begin{vmatrix} 3 & 7 \\ 2 & -3 \end{vmatrix} = (-1) \cdot \left( 3(-3) – 7(2) \right) = -(-9 – 14) = 23Cofactor(A22)=(1)3|3723|=(1)(3(3)7(2))=(914)=23
  • The cofactor of A 23 A 23 A_(23)A_{23}A23 (3) is:
    Cofactor ( A 23 ) = ( 1 ) 4 | 3 4 2 2 | = ( 3 ( 2 ) 4 ( 2 ) ) = 6 8 = 2 Cofactor ( A 23 ) = ( 1 ) 4 3 4 2 2 = 3 ( 2 ) 4 ( 2 ) = 6 8 = 2 “Cofactor”(A_(23))=(-1)^(4)*|[3,4],[2,2]|=(3(2)-4(2))=6-8=-2\text{Cofactor}(A_{23}) = (-1)^4 \cdot \begin{vmatrix} 3 & 4 \\ 2 & 2 \end{vmatrix} = \left( 3(2) – 4(2) \right) = 6 – 8 = -2