BECC-104 Solved Assignment

July 2023-January 2024

1. Consider the following two matrices
   $A=\left(\begin{array}{ccc}-1& 1& 2\\ 1& -1& -2\\ -2& 2& 4\end{array}\right)$$A=\left(\begin{array}{ccc}-1& 1& 2\\ 1& -1& -2\\ -2& 2& 4\end{array}\right)$A=([-1,1,2],[1,-1,-2],[-2,2,4])A=\left(\begin{array}{ccc}-1 & 1 & 2 \\ 1 & -1 & -2 \\ -2 & 2 & 4\end{array}\right)$A=\left(\begin{array}{ccc}-1& 1& 2\\ 1& -1& -2\\ -2& 2& 4\end{array}\right)$
   $\mathrm{B}=\left(\begin{array}{ccc}1& 0& -1\\ -1& 1& 1\\ 1& -1& -1\end{array}\right)$$\mathrm{B}=\left(\begin{array}{ccc}1& 0& -1\\ -1& 1& 1\\ 1& -1& -1\end{array}\right)$B=([1,0,-1],[-1,1,1],[1,-1,-1])\mathrm{B}=\left(\begin{array}{ccc}1 & 0 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & -1\end{array}\right)$\mathrm{B}=\left(\begin{array}{ccc}1& 0& -1\\ -1& 1& 1\\ 1& -1& -1\end{array}\right)$
   (i) Find the rank of ‘ A ‘ and ‘ B ‘
   (ii) Show that $(\mathrm{AB}{)}^{-1}={\mathrm{B}}^{-1}{\mathrm{A}}^{-1}$$(\mathrm{AB}{)}^{-1}={\mathrm{B}}^{-1}{\mathrm{A}}^{-1}$(AB)^(-1)=B^(-1)A^(-1)(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}$(\mathrm{AB}{)}^{-1}={\mathrm{B}}^{-1}{\mathrm{A}}^{-1}$
   (iii) Show that ${\left({\mathrm{A}}^{-1}\right)}^{-1}=\mathrm{A}$${\left({\mathrm{A}}^{-1}\right)}^{-1}=\mathrm{A}$(A^(-1))^(-1)=A\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}${\left({\mathrm{A}}^{-1}\right)}^{-1}=\mathrm{A}$
   (iv) Show that ${\left({\mathrm{B}}^{-1}\right)}^{-1}=\mathrm{B}$${\left({\mathrm{B}}^{-1}\right)}^{-1}=\mathrm{B}$(B^(-1))^(-1)=B\left(\mathrm{B}^{-1}\right)^{-1}=\mathrm{B}${\left({\mathrm{B}}^{-1}\right)}^{-1}=\mathrm{B}$
2. An individual consumer consumes two commodities $X_1\mathrm{\&}{X}_2$$X_1\mathrm{\&}{X}_2$X_(1)&X_(2)X_1 \& X_2$X_1\mathrm{\&}{X}_2$. The utility function is

3. Let $Z=f(x,y)=3x^3-5y^2-225x+70y+23$$Z=f(x,y)=3x^3-5y^2-225x+70y+23$Z=f(x,y)=3x^(3)-5y^(2)-225 x+70 y+23Z=f(x, y)=3 x^3-5 y^2-225 x+70 y+23$Z=f(x,y)=3x^3-5y^2-225x+70y+23$.
   (i) Find the stationary points of $z$$z$zz$z$.
   (ii) Determine if at these points the function is at a relative maximum, relative minimum, infixion point, or saddle point.
4. Solve the following differential equation

5. If $Z=f(x,y)=xy$$Z=f(x,y)=xy$Z=f(x,y)=xyZ=f(x, y)=x y$Z=f(x,y)=xy$

9. Let the production function by $Q=A{L}^a{K}^b$$Q=A{L}^a{K}^b$Q=AL^(a)K^(b)Q=A L^a K^b$Q=A{L}^a{K}^b$. Find the elasticity of production with respect to labour (L).
10. Denote by $\mathbf{a},\mathbf{b}$$\mathbf{a},\mathbf{b}$a,b\mathbf{a}, \mathbf{b}$\mathbf{a},\mathbf{b}$ and $\mathbf{c}$$\mathbf{c}$c\mathbf{c}$\mathbf{c}$ the column vectors

Expert Answer:

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Answer:

Answer:

1. Set up the utility maximization problem.
2. Formulate the budget constraint.
3. Use the Lagrangian method to find the optimal consumption bundle.
4. Derive the demand curves for $X_1$$X_1$X_(1)X_1$X_1$ and $X_2$$X_2$X_(2)X_2$X_2$.

1. Utility Function and Budget Constraint

• Price of $X_1$$X_1$X_(1)X_1$X_1$: $P_1=3$$P_1=3$P_(1)=3P_1 = 3$P_1=3$
• Price of $X_2$$X_2$X_(2)X_2$X_2$: $P_2=4$$P_2=4$P_(2)=4P_2 = 4$P_2=4$
• Income: $I=108$$I=108$I=108I = 108$I=108$

2. Lagrangian Function

3. First-Order Conditions

4. Solve for $X_1$$X_1$X_(1)X_1$X_1$ and $X_2$$X_2$X_(2)X_2$X_2$

Utility-Maximizing Quantities

• $X_1=14.4$$X_1=14.4$X_(1)=14.4X_1 = 14.4$X_1=14.4$
• $X_2=16.2$$X_2=16.2$X_(2)=16.2X_2 = 16.2$X_2=16.2$

5. Derive the Demand Curves

• Demand for $X_1$$X_1$X_(1)X_1$X_1$: $X_1=\frac{43.2}{P_1}$$X_1=\frac{43.2}{P_1}$X_(1)=(43.2)/(P_(1))X_1 = \frac{43.2}{P_1}$X_1=\frac{43.2}{P_1}$
• Demand for $X_2$$X_2$X_(2)X_2$X_2$: $X_2=\frac{64.8}{P_2}$$X_2=\frac{64.8}{P_2}$X_(2)=(64.8)/(P_(2))X_2 = \frac{64.8}{P_2}$X_2=\frac{64.8}{P_2}$

Answer:

1. Find the stationary points by setting the first partial derivatives equal to zero.
2. Classify these points using the second derivative test.

(i) Finding the Stationary Points

Step 1: Compute the First Partial Derivatives

Step 2: Set the Partial Derivatives to Zero

1. $$9x^2-225=0$$$$9x^2-225=0$$9x^(2)-225=09x^2 – 225 = 0$$9x^2-225=0$$
2. $$-10y+70=0$$$$-10y+70=0$$-10 y+70=0-10y + 70 = 0$$-10y+70=0$$

Stationary Points

(ii) Classifying the Stationary Points

Step 1: Compute the Second Partial Derivatives

Step 2: Calculate the Hessian Determinant

[
H = $$\left[\begin{array}{cc}\frac{{\partial }^{2}z}{\partial x^2}& \frac{{\partial }^{2}z}{\partial x\partial y}\\ \frac{{\partial }^{2}z}{\partial x\partial y}& \frac{{\partial }^{2}z}{\partial y^2}\end{array}\right]$$$$\left[\begin{array}{cc}\frac{{\partial }^{2}z}{\partial x^2}& \frac{{\partial }^{2}z}{\partial x\partial y}\\ \frac{{\partial }^{2}z}{\partial x\partial y}& \frac{{\partial }^{2}z}{\partial y^2}\end{array}\right]$$[[(del^(2)z)/(delx^(2)),(del^(2)z)/(del x del y)],[(del^(2)z)/(del x del y),(del^(2)z)/(dely^(2))]]\begin{bmatrix} \frac{\partial^2 z}{\partial x^2} & \frac{\partial^2 z}{\partial x \partial y} \\ \frac{\partial^2 z}{\partial x \partial y} & \frac{\partial^2 z}{\partial y^2} \end{bmatrix}$$\left[\begin{array}{cc}\frac{{\partial }^{2}z}{\partial x^2}& \frac{{\partial }^{2}z}{\partial x\partial y}\\ \frac{{\partial }^{2}z}{\partial x\partial y}& \frac{{\partial }^{2}z}{\partial y^2}\end{array}\right]$$

Step 3: Evaluate the Hessian at the Stationary Points

Summary of Results

• The stationary points are $(5,7)$$(5,7)$(5,7)(5, 7)$(5,7)$ and $(-5,7)$$(-5,7)$(-5,7)(-5, 7)$(-5,7)$.
• $(5,7)$$(5,7)$(5,7)(5, 7)$(5,7)$ is a saddle point.
• $(-5,7)$$(-5,7)$(-5,7)(-5, 7)$(-5,7)$ is a relative maximum.

Answer:

1. Find the general solution to the homogeneous differential equation.
2. Apply the initial conditions to determine the constants.

Step 1: Solve the Characteristic Equation

Step 2: Write the General Solution

Step 3: Apply the Initial Conditions

1. Initial condition: $y(0)=4$$y(0)=4$y(0)=4y(0) = 4$y(0)=4$

2. Initial condition: $\frac{dy}{dx}(0)=1$$\frac{dy}{dx}(0)=1$(dy)/(dx)(0)=1\frac{dy}{dx}(0) = 1$\frac{dy}{dx}(0)=1$

Step 4: Write the Final Solution

Final Answer

Answer:

1. By Substitution
2. By Using the Lagrangian Multiplier Method

1. Method of Substitution

2. Lagrangian Multiplier Method

Conclusion

Answer:

Definition of the Adjugate of a Matrix

Steps to Find the Adjugate of a Matrix:

1. Cofactor Matrix: Compute the cofactor matrix $C$$C$CC$C$ of $A$$A$AA$A$. The cofactor $C_{ij}$$C_{ij}$C_(ij)C_{ij}$C_{ij}$ of an element $a_{ij}$$a_{ij}$a_(ij)a_{ij}$a_{ij}$ in $A$$A$AA$A$ is given by:
   $$C_{ij}=(-1{)}^{i+j}{M}_{ij}$$$$C_{ij}=(-1{)}^{i+j}{M}_{ij}$$C_(ij)=(-1)^(i+j)M_(ij)C_{ij} = (-1)^{i+j} M_{ij}$$C_{ij}=(-1{)}^{i+j}{M}_{ij}$$
   where $M_{ij}$$M_{ij}$M_(ij)M_{ij}$M_{ij}$ is the minor of the element $a_{ij}$$a_{ij}$a_(ij)a_{ij}$a_{ij}$. The minor $M_{ij}$$M_{ij}$M_(ij)M_{ij}$M_{ij}$ is the determinant of the $(n-1)\times (n-1)$$(n-1)\times (n-1)$(n-1)xx(n-1)(n-1) \times (n-1)$(n-1)\times (n-1)$ matrix that results from deleting the $i$$i$ii$i$-th row and $j$$j$jj$j$-th column from $A$$A$AA$A$.
2. Transpose of the Cofactor Matrix: Once all cofactors $C_{ij}$$C_{ij}$C_(ij)C_{ij}$C_{ij}$ are calculated and arranged into a matrix $C$$C$CC$C$, take the transpose of this matrix to get the adjugate matrix $\text{adj}(A)$$\text{adj}(A)$“adj”(A)\text{adj}(A)$\text{adj}(A)$:
   $$\text{adj}(A)=C^T$$$$\text{adj}(A)=C^T$$“adj”(A)=C^(T)\text{adj}(A) = C^T$$\text{adj}(A)=C^T$$

Example

1. Cofactor Matrix $C$$C$CC$C$:
   • The cofactor of $a_{11}=a$$a_{11}=a$a_(11)=aa_{11} = a$a_{11}=a$ is $C_{11}=d$$C_{11}=d$C_(11)=dC_{11} = d$C_{11}=d$ (minor is $d$$d$dd$d$, no sign change as $(-1{)}^{1+1}=1$$(-1{)}^{1+1}=1$(-1)^(1+1)=1(-1)^{1+1} = 1$(-1{)}^{1+1}=1$).
   • The cofactor of $a_{12}=b$$a_{12}=b$a_(12)=ba_{12} = b$a_{12}=b$ is $C_{12}=-c$$C_{12}=-c$C_(12)=-cC_{12} = -c$C_{12}=-c$ (minor is $c$$c$cc$c$, sign change as $(-1{)}^{1+2}=-1$$(-1{)}^{1+2}=-1$(-1)^(1+2)=-1(-1)^{1+2} = -1$(-1{)}^{1+2}=-1$).
   • The cofactor of $a_{21}=c$$a_{21}=c$a_(21)=ca_{21} = c$a_{21}=c$ is $C_{21}=-b$$C_{21}=-b$C_(21)=-bC_{21} = -b$C_{21}=-b$ (minor is $b$$b$bb$b$, sign change as $(-1{)}^{2+1}=-1$$(-1{)}^{2+1}=-1$(-1)^(2+1)=-1(-1)^{2+1} = -1$(-1{)}^{2+1}=-1$).
   • The cofactor of $a_{22}=d$$a_{22}=d$a_(22)=da_{22} = d$a_{22}=d$ is $C_{22}=a$$C_{22}=a$C_(22)=aC_{22} = a$C_{22}=a$ (minor is $a$$a$aa$a$, no sign change as $(-1{)}^{2+2}=1$$(-1{)}^{2+2}=1$(-1)^(2+2)=1(-1)^{2+2} = 1$(-1{)}^{2+2}=1$).
   So, the cofactor matrix $C$$C$CC$C$ is:
   $$C=\left[\begin{array}{cc}d& -c\\ -b& a\end{array}\right]$$$$C=\left[\begin{array}{cc}d& -c\\ -b& a\end{array}\right]$$C=[[d,-c],[-b,a]]C = \begin{bmatrix} d & -c \\ -b & a \end{bmatrix}$$C=\left[\begin{array}{cc}d& -c\\ -b& a\end{array}\right]$$
2. Adjugate Matrix $\text{adj}(A)$$\text{adj}(A)$“adj”(A)\text{adj}(A)$\text{adj}(A)$:
   The adjugate of $A$$A$AA$A$ is the transpose of $C$$C$CC$C$:
   $$\text{adj}(A)=C^T=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$$$\text{adj}(A)=C^T=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$“adj”(A)=C^(T)=[[d,-b],[-c,a]]\text{adj}(A) = C^T = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$\text{adj}(A)=C^T=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$

Importance of the Adjugate

Answer:

Formal Definition

• $B$$B$BB$B$ and $D$$D$DD$D$ are square matrices (of sizes $k\times k$$k\times k$k xx kk \times k$k\times k$ and $(n-k)\times (n-k)$$(n-k)\times (n-k)$(n-k)xx(n-k)(n-k) \times (n-k)$(n-k)\times (n-k)$ respectively, for some $0<k<n$$0<k<n$0 < k < n0 < k < n$0<k<n$),
• $C$$C$CC$C$ is a matrix (of size $k\times (n-k)$$k\times (n-k)$k xx(n-k)k \times (n-k)$k\times (n-k)$),
• The zero block is a matrix of zeros (of size $(n-k)\times k$$(n-k)\times k$(n-k)xx k(n-k) \times k$(n-k)\times k$).

Key Points

• Permutation Matrix: A permutation matrix is a square binary matrix (containing only 0s and 1s) that has exactly one entry of 1 in each row and each column, with all other entries being 0. It represents a reordering of the rows or columns of the identity matrix.
• Block Upper Triangular Form: This form has all elements below the main diagonal blocks as zeros, which indicates that the matrix can be divided into independent submatrices. This is significant in many computational and theoretical contexts, such as solving systems of linear equations or in analyzing the structure of matrices.

Example

Applications

• Graph Theory: In graph theory, a matrix is decomposable if the corresponding directed graph is not strongly connected. The concept of decomposability is related to finding strongly connected components in graphs.
• Linear Algebra: Decomposable matrices are useful in simplifying the solution of linear systems, eigenvalue problems, and other matrix-related computations by breaking down complex matrices into simpler components.

Answer:

Characteristics of a Singular Matrix

1. Determinant is Zero: The most straightforward criterion for a matrix to be singular is that its determinant is zero. For a square matrix $A$$A$AA$A$, if $det(A)=0$$det(A)=0$det(A)=0\det(A) = 0$det(A)=0$, then $A$$A$AA$A$ is singular. Conversely, if $det(A)\ne 0$$det(A)\ne 0$det(A)!=0\det(A) \neq 0$det(A)\ne 0$, the matrix is said to be nonsingular (or invertible).
2. No Inverse: A singular matrix does not have an inverse matrix. If you attempt to compute the inverse of a singular matrix using standard algorithms (like Gaussian elimination), you will not be able to complete the operation because of a division by zero or the inability to achieve row reduction to an identity matrix.
3. Linearly Dependent Rows or Columns: A matrix is singular if its rows or columns are linearly dependent. This means that at least one row (or column) can be expressed as a linear combination of the other rows (or columns). When this happens, the matrix loses full rank, and its determinant becomes zero.
4. Rank Deficiency: The rank of a matrix is the maximum number of linearly independent rows or columns. A square matrix $A$$A$AA$A$ of size $n\times n$$n\times n$n xx nn \times n$n\times n$ is singular if its rank is less than $n$$n$nn$n$. A nonsingular matrix, by contrast, has full rank $n$$n$nn$n$.

Example of a Singular Matrix

Implications of a Singular Matrix

• Systems of Linear Equations: A singular matrix implies that a system of linear equations represented by this matrix has either no solutions or infinitely many solutions. This is because there isn’t a unique solution when the determinant is zero.
• Computational Issues: In numerical computations, singular matrices can cause problems because algorithms that depend on matrix inversion will fail. Techniques such as regularization or pseudo-inverse methods (like the Moore-Penrose inverse) are used to handle singular or nearly singular matrices.

Answer:

Step 1: Substitution

Step 2: Distribute and Integrate

Step 3: Substitute Back in Terms of $x$$x$xx$x$

Final Answer

Answer:

Understanding the Maximum Value of a Function

1. Definition:
   • The maximum value of a function $f(x)$$f(x)$f(x)f(x)$f(x)$ is the highest output value that the function achieves within a particular domain (a specified set of input values).
2. Local vs. Global Maximum:
   • Local Maximum: A point $x=c$$x=c$x=cx = c$x=c$ is a local maximum if, within some small neighborhood around $c$$c$cc$c$, $f(c)$$f(c)$f(c)f(c)$f(c)$ is greater than or equal to all other function values. Formally, $f(c)$$f(c)$f(c)f(c)$f(c)$ is a local maximum if there exists an interval $(a,b)$$(a,b)$(a,b)(a, b)$(a,b)$ containing $c$$c$cc$c$ such that $f(x)\le f(c)$$f(x)\le f(c)$f(x) <= f(c)f(x) \leq f(c)$f(x)\le f(c)$ for all $x\in (a,b)$$x\in (a,b)$x in(a,b)x \in (a, b)$x\in (a,b)$.
   • Global Maximum: A point $x=c$$x=c$x=cx = c$x=c$ is a global (or absolute) maximum if $f(c)$$f(c)$f(c)f(c)$f(c)$ is greater than or equal to all other function values over the entire domain of $f(x)$$f(x)$f(x)f(x)$f(x)$. Formally, $f(c)$$f(c)$f(c)f(c)$f(c)$ is a global maximum if $f(x)\le f(c)$$f(x)\le f(c)$f(x) <= f(c)f(x) \leq f(c)$f(x)\le f(c)$ for all $x$$x$xx$x$ in the domain of $f$$f$ff$f$.
3. Finding Maximum Values:
   • To find the maximum value of a function, one can use various methods depending on the nature of the function:
     • Calculus (Derivative Test): For a differentiable function, you find the critical points by setting the first derivative $f^{\prime }(x)$$f^{\prime }(x)$f^(‘)(x)f'(x)$f^{\prime }(x)$ equal to zero or where $f^{\prime }(x)$$f^{\prime }(x)$f^(‘)(x)f'(x)$f^{\prime }(x)$ does not exist. The critical points are then tested using the second derivative test or first derivative test to determine if they are maxima.
     • Closed Interval Method: If the function is defined on a closed interval $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$, evaluate the function at critical points and endpoints $a$$a$aa$a$ and $b$$b$bb$b$. The greatest value among these is the maximum value of the function on that interval.
     • Graphical Analysis: Sometimes visualizing the function can help identify maximum points, especially for more intuitive understanding or for functions that are not easily differentiable.
4. Applications:
   • Optimization Problems: Finding maximum values is central to optimization problems where one seeks to maximize profit, efficiency, yield, or other desirable outcomes.
   • Economics: In economics, the maximum value of a utility function, production function, or profit function represents the optimal point where a firm or consumer achieves the highest satisfaction, output, or profit.
   • Physics and Engineering: Maximum values can represent peak stress points, maximum energy states, or other critical conditions in physical systems.

Examples

1. Quadratic Function:
   Consider $f(x)=-x^2+4x+1$$f(x)=-x^2+4x+1$f(x)=-x^(2)+4x+1f(x) = -x^2 + 4x + 1$f(x)=-x^2+4x+1$. To find its maximum value:
   • The first derivative is $f^{\prime }(x)=-2x+4$$f^{\prime }(x)=-2x+4$f^(‘)(x)=-2x+4f'(x) = -2x + 4$f^{\prime }(x)=-2x+4$.
   • Set $f^{\prime }(x)=0$$f^{\prime }(x)=0$f^(‘)(x)=0f'(x) = 0$f^{\prime }(x)=0$ to find the critical point: $-2x+4=0$$-2x+4=0$-2x+4=0-2x + 4 = 0$-2x+4=0$ implies $x=2$$x=2$x=2x = 2$x=2$.
   • The second derivative $f^{″}(x)=-2$$f^{″}(x)=-2$f^(″)(x)=-2f”(x) = -2$f^{″}(x)=-2$ is negative, indicating a local maximum.
   • Thus, the maximum value occurs at $x=2$$x=2$x=2x = 2$x=2$ and $f(2)=-2^2+4\times 2+1=5$$f(2)=-2^2+4\times 2+1=5$f(2)=-2^(2)+4xx2+1=5f(2) = -2^2 + 4 \times 2 + 1 = 5$f(2)=-2^2+4\times 2+1=5$. So, the maximum value is $5$$5$55$5$.
2. Constraint Optimization:
   Consider a function $f(x,y)=xy$$f(x,y)=xy$f(x,y)=xyf(x, y) = xy$f(x,y)=xy$ subject to a constraint $x+y=1$$x+y=1$x+y=1x + y = 1$x+y=1$. Using substitution or the Lagrangian method, one can find the maximum value of $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ given the constraint, which, in this case, is $\frac{1}{4}$$\frac{1}{4}$(1)/(4)\frac{1}{4}$\frac{1}{4}$.

Conclusion

Answer:

• $Q$$Q$QQ$Q$ is the total output,
• $A$$A$AA$A$ is a constant representing total factor productivity,
• $L$$L$LL$L$ is the quantity of labor,
• $K$$K$KK$K$ is the quantity of capital,
• $a$$a$aa$a$ and $b$$b$bb$b$ are the output elasticities of labor and capital, respectively.

Elasticity of Production with Respect to Labor

Step-by-Step Calculation

1. Find the Partial Derivative of $Q$$Q$QQ$Q$ with Respect to $L$$L$LL$L$:
   Differentiate $Q=A{L}^a{K}^b$$Q=A{L}^a{K}^b$Q=AL^(a)K^(b)Q = A L^a K^b$Q=A{L}^a{K}^b$ with respect to $L$$L$LL$L$:
   $$\frac{\partial Q}{\partial L}=A\cdot a\cdot L^{a-1}\cdot K^b$$$$\frac{\partial Q}{\partial L}=A\cdot a\cdot L^{a-1}\cdot K^b$$(del Q)/(del L)=A*a*L^(a-1)*K^(b)\frac{\partial Q}{\partial L} = A \cdot a \cdot L^{a-1} \cdot K^b$$\frac{\partial Q}{\partial L}=A\cdot a\cdot L^{a-1}\cdot K^b$$
2. Substitute into the Elasticity Formula:
   Now, plug $\frac{\partial Q}{\partial L}$$\frac{\partial Q}{\partial L}$(del Q)/(del L)\frac{\partial Q}{\partial L}$\frac{\partial Q}{\partial L}$ and $Q=A{L}^a{K}^b$$Q=A{L}^a{K}^b$Q=AL^(a)K^(b)Q = A L^a K^b$Q=A{L}^a{K}^b$ into the elasticity formula:
   $$E_L=(A\cdot a\cdot L^{a-1}\cdot K^b)\cdot \frac{L}{A{L}^a{K}^b}$$$$E_L=\left(A\cdot a\cdot L^{a-1}\cdot K^b\right)\cdot \frac{L}{A{L}^a{K}^b}$$E_(L)=(A*a*L^(a-1)*K^(b))*(L)/(AL^(a)K^(b))E_L = \left( A \cdot a \cdot L^{a-1} \cdot K^b \right) \cdot \frac{L}{A L^a K^b}$$E_L=(A\cdot a\cdot L^{a-1}\cdot K^b)\cdot \frac{L}{A{L}^a{K}^b}$$
3. Simplify the Expression:
   Simplify the expression by canceling out common terms:
   $$E_L=a\cdot \frac{L^{a-1}\cdot L}{L^a}=a\cdot \frac{L^a}{L^a}=a$$$$E_L=a\cdot \frac{L^{a-1}\cdot L}{L^a}=a\cdot \frac{L^a}{L^a}=a$$E_(L)=a*(L^(a-1)*L)/(L^(a))=a*(L^(a))/(L^(a))=aE_L = a \cdot \frac{L^{a-1} \cdot L}{L^a} = a \cdot \frac{L^a}{L^a} = a$$E_L=a\cdot \frac{L^{a-1}\cdot L}{L^a}=a\cdot \frac{L^a}{L^a}=a$$

Conclusion

Answer:

1. Calculate $2\mathbf{a}-5\mathbf{b}$$2\mathbf{a}-5\mathbf{b}$2a-5b2 \mathbf{a} – 5 \mathbf{b}$2\mathbf{a}-5\mathbf{b}$

2. Calculate $2\mathbf{a}-5\mathbf{b}+\mathbf{c}$$2\mathbf{a}-5\mathbf{b}+\mathbf{c}$2a-5b+c2 \mathbf{a} – 5 \mathbf{b} + \mathbf{c}$2\mathbf{a}-5\mathbf{b}+\mathbf{c}$

3. Calculate $\mathbf{a}\cdot \mathbf{b}$$\mathbf{a}\cdot \mathbf{b}$a*b\mathbf{a} \cdot \mathbf{b}$\mathbf{a}\cdot \mathbf{b}$ (Dot Product)

Summary of Results

• $2\mathbf{a}-5\mathbf{b}=\left(\begin{array}{c}12\\ -1\\ 21\end{array}\right)$$2\mathbf{a}-5\mathbf{b}=\left(\begin{array}{c}12\\ -1\\ 21\end{array}\right)$2a-5b=([12],[-1],[21])2 \mathbf{a} – 5 \mathbf{b} = \begin{pmatrix} 12 \\ -1 \\ 21 \end{pmatrix}$2\mathbf{a}-5\mathbf{b}=\left(\begin{array}{c}12\\ -1\\ 21\end{array}\right)$
• $2\mathbf{a}-5\mathbf{b}+\mathbf{c}=\left(\begin{array}{c}10\\ -2\\ 22\end{array}\right)$$2\mathbf{a}-5\mathbf{b}+\mathbf{c}=\left(\begin{array}{c}10\\ -2\\ 22\end{array}\right)$2a-5b+c=([10],[-2],[22])2 \mathbf{a} – 5 \mathbf{b} + \mathbf{c} = \begin{pmatrix} 10 \\ -2 \\ 22 \end{pmatrix}$2\mathbf{a}-5\mathbf{b}+\mathbf{c}=\left(\begin{array}{c}10\\ -2\\ 22\end{array}\right)$
• $\mathbf{a}\cdot \mathbf{b}=-9$$\mathbf{a}\cdot \mathbf{b}=-9$a*b=-9\mathbf{a} \cdot \mathbf{b} = -9$\mathbf{a}\cdot \mathbf{b}=-9$