Samacheer Kalvi Class 11 Business Maths Solutions Chapter 1 Matrices and Determinants Exercise 1.4

 

Question 1. The technology matrix of an economic system of two industries is \[ \left[\begin{array}{cc} 0.50 & 0.30 \\ 0.41 & 0.33 \end{array}\right] \]. Test whether the system is viable as per Hawkins Simon conditions.
Answer:
To test the viability of the system using Hawkins-Simon conditions, we first need the technology matrix B and then calculate \( I - B \). The technology matrix B is given as: \[ B = \left[\begin{array}{cc} 0.50 & 0.30 \\ 0.41 & 0.33 \end{array}\right] \] Next, we find the Leontief matrix \( I - B \), where I is the identity matrix: \[ I - B = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 0.50 & 0.30 \\ 0.41 & 0.33 \end{array}\right] \] Subtracting the matrices gives: \[ I - B = \left[\begin{array}{rr} 1-0.50 & 0-0.30 \\ 0-0.41 & 1-0.33 \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 0.50 & -0.30 \\ -0.41 & 0.67 \end{array}\right] \] Now, we check the Hawkins-Simon conditions: 1. All main diagonal elements of \( I - B \) must be positive. Here, the main diagonal elements are \( 0.50 \) and \( 0.67 \), both of which are positive. 2. The determinant of \( I - B \) must be positive (\( |I - B| > 0 \)). \[ |I - B| = \left|\begin{array}{rr} 0.50 & -0.30 \\ -0.41 & 0.67 \end{array}\right| \]
\( \implies \) \( |I - B| = (0.50)(0.67) - (-0.30)(-0.41) \)
\( \implies \) \( |I - B| = 0.335 - 0.123 \)
\( \implies \) \( |I - B| = 0.212 \) Since \( 0.212 \) is positive, both Hawkins-Simon conditions are satisfied. This indicates that the economic system is viable.In simple words: First, we subtracted the technology matrix from an identity matrix. Then, we checked two things: if the numbers on the main diagonal of the new matrix were positive, and if the overall value of this matrix (its determinant) was also positive. Since both were true, the economic system is considered healthy and can keep producing.

🎯 Exam Tip: Remember to check both conditions (positive diagonal elements and positive determinant) for \( I - B \) to confirm system viability. Showing each step of the matrix subtraction and determinant calculation is crucial.

 

Question 2. The technology matrix of ah economic system of two industries is \[ \left[\begin{array}{rr} 0.6 & 0.9 \\ 0.20 & 0.80 \end{array}\right] \]. Test whether the system is viable as per Hawkins-Simon conditions.
Answer:
To determine the viability, we start with the technology matrix B: \[ B = \left[\begin{array}{cc} 0.60 & 0.9 \\ 0.20 & 0.80 \end{array}\right] \] Next, we calculate the Leontief matrix \( I - B \): \[ I - B = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} 0.60 & 0.9 \\ 0.20 & 0.80 \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 1-0.60 & 0-0.9 \\ 0-0.20 & 1-0.80 \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 0.4 & -0.9 \\ -0.20 & 0.20 \end{array}\right] \] Now, let's check the Hawkins-Simon conditions: 1. All main diagonal elements of \( I - B \) must be positive. Here, the main diagonal elements are \( 0.4 \) and \( 0.20 \), both of which are positive. 2. The determinant of \( I - B \) must be positive (\( |I - B| > 0 \)). \[ |I - B| = \left|\begin{array}{rr} 0.4 & -0.9 \\ -0.20 & 0.20 \end{array}\right| \]
\( \implies \) \( |I - B| = (0.4)(0.20) - (-0.9)(-0.20) \)
\( \implies \) \( |I - B| = 0.08 - 0.18 \)
\( \implies \) \( |I - B| = -0.10 \) Since the determinant \( |I - B| \) is \( -0.10 \), which is negative, the second Hawkins-Simon condition is not satisfied. This means that the given economic system is not viable.In simple words: We calculated a special matrix called \( I - B \) and then looked at its main diagonal numbers, which were fine. But when we found its overall value (determinant), it was a negative number. Because of this negative value, the system is not considered strong enough to function properly.

🎯 Exam Tip: If even one of the Hawkins-Simon conditions is not met, the system is immediately declared non-viable. Ensure your arithmetic for the determinant is accurate, as a small error can change the viability conclusion.

 

Question 3. The technology matrix of an economic system of two industries is \[ \left[\begin{array}{ll} 0.50 & 0.25 \\ 0.40 & 0.67 \end{array}\right] \]. Test whether the system is viable as per Hawkins-Simon conditions.
Answer:
We are given the technology matrix B: \[ B = \left[\begin{array}{ll} 0.50 & 0.25 \\ 0.40 & 0.67 \end{array}\right] \] First, we compute the Leontief matrix \( I - B \): \[ I - B = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 0.50 & 0.25 \\ 0.40 & 0.67 \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 1-0.50 & 0-0.25 \\ 0-0.40 & 1-0.67 \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 0.50 & -0.25 \\ -0.40 & 0.33 \end{array}\right] \] Now we apply the Hawkins-Simon conditions: 1. All main diagonal elements of \( I - B \) must be positive. Here, the main diagonal elements are \( 0.50 \) and \( 0.33 \), both of which are positive. 2. The determinant of \( I - B \) must be positive (\( |I - B| > 0 \)). \[ |I - B| = \left|\begin{array}{rr} 0.50 & -0.25 \\ -0.40 & 0.33 \end{array}\right| \]
\( \implies \) \( |I - B| = (0.50)(0.33) - (-0.25)(-0.40) \)
\( \implies \) \( |I - B| = 0.165 - 0.100 \)
\( \implies \) \( |I - B| = 0.065 \) Since \( 0.065 \) is positive, both Hawkins-Simon conditions are met. Therefore, the given economic system is viable. This means the system can support itself and meet its demands.In simple words: After finding the \( I - B \) matrix, we saw that its main diagonal numbers were positive. Then, we calculated its determinant, which also came out positive. Since both checks passed, the economic system is considered viable.

🎯 Exam Tip: Pay close attention to the signs when calculating the determinant, especially with negative values. A common mistake is an error in multiplying negative numbers.

 

Question 4. Two commodities A and B are produced such that 0.4 tonne of A and 0.7 tonnes of B are required to produce a tonnes of A. Similarly 0.1 tonne of A and 0.7 tonne of B are needed to produce a tonnes of B. Write down the technology matrix. If 6.8 tonnes of A and 10.2 tones of B are required, find the gross production of both of them.
Answer:
First, we need to construct the technology matrix B and the final demand matrix D. The input-output requirements are: To produce 1 tonne of A, 0.4 tonne of A and 0.7 tonne of B are needed. To produce 1 tonne of B, 0.1 tonne of A and 0.7 tonne of B are needed. This gives us the technology matrix B: \[ B = \left[\begin{array}{cc} 0.4 & 0.1 \\ 0.7 & 0.7 \end{array}\right] \] The final demand matrix D is given as: \[ D = \left[\begin{array}{l} 6.8 \\ 10.2 \end{array}\right] \] Next, we calculate the Leontief matrix \( I - B \): \[ I - B = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 0.4 & 0.1 \\ 0.7 & 0.7 \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 1-0.4 & 0-0.1 \\ 0-0.7 & 1-0.7 \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 0.6 & -0.1 \\ -0.7 & 0.3 \end{array}\right] \] We check the Hawkins-Simon conditions: 1. Main diagonal elements of \( I - B \) (\( 0.6 \) and \( 0.3 \)) are both positive. 2. The determinant of \( I - B \): \[ |I - B| = \left|\begin{array}{rr} 0.6 & -0.1 \\ -0.7 & 0.3 \end{array}\right| \]
\( \implies \) \( |I - B| = (0.6)(0.3) - (-0.1)(-0.7) \)
\( \implies \) \( |I - B| = 0.18 - 0.07 \)
\( \implies \) \( |I - B| = 0.11 \) Since \( |I - B| = 0.11 \) is positive, both conditions are satisfied, and the system is viable. To find the gross production X, we use the formula \( X = (I - B)^{-1}D \). First, find the adjoint of \( I - B \): \[ \text{adj}(I - B) = \left[\begin{array}{rr} 0.3 & 0.1 \\ 0.7 & 0.6 \end{array}\right] \] Now, calculate the inverse \( (I - B)^{-1} \): \[ (I - B)^{-1} = \frac{1}{|I - B|} \text{adj}(I - B) \]
\( \implies \) \[ (I - B)^{-1} = \frac{1}{0.11} \left[\begin{array}{rr} 0.3 & 0.1 \\ 0.7 & 0.6 \end{array}\right] \] Finally, compute the gross production X: \[ X = (I - B)^{-1}D \]
\( \implies \) \[ X = \frac{1}{0.11} \left[\begin{array}{rr} 0.3 & 0.1 \\ 0.7 & 0.6 \end{array}\right] \left[\begin{array}{l} 6.8 \\ 10.2 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{0.11} \left[\begin{array}{l} (0.3)(6.8) + (0.1)(10.2) \\ (0.7)(6.8) + (0.6)(10.2) \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{0.11} \left[\begin{array}{l} 2.04 + 1.02 \\ 4.76 + 6.12 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{0.11} \left[\begin{array}{l} 3.06 \\ 10.88 \end{array}\right] \]
\( \implies \) \[ X = \left[\begin{array}{l} \frac{3.06}{0.11} \\ \frac{10.88}{0.11} \end{array}\right] \]
\( \implies \) \[ X = \left[\begin{array}{l} 27.8181 \\ 98.9090 \end{array}\right] \] Rounding to two decimal places: \[ X = \left[\begin{array}{l} 27.82 \\ 98.91 \end{array}\right] \] Therefore, the gross production of commodity A should be 27.82 tonnes, and the gross production of commodity B should be 98.91 tonnes to meet the final demand.In simple words: We first set up a matrix for how much of each product is needed to make the other. Then, using the given final demand, we did some matrix math to figure out the total amount of each product that needs to be made. This ensures enough is produced for both internal use and the final demand.

🎯 Exam Tip: When calculating gross production, remember the formula \( X = (I - B)^{-1}D \). Make sure to correctly find the inverse matrix and then perform the matrix multiplication with the demand matrix.

 

Question 5. Suppose the inter-industry flow of the product of two industries is given as under.

Determine the technology matrix and test Hawkin's-Simon conditions for the viability of the system. If the domestic changes to 80 and 40 units respectively, what should be the gross output of each sector in order to meet the new demands.
Answer:
From the given transaction matrix, we can extract the input-output coefficients \( b_{ij} \) to form the technology matrix B. The values are: \( a_{11} = 30 \), \( a_{12} = 40 \), \( x_1 = 120 \) \( a_{21} = 20 \), \( a_{22} = 10 \), \( x_2 = 60 \) The technical coefficients are calculated as: \( b_{11} = \frac{a_{11}}{x_1} = \frac{30}{120} = \frac{1}{4} \) \( b_{12} = \frac{a_{12}}{x_2} = \frac{40}{60} = \frac{2}{3} \) \( b_{21} = \frac{a_{21}}{x_1} = \frac{20}{120} = \frac{1}{6} \) \( b_{22} = \frac{a_{22}}{x_2} = \frac{10}{60} = \frac{1}{6} \) So, the technology matrix B is: \[ B = \left[\begin{array}{cc} \frac{1}{4} & \frac{2}{3} \\ \frac{1}{6} & \frac{1}{6} \end{array}\right] \] Now, we calculate the Leontief matrix \( I - B \): \[ I - B = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} \frac{1}{4} & \frac{2}{3} \\ \frac{1}{6} & \frac{1}{6} \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 1-\frac{1}{4} & 0-\frac{2}{3} \\ 0-\frac{1}{6} & 1-\frac{1}{6} \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} \frac{3}{4} & -\frac{2}{3} \\ -\frac{1}{6} & \frac{5}{6} \end{array}\right] \] We check the Hawkins-Simon conditions: 1. All main diagonal elements of \( I - B \) are positive (\( \frac{3}{4} \) and \( \frac{5}{6} \)). This condition is satisfied. 2. The determinant of \( I - B \) must be positive (\( |I - B| > 0 \)). \[ |I - B| = \left|\begin{array}{rr} \frac{3}{4} & -\frac{2}{3} \\ -\frac{1}{6} & \frac{5}{6} \end{array}\right| \]
\( \implies \) \( |I - B| = \left(\frac{3}{4}\right)\left(\frac{5}{6}\right) - \left(-\frac{2}{3}\right)\left(-\frac{1}{6}\right) \)
\( \implies \) \( |I - B| = \frac{15}{24} - \frac{2}{18} \)
\( \implies \) \( |I - B| = \frac{5}{8} - \frac{1}{9} \)
\( \implies \) \( |I - B| = \frac{45 - 8}{72} \)
\( \implies \) \( |I - B| = \frac{37}{72} \) Since \( \frac{37}{72} \) is positive, both Hawkins-Simon conditions are satisfied, so the system is viable. Now, we need to find the gross output for new domestic demands: New domestic demand \( D = \left[\begin{array}{l} 80 \\ 40 \end{array}\right] \) We need to calculate \( X = (I - B)^{-1}D \). First, find the adjoint of \( I - B \): \[ \text{adj}(I - B) = \left[\begin{array}{rr} \frac{5}{6} & \frac{2}{3} \\ \frac{1}{6} & \frac{3}{4} \end{array}\right] \] Now, calculate the inverse \( (I - B)^{-1} \): \[ (I - B)^{-1} = \frac{1}{|I - B|} \text{adj}(I - B) \]
\( \implies \) \[ (I - B)^{-1} = \frac{1}{\frac{37}{72}} \left[\begin{array}{rr} \frac{5}{6} & \frac{2}{3} \\ \frac{1}{6} & \frac{3}{4} \end{array}\right] \]
\( \implies \) \[ (I - B)^{-1} = \frac{72}{37} \left[\begin{array}{rr} \frac{5}{6} & \frac{2}{3} \\ \frac{1}{6} & \frac{3}{4} \end{array}\right] \]
\( \implies \) \[ (I - B)^{-1} = \left[\begin{array}{rr} \frac{72}{37} \times \frac{5}{6} & \frac{72}{37} \times \frac{2}{3} \\ \frac{72}{37} \times \frac{1}{6} & \frac{72}{37} \times \frac{3}{4} \end{array}\right] \]
\( \implies \) \[ (I - B)^{-1} = \left[\begin{array}{rr} \frac{12 \times 5}{37} & \frac{24 \times 2}{37} \\ \frac{12 \times 1}{37} & \frac{18 \times 3}{37} \end{array}\right] \]
\( \implies \) \[ (I - B)^{-1} = \left[\begin{array}{rr} \frac{60}{37} & \frac{48}{37} \\ \frac{12}{37} & \frac{54}{37} \end{array}\right] \] Finally, calculate the gross output X: \[ X = (I - B)^{-1}D \]
\( \implies \) \[ X = \left[\begin{array}{rr} \frac{60}{37} & \frac{48}{37} \\ \frac{12}{37} & \frac{54}{37} \end{array}\right] \left[\begin{array}{l} 80 \\ 40 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{37} \left[\begin{array}{l} (60)(80) + (48)(40) \\ (12)(80) + (54)(40) \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{37} \left[\begin{array}{l} 4800 + 1920 \\ 960 + 2160 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{37} \left[\begin{array}{l} 6720 \\ 3120 \end{array}\right] \]
\( \implies \) \[ X = \left[\begin{array}{l} \frac{6720}{37} \\ \frac{3120}{37} \end{array}\right] \]
\( \implies \) \[ X = \left[\begin{array}{l} 181.6216 \\ 84.3243 \end{array}\right] \] Rounding to two decimal places: \[ X = \left[\begin{array}{l} 181.62 \\ 84.32 \end{array}\right] \] So, the gross output of industry X should be 181.62 units and industry Y should be 84.32 units to meet the new demands.In simple words: We first built a matrix showing how much each industry needs from itself and the other to produce goods. Then, we checked if this system was strong enough (viable). Since it was, we used a formula with the new demands to find out the total amount each industry needs to produce.

🎯 Exam Tip: When dealing with fractions in matrix calculations, simplify them at each step where possible to avoid errors. Carefully multiply and add fractions before finding the common denominator.

 

Question 6. You are given the following transaction matrix for a two-sector economy.

(i) Write the technology matrix? (ii) Determine the output when the final demand for the output sector 1 alone S.
Answer:
(i) To write the technology matrix B, we use the formula \( b_{ij} = \frac{a_{ij}}{x_j} \), where \( a_{ij} \) is the input from sector i to sector j, and \( x_j \) is the total output of sector j. From the table: For sector 1: \( a_{11} = 4 \), \( a_{12} = 3 \), \( x_1 = 20 \) For sector 2: \( a_{21} = 5 \), \( a_{22} = 4 \), \( x_2 = 12 \) Calculate the technical coefficients: \( b_{11} = \frac{a_{11}}{x_1} = \frac{4}{20} = \frac{1}{5} \) \( b_{12} = \frac{a_{12}}{x_2} = \frac{3}{12} = \frac{1}{4} \) \( b_{21} = \frac{a_{21}}{x_1} = \frac{5}{20} = \frac{1}{4} \) \( b_{22} = \frac{a_{22}}{x_2} = \frac{4}{12} = \frac{1}{3} \) So, the technology matrix B is: \[ B = \left[\begin{array}{cc} \frac{1}{5} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{3} \end{array}\right] \] (ii) Determine the output when the final demand for the output sector 1 alone. This phrase from the question seems incomplete "sector 1 alone S.". Assuming it means that the final demand for sector 1 is some value 'S' and sector 2's demand is 0, or it refers to a specific numerical demand which is not explicitly stated. Based on common input-output problems, this usually implies finding total output X when the demand changes. Given the next page of calculations, it seems the question implies finding the gross output when a specific final demand is given (likely from another problem or implicitly understood as a unit demand for sector 1 or a derived demand). Without a specified value for the demand, we cannot provide a numerical answer for this part, or the source text has an implicit value. Let's continue with the calculation steps that are typically performed in such a problem, which involves finding \( (I-B)^{-1} \). First, calculate \( I - B \): \[ I - B = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} \frac{1}{5} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{3} \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 1-\frac{1}{5} & 0-\frac{1}{4} \\ 0-\frac{1}{4} & 1-\frac{1}{3} \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} \frac{4}{5} & -\frac{1}{4} \\ -\frac{1}{4} & \frac{2}{3} \end{array}\right] \] Check Hawkins-Simon conditions: 1. The main diagonal elements \( \frac{4}{5} \) and \( \frac{2}{3} \) are both positive. 2. Calculate the determinant of \( I - B \): \[ |I - B| = \left|\begin{array}{rr} \frac{4}{5} & -\frac{1}{4} \\ -\frac{1}{4} & \frac{2}{3} \end{array}\right| \]
\( \implies \) \( |I - B| = \left(\frac{4}{5}\right)\left(\frac{2}{3}\right) - \left(-\frac{1}{4}\right)\left(-\frac{1}{4}\right) \)
\( \implies \) \( |I - B| = \frac{8}{15} - \frac{1}{16} \)
\( \implies \) \( |I - B| = \frac{(8 \times 16) - (1 \times 15)}{15 \times 16} \)
\( \implies \) \( |I - B| = \frac{128 - 15}{240} \)
\( \implies \) \( |I - B| = \frac{113}{240} \) Since \( \frac{113}{240} \) is positive, the system is viable. Now, find the adjoint of \( I - B \): \[ \text{adj}(I - B) = \left[\begin{array}{rr} \frac{2}{3} & \frac{1}{4} \\ \frac{1}{4} & \frac{4}{5} \end{array}\right] \] Next, calculate the inverse \( (I - B)^{-1} \): \[ (I - B)^{-1} = \frac{1}{|I - B|} \text{adj}(I - B) \]
\( \implies \) \[ (I - B)^{-1} = \frac{1}{\frac{113}{240}} \left[\begin{array}{rr} \frac{2}{3} & \frac{1}{4} \\ \frac{1}{4} & \frac{4}{5} \end{array}\right] \]
\( \implies \) \[ (I - B)^{-1} = \frac{240}{113} \left[\begin{array}{rr} \frac{2}{3} & \frac{1}{4} \\ \frac{1}{4} & \frac{4}{5} \end{array}\right] \] The question then implies a specific demand matrix D. From the next page content, it appears D is \( \left[\begin{array}{l} 23 \\ 3 \end{array}\right] \). Let's use that for the calculation. \[ D = \left[\begin{array}{l} 23 \\ 3 \end{array}\right] \] Now, calculate \( X = (I - B)^{-1}D \): \[ X = \frac{240}{113} \left[\begin{array}{rr} \frac{2}{3} & \frac{1}{4} \\ \frac{1}{4} & \frac{4}{5} \end{array}\right] \left[\begin{array}{l} 23 \\ 3 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{113} \left[\begin{array}{rr} 240 \times \frac{2}{3} & 240 \times \frac{1}{4} \\ 240 \times \frac{1}{4} & 240 \times \frac{4}{5} \end{array}\right] \left[\begin{array}{l} 23 \\ 3 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{113} \left[\begin{array}{rr} 160 & 60 \\ 60 & 192 \end{array}\right] \left[\begin{array}{l} 23 \\ 3 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{113} \left[\begin{array}{l} (160)(23) + (60)(3) \\ (60)(23) + (192)(3) \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{113} \left[\begin{array}{l} 3680 + 180 \\ 1380 + 576 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{113} \left[\begin{array}{l} 3860 \\ 1956 \end{array}\right] \]
\( \implies \) \[ X = \left[\begin{array}{l} \frac{3860}{113} \\ \frac{1956}{113} \end{array}\right] \]
\( \implies \) \[ X = \left[\begin{array}{l} 34.159 \\ 17.3097 \end{array}\right] \] Rounding to two decimal places: \[ X = \left[\begin{array}{l} 34.16 \\ 17.31 \end{array}\right] \] Thus, the output of sector 1 should be 34.16 units and sector 2 should be 17.31 units.In simple words: First, we built a matrix showing how much each industry depends on the other sectors to produce its goods. Then, we checked if the system was stable enough to work. After that, we used a formula with the specific demand for each sector to figure out the exact amount each sector needs to produce in total.

🎯 Exam Tip: When the final demand is not explicitly stated in the question, look for it in the context of the solution provided. Ensure all fraction arithmetic is done carefully to avoid calculation errors.

 

Question 7. Suppose the inter-industry flow of the product of two Sectors X and Y are given as under.

Find the gross output when the domestic demand changes to 12 for X and 18 for Y.
Answer:
From the given table, we first identify the inputs and total outputs for each sector. For sector X: \( a_{11} = 15 \), \( a_{12} = 10 \), \( x_1 = 35 \) For sector Y: \( a_{21} = 20 \), \( a_{22} = 30 \), \( x_2 = 65 \) Now, we calculate the technical coefficients to form the technology matrix B: \( b_{11} = \frac{a_{11}}{x_1} = \frac{15}{35} = \frac{3}{7} \) \( b_{12} = \frac{a_{12}}{x_2} = \frac{10}{65} = \frac{2}{13} \) \( b_{21} = \frac{a_{21}}{x_1} = \frac{20}{35} = \frac{4}{7} \) \( b_{22} = \frac{a_{22}}{x_2} = \frac{30}{65} = \frac{6}{13} \) So, the technology matrix B is: \[ B = \left[\begin{array}{cc} \frac{3}{7} & \frac{2}{13} \\ \frac{4}{7} & \frac{6}{13} \end{array}\right] \] Next, we compute the Leontief matrix \( I - B \): \[ I - B = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} \frac{3}{7} & \frac{2}{13} \\ \frac{4}{7} & \frac{6}{13} \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} 1-\frac{3}{7} & 0-\frac{2}{13} \\ 0-\frac{4}{7} & 1-\frac{6}{13} \end{array}\right] \]
\( \implies \) \[ I - B = \left[\begin{array}{rr} \frac{4}{7} & -\frac{2}{13} \\ -\frac{4}{7} & \frac{7}{13} \end{array}\right] \] We check the Hawkins-Simon conditions: 1. The main diagonal elements \( \frac{4}{7} \) and \( \frac{7}{13} \) are both positive. 2. Calculate the determinant of \( I - B \): \[ |I - B| = \left|\begin{array}{rr} \frac{4}{7} & -\frac{2}{13} \\ -\frac{4}{7} & \frac{7}{13} \end{array}\right| \]
\( \implies \) \( |I - B| = \left(\frac{4}{7}\right)\left(\frac{7}{13}\right) - \left(-\frac{2}{13}\right)\left(-\frac{4}{7}\right) \)
\( \implies \) \( |I - B| = \frac{28}{91} - \frac{8}{91} \)
\( \implies \) \( |I - B| = \frac{20}{91} \) Since \( \frac{20}{91} \) is positive, both Hawkins-Simon conditions are satisfied, so the system is viable and has a solution. The new domestic demand matrix D is given as: \[ D = \left[\begin{array}{l} 12 \\ 18 \end{array}\right] \] To find the gross output X, we use the formula \( X = (I - B)^{-1}D \). First, find the adjoint of \( I - B \): \[ \text{adj}(I - B) = \left[\begin{array}{rr} \frac{7}{13} & \frac{2}{13} \\ \frac{4}{7} & \frac{4}{7} \end{array}\right] \] Now, calculate the inverse \( (I - B)^{-1} \): \[ (I - B)^{-1} = \frac{1}{|I - B|} \text{adj}(I - B) \]
\( \implies \) \[ (I - B)^{-1} = \frac{1}{\frac{20}{91}} \left[\begin{array}{rr} \frac{7}{13} & \frac{2}{13} \\ \frac{4}{7} & \frac{4}{7} \end{array}\right] \]
\( \implies \) \[ (I - B)^{-1} = \frac{91}{20} \left[\begin{array}{rr} \frac{7}{13} & \frac{2}{13} \\ \frac{4}{7} & \frac{4}{7} \end{array}\right] \]
\( \implies \) \[ (I - B)^{-1} = \left[\begin{array}{rr} \frac{91}{20} \times \frac{7}{13} & \frac{91}{20} \times \frac{2}{13} \\ \frac{91}{20} \times \frac{4}{7} & \frac{91}{20} \times \frac{4}{7} \end{array}\right] \]
\( \implies \) \[ (I - B)^{-1} = \left[\begin{array}{rr} \frac{7 \times 7}{20} & \frac{7 \times 2}{20} \\ \frac{13 \times 4}{20} & \frac{13 \times 4}{20} \end{array}\right] \]
\( \implies \) \[ (I - B)^{-1} = \left[\begin{array}{rr} \frac{49}{20} & \frac{14}{20} \\ \frac{52}{20} & \frac{52}{20} \end{array}\right] \] Finally, calculate the gross output X: \[ X = (I - B)^{-1}D \]
\( \implies \) \[ X = \left[\begin{array}{rr} \frac{49}{20} & \frac{14}{20} \\ \frac{52}{20} & \frac{52}{20} \end{array}\right] \left[\begin{array}{l} 12 \\ 18 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{20} \left[\begin{array}{l} (49)(12) + (14)(18) \\ (52)(12) + (52)(18) \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{20} \left[\begin{array}{l} 588 + 252 \\ 624 + 936 \end{array}\right] \]
\( \implies \) \[ X = \frac{1}{20} \left[\begin{array}{l} 840 \\ 1560 \end{array}\right] \]
\( \implies \) \[ X = \left[\begin{array}{l} \frac{840}{20} \\ \frac{1560}{20} \end{array}\right] \]
\( \implies \) \[ X = \left[\begin{array}{l} 42 \\ 78 \end{array}\right] \] Therefore, the gross output for sector X should be 42 units and for sector Y should be 78 units to meet the new demands.In simple words: We calculated how much each sector needs from itself and the other to produce goods. Then, we used a formula with the new demand numbers (12 for X and 18 for Y) to find out the total amount each sector must produce. This ensures all needs, both internal and external, are met.

🎯 Exam Tip: Always double-check your fraction calculations and matrix multiplication steps. Simplifying fractions where possible can make the calculations easier and reduce the chance of errors.