# In Exercises 5–12, you are given a technology matrix A and an external demand vector D. Find the corresponding production vector X. [ HinT: See Quick Example 1.] A = [ 0.5 0.1 0 0 0.5 0.1 0 0 0.5 ] , D = [ 3 , 000 3 , 800 2 , 000 ]

### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
Publisher: Cengage Learning
ISBN: 9781337274203

### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
Publisher: Cengage Learning
ISBN: 9781337274203

#### Solutions

Chapter
Section
Chapter 5.5, Problem 10E
Textbook Problem

## In Exercises 5–12, you are given a technology matrix A and an external demand vector D. Find the corresponding production vector X. [HinT: See Quick Example 1.] A = [ 0.5 0.1 0 0 0.5 0.1 0 0 0.5 ] , D = [ 3 , 000 3 , 800 2 , 000 ]

Expert Solution
To determine

To calculate: The production vector where the technology matrix is A=[0.50.1000.50.1000.5] and demand vector is D=[3,0003,8002,000].

### Explanation of Solution

Given Information:

The technology matrix is A=[0.50.1000.50.1000.5] and demand vector is D=[3,0003,8002,000].

Formula used:

According to the product rule of matrices, the matrix X with dimension m×n and the matrix Y with dimension n×k, the product XY is the matrix of dimension m×k and ijth entry of AB is the sum of product of corresponding entries of row i of X and column j of Y.

When a matrix has m rows and n columns, then the matrix is said to have dimension of m×n.

For two matrices A and B with equal dimension, the sum of difference is simplified as,

(a11a12a21a22)±(b11b12b21b22)=(a11±b11a12±b12a21±b21a22±b22)

If the matrix obtained after row reduction has identity matrix on the left side then the matrix obtained on the right is the required inverse of A otherwise A is singular.

Calculation:

Consider technology matrix is A=[0.50.1000.50.1000.5] and demand vector is D=[3,0003,8002,000].

As, the production vector X satisfies X=(IA)1D, where A is the technology matrix, D is the demand vector and I is the identity matrix.

Substitute [0.50.1000.50.1000.5] for A, [100010001] for I, and [3,0003,8002,000] for D in X=(IA)1D.

X=(IA)1D=([100010001][0.50.1000.50.1000.5])1[3,0003,8002,000]

It is known that a matrix with m rows and n columns is of dimension m×n, where m and n are positive integers.

Since, number of rows in the matrix [0.50.1000.50.1000.5] is 3 and number of columns is 3,

Substitute 3 for m and 3 for n in m×n.

Thus, dimension of the matrix [0.50.1000.50.1000.5] is 3×3.

Since, number of rows in the matrix [100010001] is 3 and number of columns is 3,

Substitute 3 for m and 3 for n in m×n.

Thus, dimension of the matrix [100010001] is 3×3.

Thus, both the matrices have same dimensions.

Since, number of rows in the matrix [3,0003,8002,000] is 3 and number of columns is 1,

Substitute 3 for m and 1 for n in m×n.

Thus, dimension of [3,0003,8002,000] is 3×1.

It is known that for two matrices A and B with equal dimension, AB is the and difference of the corresponding entries of the matrices.

Simplify [100010001][0.50.1000.50.1000.5].

[100010001][0.50.1000.50.1000.5]=[0.50.1000.50.1000.5]

Consider, X=[20.40.08020.4002][3,0003,8002,000]

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