Solve Using an Inverse Matrix

Problem

{[3*a+5*b+2*d=0],[4*a+7*b+5*d=0],[a+b−4*d=0],[2*a+9*b+6*d=0])

Solution

1. Represent the system as a matrix equation A*X=B Since there are 4 equations and 3 variables (a,b,d, the matrix A is a 4×3 matrix.

A=([3,5,2],[4,7,5],[1,1,−4],[2,9,6])

X=([a],[b],[d])

B=([0],[0],[0],[0])

2. Identify the nature of the system. This is a homogeneous system of linear equations where the constant vector B is the zero vector.

3. Determine if a unique solution exists. For a homogeneous system, the trivial solution a=0,b=0,d=0 is always a solution. An inverse matrix can only be used directly if A is a square n×n matrix and is non-singular.

4. Analyze the rank of the matrix A We perform row reduction to see if there are non-trivial solutions.

([1,1,−4],[3,5,2],[4,7,5],[2,9,6])

5. Eliminate entries in the first column.

(R_2)−3*(R_1)⇒([1,1,−4],[0,2,14],[4,7,5],[2,9,6])

(R_3)−4*(R_1)⇒([1,1,−4],[0,2,14],[0,3,21],[2,9,6])

(R_4)−2*(R_1)⇒([1,1,−4],[0,2,14],[0,3,21],[0,7,14])

6. Simplify the second and third rows.

1/2*(R_2)⇒([1,1,−4],[0,1,7],[0,3,21],[0,7,14])

(R_3)−3*(R_2)⇒([1,1,−4],[0,1,7],[0,0,0],[0,7,14])

7. Eliminate the entry in the fourth row.

(R_4)−7*(R_2)⇒([1,1,−4],[0,1,7],[0,0,0],[0,0,−35])

8. Conclude the rank. Since there are 3 pivots (in columns 1, 2, and 3), the rank of A is 3. Because the rank equals the number of variables, the only solution is the trivial solution.

Final Answer

a=0,b=0,d=0

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