Suppose the solutions of a homogeneous system of four linear equations in five unknowns are all multiples of one nonzero solution. Will the system necessarily have have a solution for every possible choice of constants on the right sides of the equations? Explain.

Consider a homogeneous machine of four linear equations in five unknowns are all multiples of 1 non-0 solution. Objective is to give an explanation for the gadget have an answer for each viable preference of constants on the proper facets of the equations.

Yes, it's miles true.

Consider the machine as Ax = 0. in which A is 4x5 matrix.

From given dim Nul A=1. Since, the rank theorem states that

The dimensions of the column space and the row space of a mxn matrix A are equal. This not unusual size, the rank of matrix A, additionally equals the number of pivot positions in A and satisfies the equation

rank A+ dim NulA = n

dim NulA =n- rank A

Rank A = 5 - dim Nul A

Rank A = 4

Thus, the measurement of dim Col A = rank A = five

And since Col A is a subspace of R^4, Col A = R^4.

So, every vector b in R^4 also in Col A, and Ax = b, has an answer for all b. Hence, the structures have an answer for every viable preference of constants on the right aspects of the equations.

the amount you'd have after the work is $1.55

step-by-step explanation:

you subtract 95% from $31.