The given system of equations can be represented by the following augmented matrix:
⎡1 0 2 a² 3 | b ⎤
⎢0 a b 0 1 | a ⎥
⎢0 0 0 a 0 | b²-1⎥
For the system to be consistent, the third row of the augmented matrix must be all zeros. This means that a must be equal to 0 and b²-1 must also be equal to 0. Solving for b, we find that b = ±1.
Substituting these values of a and b into the first two rows of the augmented matrix, we get:
⎡1 0 2 0 3 | ±1⎤
⎢0 0 ±1 0 1 | 0 ⎥
From the second row, we can see that the third variable must be equal to ±1. Let’s call this variable x3. Substituting this value into the first row, we get:
1x1 + 2x3 + 3x5 = ±1
x1 + 2(±1) + 3x5 = ±1
x1 = -3x5 ∓ 2
Thus, the general solution to the system is given by:
x1 = -3x5 ∓ 2
x2 = free variable
x3 = ±1
x4 = free variable
x5 = free variable
where x2, x4, and x5 can take on any value.
The given system of equations can be represented by the following augmented matrix: ⎡1 0 2 a² 3 | b ⎤ ⎢0 a b 0 1 | a ⎥ ⎢0 0 0 a 0 | b²-1⎥ For the system to be inconsistent, there must be a row in the augmented matrix where all the coefficients of the variables are zero, but the constant term is non-zero. In other words, there must be a row of the form [0 0 0 0 | k], where k is a non-zero constant. Looking at the third row of the augmented matrix, we can see that if a = 0 and b²-1 ≠ 0, then the system will be inconsistent. This means that a = 0 and b ≠ ±1. Thus, for all values of a = 0 and b ≠ ±1, the system is inconsistent.
Not for all types of equations. But always in second degree equations they do. Consider a third degree equation with 3 different roots. Obviously, one of the roots can not be in a pair.
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Data types
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You do not give "which of the following" and we can not help you.