Consider the following system of linear equations \begin{bmatrix}2 & 1 & -4…

To find the number of values of α for which the system has infinitely many solutions, we will use Gaussian elimination to reduce the augmented matrix into Row Echelon Form.

The system can be written as the augmented matrix:

[ 2   1  -4 | alpha ]
[ 4   3 -12 |   5   ]
[ 1   2  -8 |   7   ]

Step 1: Row Operations to Row Echelon Form

First, let's swap Row 1 (R1) and Row 3 (R3) to make the top-left pivot 1. This makes the arithmetic much simpler.

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R1 <--> R3

[ 1   2  -8 |   7   ]
[ 4   3 -12 |   5   ]
[ 2   1  -4 | alpha ]

Now, eliminate the entries below the pivot in Column 1:

• To eliminate the 4 in Row 2: R2 = R2 - 4*R1

• To eliminate the 2 in Row 3: R3 = R3 - 2*R1

Our matrix now looks like this:

[ 1   2  -8 |    7     ]
[ 0  -5  20 |   -23    ]
[ 0  -3  12 | alpha-14 ]

Next, let's look at Row 2 and Row 3. Notice that the coefficient parts are multiples of each other:

• If we multiply Row 2 by 3: [0, -15, 60]

• If we multiply Row 3 by 5: [0, -15, 60]

Let's apply the row operation: R3 = 5*R3 - 3*R2

Updating the right-hand side of Row 3

Our fully reduced Row Echelon matrix is:

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[ 1   2  -8 |    7     ]
[ 0  -5  20 |   -23    ]
[ 0   0   0 | 5*alpha-1]

Step 2: Condition for Infinitely Many Solutions

For a system of linear equations to have infinitely many solutions, two conditions must be met:

1. The system must be consistent. This means we cannot have a row of the form [0 0 0 | non-zero value].

2. The rank of the matrix must be less than the number of variables (3 variables: x, y, z).

The third row has zeros for all coefficients (0x + 0y + 0z). For the system to be consistent and have solutions, the right-hand side of that row must also be exactly 0.

Set the last entry to 0:

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5*alpha - 1 = 0
5*alpha = 1
alpha = 1/5

Conclusion

• If alpha = 1/5 (which is 0.2), the last row becomes 0 = 0. The system is consistent with a rank of 2, which is less than the number of variables, yielding infinitely many solutions.

• If alpha is any other number, the last row produces a contradiction like 0 = non-zero, resulting in zero solutions.

Therefore, there is exactly 1 value of α (specifically, α = 1/5) that gives the system infinitely many solutions.