Sure! Here’s a step-by-step solution to the problem:
First, let’s write the matrix equation that represents the problem: Ax = b, where A is the given matrix, x is the vector we’re trying to find, and b is the given vector.
We can solve this equation by augmenting matrix A with vector b and then performing row operations to put the augmented matrix in reduced row echelon form (RREF).
The augmented matrix is:
[ 1 0 -3 | -2 ]
[-3 1 6 | 3 ]
[ 2 -2 -1 | -1 ]
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We can start by adding 3 times the first row to the second row to eliminate the -3 in the second row:
[ 1 0 -3 | -2 ]
[ 0 1 3 | -3 ]
[ 2 -2 -1 | -1 ]
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Next, we can subtract 2 times the first row from the third row to eliminate the 2 in the third row:
[ 1 0 -3 | -2 ]
[ 0 1 3 | -3 ]
[ 0 -2 5 | 3 ]
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Now we can add twice the second row to the third row to eliminate the -2 in the third row:
[ 1 0 -3 | -2 ]
[ 0 1 3 | -3 ]
[ 0 0 11 | -3 ]
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Finally, we can divide the third row by 11 to get a 1 in the third column:
[1 0 -3 | -2]
[0 1 3 | -3]
[0 0 1 | -3/11]
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Now that we have an RREF matrix, we can solve for x. From the last row of our RREF matrix, we see that x_3 = -3/11. Substituting this value into the second row of our RREF matrix gives us x_2 + (3)(-3/11) = x_2 -9/11 =-3, so x_2 =-24/11. Finally, substituting these values into the first row of our RREF matrix gives us x_1 + (-3)(-3/11) = x_1 +9/11 =-2, so x_1 =-31/11.
Therefore, our solution for x is:
x = [-31/11
-24/11
-3/11]
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Since we were able to solve for a unique solution for x, it means that x is unique.
I hope this helps