Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Solve for a
a=2
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To solve the given system of linear equations using the matrix method, we can represent the system of equations in matrix form.
First, let's rewrite the system of equations:
1a + 3b + 2c = 3 (Equation 1)
2a - 1b - 3c = -8 (Equation 2)
5a + 2b + 1c = 9 (Equation 3)
Now, we can write the augmented matrix by rearranging the coefficients and the constants on the right-hand side:
[ 1 3 2 | 3 ]
[ 2 -1 -3 | -8 ]
[ 5 2 1 | 9 ]
To solve for 'a', we need to perform row operations to get the augmented matrix into row-echelon form or reduced row-echelon form. Let's start by using row operations to eliminate the 'a' variable from the second and third equations.
We'll multiply Equation 1 by 2 and subtract Equation 2 multiplied by 1 from it. Then, we'll multiply Equation 1 by 5 and subtract Equation 3 multiplied by 1 from it.
Row operation: Row2 ← (2 * Row1) - Row2
[ 1 3 2 | 3 ]
[ 0 -7 -7 | -14 ]
[ 5 2 1 | 9 ]
Row operation: Row3 ← (5 * Row1) - Row3
[ 1 3 2 | 3 ]
[ 0 -7 -7 | -14 ]
[ 0 -13 -9 | -12 ]
Now, we'll proceed with row operations to further reduce the matrix to row-echelon or reduced row-echelon form. Since the coefficient matrix does not have a strictly triangular form, we'll interchange Row2 and Row3 to simplify the calculations.
Row operation: Row2 ↔ Row3
[ 1 3 2 | 3 ]
[ 0 -13 -9 | -12 ]
[ 0 -7 -7 | -14 ]
Next, we'll multiply Row2 by -1/13 to make the leading coefficient (first non-zero entry) in Row2 equal to 1.
Row operation: Row2 ← (-1/13) * Row2
[ 1 3 2 | 3 ]
[ 0 1 9/13 | 12/13 ]
[ 0 -7 -7 | -14 ]
Now, we'll subtract 3 times Row2 from Row1 to eliminate the 'b' variable from the first equation.
Row operation: Row1 ← Row1 - (3 * Row2)
[ 1 0 -23/13 | -33/13 ]
[ 0 1 9/13 | 12/13 ]
[ 0 -7 -7 | -14 ]
At this point, our augmented matrix is in row-echelon form. The variables 'a' and 'b' have been eliminated from the first two rows.
Now, let's back-substitute to find the solution.
From the first equation, we have:
1a + 0b - (23/13)c = -33/13
a - (23/13)c = -33/13
We can rearrange the equation to solve for 'a':
a = -33/13 + (23/13)c
a = (-33 + 23c)/13
Therefore, the solution for 'a' is (-33 + 23c)/13.