If the system of linear equations: x+y+2z2x+3y+az−x−3y+bz=6=a+1=2b where a,b∈R, has infinitely many solutions, then 7a + 3b is equal to:
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Step-by-Step Solution
Step 1: Write the augmented matrix
To solve a system of linear equations, we first write its augmented matrix. This matrix combines the coefficients of the variables and the constants on the right-hand side of the equations. Each row represents an equation, and each column (before the vertical bar) corresponds to a variable.
Step 2: Perform row operations to simplify the matrix
We perform elementary row operations to transform the matrix into an echelon form. First, we replace R2 with R2−2R1 and R3 with R3+R1. Then, we replace R3 with R3+2R2. These operations help us eliminate variables and simplify the system.
Step 3: Apply condition for infinitely many solutions
For a system of linear equations to have infinitely many solutions, the last row of the echelon form of the augmented matrix must be entirely zeros. This means both the coefficient of z and the constant term in the last row must be zero.
Step 4: Solve the system of equations for a and b
We now have a system of two linear equations with two variables, a and b: a+b=10 and a+2b=17. Subtracting the first equation from the second gives b=7. Substituting b=7 into the first equation gives a+7=10, so a=3.
Step 5: Calculate 7a + 3b
Finally, we substitute the values of a=3 and b=7 into the expression 7a + 3b to find the required value.