The following augmented matrix is in row-echelon form and represents a linear system. Solve the system by using back-substitution, if possible.
[1 2|3 ]
[0 1|-3]
To solve the system using back-substitution, we start from the bottom row and work our way up.
Step 1: Write the last equation in terms of the variables.
From the last row of the augmented matrix, we have 0x + 1y = -3.
This equation can be rewritten as y = -3.
Step 2: Substitute the value of y into the equation above it.
From the first row of the augmented matrix, we have 1x + 2y = 3.
Substituting y = -3, we get 1x + 2(-3) = 3.
This simplifies to 1x - 6 = 3.
Step 3: Solve for x.
Adding 6 to both sides of the equation, we get 1x = 3 + 6.
This simplifies to 1x = 9.
Therefore, x = 9.
Step 4: Substitute the values of x and y back into the original equations to check.
From the first equation: 1(9) + 2( - 3) = 3.
This simplifies to 9 - 6 = 3, which is true.
From the second equation: 0(9) + 1( - 3) = -3.
This simplifies to -3 = -3, which is also true.
Therefore, the solution to the system is x = 9 and y = -3.