Explain me how to solve system of equations in matrix in easy.

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Let's learn how to solve systems of equations using matrices. A system of linear equations like 2x plus 3y equals 7, and 4x minus y equals 1, can be written in matrix form as A times x equals b. Here, A is the coefficient matrix containing the numbers in front of our variables, x is the variable vector, and b is the constant vector on the right side of our equations. Now let's see how to set up any system in matrix form using a 3 by 3 example. We have three equations with three unknowns: x, y, and z. Step 1: Extract the coefficient matrix A by taking all the numbers in front of our variables, maintaining their positions. Step 2: Create the variable vector x containing our unknowns. Step 3: Form the constant vector b with the numbers on the right side of each equation. Notice how we maintain consistent variable ordering and include zero coefficients when terms are missing. This gives us the compact matrix form A times x equals b. Gaussian elimination is the primary method for solving matrix systems. We start with the augmented matrix that combines the coefficient matrix A and constant vector b. The goal is to transform this into row echelon form using three elementary row operations: row swapping, row multiplication, and row addition. First, we make the first pivot element equal to 1 by dividing the first row by 2. Then we eliminate all elements below this pivot by subtracting appropriate multiples of the first row. Next, we make the second pivot equal to 1, and eliminate below it. Finally, we continue this process until we have an upper triangular form with 1s on the diagonal. Now we use back substitution to find our solution from the row echelon form. We start from the bottom row and work our way up. From the third row, we see that negative z equals zero, so z equals zero. Moving to the second row, we have y minus nine sevenths times z equals 2. Substituting z equals zero, we get y equals 2. Finally, from the first row, x plus 1.5 times y minus 0.5 times z equals 4. Substituting our known values y equals 2 and z equals zero, we solve for x and get x equals 1. Therefore, our complete solution is x equals 1, y equals 2, and z equals 0. Let's work through a complete example from start to finish. We have the system: x plus 2y minus z equals 3, 2x minus y plus 3z equals 7, and negative x plus 3y plus z equals 1. First, we convert this to matrix form with the augmented matrix. Next, we apply Gaussian elimination to get row echelon form. Then we use back substitution: from the bottom row, z equals 2. From the second row, y minus z equals negative one fifth, so y equals nine fifths. From the first row, x plus 2y minus z equals 3, so x equals negative eight fifths. Finally, we verify our solution by substituting back into all original equations, and they all check out perfectly.