Solve the system of equations:0.5X+2YX−31Y=27=10
MathBot Answer:
The system of equations has one solution.X=13174Y=13132
Solve by substitution:
Solve 2X+2Y=27 for X. X=54−4Y
Substitute 54−4Y for X in X−3Y=10 and simplify. X−3Y(54−4Y)−3YY=10=10=13132
Use substitution of the numerical value of Y to get the values of X. XXX=54−4Y=54−4⋅132⋅131=13174
Solve by Gauss-Jordan Elimination:
Begin by writing the augmented matrix of the system of equations. [2112−312710]
Then use a series of elementary row operations to convert the matrix into reduced-row echelon form. The three elementary row operations are:
1. Swap the positions of any two rows.
2. Multiply any row by a nonzero scalar.
3. Multiply a row by a nonzero scalar and add it to any other row.
First, switch the rows in the matrix such that the row with the leftmost non-zero entry with the greatest magnitude is at the top.
[121−3121027]
The leading term of row 1 is already 1 so this row does not need to be multiplied by a scalar.
[121−3121027]
Multiply row 1 by scalar −21 and add it to row 2.
[10−316131022]
Multiply row 2 by scalar 136 to make the leading term 1.
[10−3111013132]
Multiply row 2 by scalar 31 and add it to row 1.
[10011317413132]
Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. 1⋅X+0⋅Y=13174X=131740⋅X+1⋅Y=13132Y=13132
Solve by matrix inversion:
In cases where the coefficient matrix of the system of equations is invertible, we can use the inverse to solve the system. Use this method with care as matrix inversion can be numerically unstable for ill-conditioned matrices.
Express the linear equations in the form A×X=B where A is the coefficient matrix, X is the matrix of unknowns, and B is the constant matrix.[2112−31]×[XY]=[2710]
The product of A and its inverse A−1 is the identity matrix. Any matrix multiplied by the identity matrix remains unchanged, so this yields the matrix of unknowns on the left hand side of the equation, and the solution matrix on the right. A×XA−1×A×XI×XX=B=A−1×B=A−1×B=A−1×B
Using a computer algebra system, calculate A−1. [1321361312−133]
Multiply both sides of the equation by the inverse. [1321361312−133]×[2112−31]×[XY]=[1321361312−133]×[2710][XY]=[1317413132]