"Convert the following system of linear equations into an augmented matrix and then solve it"x1+x2−x3−x4=72x1−x2+3x3+x4=1x1−5x2+9x3−x4=3My approach: So the augmented matrix, A, isaugA=[11−1−172−13111−59−13]Then we perform elementary row operations to reduce the matrix into row echelon form:B=[11−1−170−353−130−6100−4]C=[11−1−170−353−13000−622]Matrix C is our system of linear equations in echelon form. From the last row of matrix C,−6x4=22x4=−113Now we let x3=t for t is any real numberThen we solve the two equations from the first two rows of matrix C,x2=13(−2−5t)x1=83t+223SO this should be the solutions to the equations. However, according to the answer key the solution is the empty set. I inputted the regular matrix into a determinant calculator and indeed the determinant was zero. However, I managed to get a solution to the system of linear equations. Can someone tell me where I went wrong?

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&quot;Convert the following system of linear equations into an augmented matrix and then solve it&quot;x1+x2−x3−x4=72x1−x2+3x3+x4=1x1−5x2+9x3−x4=3My approach: So the augmented matrix, A, isaugA=[11−1−172−13111−59−13]Then we perform elementary row operations to reduce the matrix into row echelon form:B=[11−1−170−353−130−6100−4]C=[11−1−170−353−13000−622]Matrix C is our system of linear equations in echelon form. From the last row of matrix C,−6x4=22x4=−113Now we let x3=t for t is any real numberThen we solve the two equations from the first two rows of matrix C,x2=13(−2−5t)x1=83t+223SO this should be the solutions to the equations. However, according to the answer key the solution is the empty set. I inputted the regular matrix into a determinant calculator and indeed the determinant was zero. However, I managed to get a solution to the system of linear equations. Can someone tell me where I went wrong?

Jowiszowy9zb · Accepted Answer

To answer the more general question of when a matrix has no solution:  A system of equations can have one of three things: a unique solution, infinitely many solutions, and no solution.  Case One: unique solution  An augmented matrix has an unique solution when the equations are all consistent and the number of variables is equal to the number of rows. Simply put if the non-augmented matrix has a nonzero determinant, then it has a solution given by x→=A−1b→.  Case Two: Infinitely many solutions  The number of rows is less than the number of variables. Think of it this way, we need one equation to solve for one unknown variable, two equations to solve for two variables, three equations to solve for three variables, and so on... The number of rows represents(at most) the number of independent equations we have so if its

"Convert the following system of linear equations into an augmented matrix and then solve it"

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