Step 1 The solution of the system can be found using the Gauss Jordan elimination Recall that when using Gauss Jordan elimination the end goal is to obtain an identity matrix for the coefficien matrix by using elementary row operations Notice column 1 already has a 1 and then 2 zeros So next we focus on obtaining a 1 in row 2 column 2 This can be achieved by adding row 3 to row 2 and putting the result in row 2 1133 0419 0 360 H 20 3 1 1 3 37 179 0 360 0 Step 2 Next we focus on column 2 to obtain zeros in the first and third row entries To get a 0 in row 1 column 2 we can add 1 times row 2 to row 1 and put the result in row 1 Also to get a 0 in row 3 column 2 we can add 3 3 times row 2 to row 3 and put the result in row 3 10 4 01 LO O Submit Skip you cannot come mil 79 7 27 27 7 6 9 9 27 Step 3 Now we want to obtain a 1 in the third row last column To do this we need to multiply row 3 by 10 4 6 10 6 01 7 9 01 9 0 0 27 27 00 and put the result in row 3