The TI83 family of calculators includes two matrix functions that can be used to perform the row operations needed in the simplex algorithm.
The *row function is found in the list of matrix math operations:
1.
2. Use the right cursor to move to the matrix math menu.
3. Use the up cursor to scroll up from the bottom to E: *row( and tap
This function has three arguments; a number, a matrix, and a row:
*row(multiplier, matrix, row number)
The *row function multiplies the specified row in the specified matrix by the specified multiplier:
*row(3, [A], 1) multiplies the first row in matrix A by 3. *row(5, [B], 3) multiplies the third row of matrix B by 5.
Expressed algebraically:
*row(3, [A], 1) means that in matrix A, row_{1} is replaced by 3*row_{1} (3R_{1} → R_{1}) *row(5, [B], 3) means that in matrix B, row_{3} is replaced by 5*row_{3} (5R_{3} → R_{3})
The *row+ function is also found in the list of matrix math operations and is just below the *row function in the list (it is function F, the last function in the list). This function has four arguments; a number, a matrix, and two rows:
*row+(multiplier, matrix, first row number, second row number)
In the specified matrix, this function multiplies the row indicated by the first row number by the multiplier and adds the results to the corresponding values in the row indicated by the second row number:
*row+(3, [A], 1, 2) multiplies row 1 in matrix A by 3 and adds the results to row 2. *row+(5, [B], 3, 1) multiplies row 3 of matrix B by 5 and adds the results to row 1.
Expressed algebraically:
*row+(3, [A], 1, 2) means that in matrix A, 3R_{1} + R_{2} → R_{2} *row+(5, [B], 3, 1) means that in matrix B, 5R_{3} + R_{1} → R_{1}
In the matrix below, we want to perform a row operation that will yield a value of 1 in the top row of the second column followed by row operations that will yield values of zero in the other rows:
Assuming that this matrix has been entered as matrix A in your TI83 calculator, the correct sequence of row operations is given here:
Replace Row1 with 0.5*Row1: *row(0.5, [A], 1) Replace Row2 with 4*Row1 + Row2: *row+(4, Ans, 1, 2) Replace Row3 with 80*Row1 + Row3: *row+(80, Ans, 1, 3)
Notice that the first row operation is performed on matrix A but each succeeding row operation is performed on the answer from the previous row operation:
Initial State 

*row(0.5, [A], 1) yields 

*row+(4, Ans, 1, 2) yields 

*row+(80, Ans, 1, 3) yields 
The final matrix represents the second tableau:
In the matrix below, we want to perform a row operation that will yield a value of 1 in the second row of the first column followed by row operations that will yield values of zero in the other rows:
Given that the value in the second row of the first column is already 1, we need the following row operations to get a value of 0 in the remaining rows:
Replace Row1 with 0.5*Row2 + Row1: *row+(0.5, Ans, 2, 1) Replace Row3 with 10*Row2 + Row3: *row+(10, Ans, 2, 3)
Here are the stepbystep results:
Initial State 

*row+(0.5, Ans, 2, 1) yields 

*row+(10, Ans, 2, 3) yields 
The final matrix represents the third (and final) tableau: