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Section 7.1 Matrices and Systems of Linear Equations Matrices A matrix is a rectangular array of numbers written within brackets Open books to page 562 Order of a Matrix Order of a Matrix: Row x Column 2 6 4 3 9 8 2 3 4 8 10 3 2 4 2 1 8 2 3 3 4 4 6 5 6 3 rows, 2 columns 3x2 4 rows, 4 columns 1 row, 3 columns 4x4 1x3 Augmented Matrices Matrices can be used as shorthand for systems of equations. When done so, they are called augmented matrices. 3x 4 y 1 x 2y 7 3 4 1 1 2 7 Each row is an equation Vertical line represents the equal sign First column is coefficients on the x Second column is coefficients on the y Constants to the right of the vertical line Any variable not in the equation has an implied coefficient of 0 Write the system as an augmented matrix x 3 y 2z 7 2x 4z 3 0 1 3 2 7 2 0 4 3 yxz 7 x yz 2 1 1 1 7 1 1 1 2 0 1 1 1 z y 1 Row Operations (Solving Systems) Interchange any 2 rows Ri R j Multiply a row by a nonzero constant cRi Ri Add a multiple of 1 row to another cRi R j R j Perform the row operation 2 1 3 0 2 1 0 2 1 2 1 3 R1 R2 Perform the row operation 5 7 1 2 3 2 4 1 6 1 5 4 2R2 R2 5 7 1 2 6 1 5 4 Perform the row operation 1 0 2 3 0 1 2 3 4 2 1 2 R1 2R3 R1 0 1 2 3 4 2 1 2 Perform the row operation 1 0 1 2 3 1 2 3 0 1 3 1 R1 3R2 R1 3 1 2 3 0 1 3 1 Row-Echelon Form of a Matrix Rows consisting entirely of 0’s are at the bottom of the matrix For each row that does not consist entirely of 0’s, the first (leftmost) nonzero entry is 1 (called the leading 1) The leading 1 in each row must have all zeros underneath it. 1 0 0 0 3 5 6 1 2 1 0 1 5 0 0 0 Determine whether the matrices are in Row-Echelon Form 1 2 2 5 0 1 4 6 0 0 1 7 Yes 1 2 6 5 0 1 5 1 1 0 4 7 No 1 12 26 5 0 1 15 3 0 0 0 0 Yes 1 7 8 10 0 4 6 2 0 0 1 7 No Rewrite the Matrix in Row Echelon Form 3 2 2 1 1 3 R1 R2 1 1 3 3 2 2 3R1 R2 R2 1 1 3 0 1 11 1 1 3 0 1 11 1R2 R2 Solve the system using Gaussian Elimination 2 x y 8 x 3y 6 Step 1: Write as an augmented matrix 2 1 8 1 3 6 1 3 6 R1 R2 2 1 8 Step 2: Use row operations to write in row-echelon form. 1 3 6 Need a 0 below the leading 1 in row 1 0 5 20 2R1 R2 R2 …continued 1 3 6 0 5 20 1 3 6 0 1 4 x 3y 6 y4 x 34 6 x 12 6 x 6 Need a leading 1 in row 2 (turn the -5 into a 1) 1 R2 R2 5 Step 3: Write the augmented matrix as a system of equations. Step 4: Back substitute to find all other variables. x 6 y4 Solve the system x y z 2 3 x y z 4 2 x 2 y 3z 3 1 1 1 2 3 1 1 4 2 2 3 3 1 1 1 2 0 2 2 2 0 4 5 7 3R1 R2 R2 and 2 R1 R3 R3 1 1 1 2 0 2 2 2 0 4 5 7 1 1 1 2 0 1 1 1 0 4 5 7 1 1 1 2 0 1 1 1 0 0 1 3 x y z 2 y z 1 z 3 1 R2 R2 2 y 3 1 y2 4R2 R3 R3 x 1 x 2 3 2 x 1 2 x 1 y2 z 3 Infinitely Many and No Solutions 1 2 3 2 0 1 6 4 0 0 0 0 Row 3 equation would say: 0x + 0y + 0z = 0 0=0 Infinitely Many Solutions on a line 1 2 3 2 0 1 6 4 0 0 0 4 Row 3 equation would say: 0x + 0y + 0z = 4 0=4 No Solutions