8.1- one - E?— l Pic Ht waif r” H 8.1 Matrix Solutions...

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Unformatted text preview: E?— l:. Pic Ht waif} r” H 8.1 Matrix Solutions to Linear Systems In Math, a matrix (plural matrices), is a rectangular array of numbers, symbols, or expressions. The individual items in a matrix are called its elements or entries. An ‘3 31] 2 4- Matrices are used to solve systems of linear equations. The first step in solving a system of linear equations using matrices is to Write the augmented matrix. example of a matrix with six elements is [ Write the augmented matrix for a linear system. “An augmented matrix has a vertical bar separating the columns of the matrix into two groups. The coefficient of each variable are placed to the left of the vertical line and the constants of each variable are placed to the left of the vertical line and the constants are placed to the right. If any variable is missing its coefficient is 0 p I _..A '1 A, 34 2 :2 s 2 e ‘ r .e- "i ' ‘6) t w w 3'.) '2 r i 2' L ' F \2 l t" it i l l , {Drama/mid: 1? -” l 5L r —3 aw—i—sx—Bag +%:51 Z: 5 a 3Xl+a r; q- ::> 5 L D to x +5“ a ’ mi 5 0 la H... v : 5“ -5‘ —2— O bee—“57 —-13 :{ DTtLMékl; F Li an] :3” l i 6 Max 4 UL w} “H53; 'f" 2 1’33 i 1—} o *5 (35' :3 Mimi-JZLS 5 O C “l L’ r 71 Z : L}, Perform matrix row operations. Two matrices are row equivalent if one can be obtained from the other by a sequence of row operations. Matrix Row Operations The following row operations produce matrices that represent systems with the same solution set: 1. Two rows of a matrix may be interchanged .This is the same as interchanging two equations in a linear system. 2. The elements in any row may be multiplied by a nonzero number .This is the same as multiplying both sides of an equation by a non -zero number. 3. The elements of any row may be multiplied by a non— zero number and these products may be added to the corresponding elements in any other row. This is the same as multiplying both sides of the equation by a nonzero number and then adding equations to eliminate a variable. A matrix with “1’s” down the main diagonal and “0’s” below the ones is said to be in row—echelon form. We use row operations on the augmented matrix .These row operations are just like what you did when solving a linear system by the addition method. Perform each matrix row operation and write the new matrix. 1 —2 4 —1 2 3 —2 3 —2RI+R3 5“ 0—1—3 2 PVCbiiemfilLf % 4-42 6 ('7 "ER; g —;f A a I 441; 0 3:) l -% Lr 0 a o 7 it a o It Using matrices and Gaussian elimination to solve systems. Solving linear systems using Gaussian Elimination: 1. Write the augmented matrix for the system 2. Use matrix row operations to simplify the matrix to a row—equivalent matrix in row echelon form, with ones down the main diagonal from upper left to lower right, and zeros below the ones. 3. Write the systems of linear equations corresponding to the matrix in step 2 4. Use back substitution to find the systems solutions. Use Gaussian elimination with back-substitution to solve the system. 2x—y+3=—2 & -l 1 ’ml -x+y-22=—3 :3 —: i «a ’3 J 3x—2y+z=—5 3 _9‘ i *5” \jr % 2:: P8“ I “"l "5 L l —- "'"1 . _ w?) [5) -— Rplh-HRL m; p ._;_ was El 3 «.1 I "-5- Elwci : fig 3 :1“ R P l i o w—l *5“ -H -—i-‘ r—~ a”! a c n "6 u; «-—‘-Z:«-b 3 -—2., l *5 a x ._ 7%?) 2 ~13" I o “l "L’ R3 "L HgRl-Hafi o l "3 “" 8 : —?—\\ Hi "*9 o 1 <3 ‘8 \_ t ,_.. v R .w—JAR l 0 -l he“ 3 ‘* 2_., 3 0 l “3 we 8 L ’ I I; Use matrices and Gauss—Jordan elimination to solve systems. Solving linear systems, using Gauss —Jordan Elimination: 1. Write the augmented matrix for the system. 2. Use matrix row operation to simplify the matrix to a row—equivalent matrix in reduced row-echelon form, with ones down the main diagonal from upper left to lower right, and zeros above and below the one. (a) Get one in the upper left-hand comer. (b) Use the one in the first column to get zeros below it. (c) Get one in the second row, second column. (d) Use the one in the second column to make the remaining entries in the second column zero. (e) Get one in the third row, third column. (f) Use the one in the third column to make the remaining entries in the third column zero. (g) Continue this procedure as far as possible 3. Use the reduced row-echelon form of the matrix in step 2 to write the system’s solution set. (Back-substitution is not necessary.) Use Gauss—Jordan elimination to solve the s stem. ... r 13 y i o “9/4 [If — ‘ “‘“ l ’3 ,h .' 3c] '- x+2y—z—«13 :) :20} i [3 D I all? ~2x+3y+z=13 n 2— ! { r I —x+4y+2z=13 I LIL 3 7 ‘ r r: i ‘ ‘- Rzi'.‘ ilk-Hal. :: i F: g? ' _ K t -' " ff ._. y . . _- w, n 2 ’3 Rt: ZR3'*‘R| O l «41‘; 5‘?) D Find the quadratic function W) 26112 +bx+c for which m) :22, f(—2)=17, And f(1)=2. 7%) a (31% 5(B)+C Ha) ; q L~l)l+ H.134— C 3%: aq+35+¢ [email protected] '74 : 4Q-—1L>+C ,[email protected] HI): aLIJZ+E>Ci)+C_ a: CL—l‘LI-FC- “€393 R5: EAR?) 51 3 I 13‘ I 0 will? firs/’8‘ W o I 2—9113 ~‘Ht% 1‘ 0 a I . t V i L ig x7 --'1 1% O O I I r 1 l ‘ (39:3 _«- __ ~"\ an (1 R 2R “R ’ 5 (3 ‘I _ 3 I 3 0 [‘3 19 -l} C :— I C) 'L *- a'f‘ } - _ )— RlzLRL ' *5“ "4 ’9» )«fm: 3x_ax-+| 1% 0 i _— // D we 45» . 'f I I \‘ RI: 5R1+R1 t p [I],) 4/” 0 ’ 291” "Wis O -—é $8, Lf ...
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