Explanation Procedure used: 1. Form an augmented matrix [ A | I ] where a matrix A is placed along with a same dimension identity matrix I separated by a line. 2. Apply row operations on [ A | I ] in order to obtain another augmented matrix [ I | B ] where I , is again an identity matrix and B is an inverse of A. 3. If the reduction process yields a row of zeros in the left half of the augmented matrix, then the inverse of A does not exist. Calculation: Take A = [ − 1 1 0 − 2 3 − 2 2 − 2 1 ] and X = [ x y z ] and transform [ − 1 1 0 − 2 3 − 2 2 − 2 1 ] [ x y z ] = [ 3 5 8 ] as follows. A − 1 A X = A − 1 [ 3 5 8 ] ( Multiply with A − 1 on RHS of both sides ) I X = A − 1 [ 3 5 8 ] ( ∵ A − 1 A = I ) X = A − 1 [ 3 5 8 ] Obtain the augmented matrix as follows. [ − 1 1 0 1 0 0 − 2 3 − 2 0 1 0 2 − 2 1 0 0 1 ] Apply the row operation − 2 R 1 + R 2 → R 2 and 2 R 1 + R 3 → R 3 and reduce the matrix. [ − 1 1 0 1 0 0 0 1 − 2 − 2 1 0 0 0 1 2 0 1 ] Apply the row operation R 2 + 2 R 3 → R 2 and reduce the matrix

COLLEGE ALGEBRA IN CONTEXT W/ INT. REVIE

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Chapter 7.4, Problem 17E

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To find: The matrix [xyz] such that [−110−23−22−21][xyz]=[358].

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Chapter 7.4, Problem 16E

Chapter 7.4, Problem 18E

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Solve the linear system of equations attached using Gaussian elimination (not Gauss-Jordan) and back subsitution. Remember that: A matrix is in row echelon form if Any row that consists only of zeros is at the bottom of the matrix. The first non-zero entry in each other row is 1. This entry is called aleading 1. The leading 1 of each row, after the first row, lies to the right of the leading 1 of the previous row.

PRIMERA EVALUACIÓN SUMATIVA 10. Determina la medida de los ángulos in- teriores coloreados en cada poligono. ⚫ Octágono regular A 11. Calcula es número de lados qu poligono regular, si la medida quiera de sus ángulos internos • a=156° A= (-2x+80 2 156 180- 360 0 = 24-360 360=24° • a = 162° 1620-180-360 6=18-360 360=19 2=360= 18 12. Calcula las medida ternos del cuadrilá B X+5 x+10 A X+X+ Sx+6 5x=3 x=30 0 лаб • Cuadrilátero 120° 110° • α = 166° 40' 200=180-360 0 = 26-360 360=20 ひ=360 20 18 J 60° ⚫a=169° 42' 51.43" 169.4143180-340 0 = 10.29 54-360 360 10.2857 2=360 10.2857 @Sa

Please help I'm a working mom trying to help my son last minute (6th grader)! Need help with the blank ones and check the ones he got with full calculation so we can use it to study! Especially the mixed number fractions cause I'm rusty. Thanks in advance!

Chapter 7 Solutions

COLLEGE ALGEBRA IN CONTEXT W/ INT. REVIE

Ch. 7.1 - In Exercises 14, solve the systems of equations....Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

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The matrix [ x y z ] such that [ − 1 1 0 − 2 3 − 2 2 − 2 1 ] [ x y z ] = [ 3 5 8 ] .

Solution Summary: The author explains how to form an augmented matrix left[A|Iright], where a matrix A is placed along with an identity matrix and inverse of A. If the reduction process yields

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To find: The matrix [xyz] such that [−110−23−22−21][xyz]=[358].

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Chapter 7 Solutions