Question #58060

1 Solve the set of linear equations by Guassian elimination method : a+2b+3c=5, 3a-b+2c=8, 4a-6b-4c=-2. Find c
1
5
4
10

2 Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Sove for b
9
-3
5
-4

3 Solve the set of linear equations by Guassian elimination method : a+2b+3c=5, 3a-b+2c=8, 4a-6b-4c=-2. Find b
4
-5
-3
5

4 Solve the set of linear equations by Guassian elimination method : x+2y+3z=5, 3x-y+2z=8, 4x-6y-4=-2. Find a
-1
4
5
-11

5 Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Sove for a
2
4
7
3

6 For a relation R in A. if , then R is a ______ relation.
transitive
reflexive
symmentric
associative

7 For a relation R in A.[Math Processing Error],it implies that a =b, then R is an ______ relation.
associative
anti- symmetric
complex
transitive

Expert's answer

Answer on Question #58060 – Math – Linear Algebra

Question

1. Solve the set of linear equations by Gaussian elimination method, find cc.


{a+2b+3c=5,3a−b+2c=8,4a−6b−4c=−2.\left\{ \begin{array}{l} a + 2b + 3c = 5, \\ 3a - b + 2c = 8, \\ 4a - 6b - 4c = -2. \end{array} \right.

Solution

(12353−1284−6−4−2)∼(12350−7−7−70−14−16−22)∼(123501110−14−16−22)∼(1235011100−2−8)\left(\begin{array}{cccc} 1 & 2 & 3 & 5 \\ 3 & -1 & 2 & 8 \\ 4 & -6 & -4 & -2 \end{array}\right) \sim \left(\begin{array}{cccc} 1 & 2 & 3 & 5 \\ 0 & -7 & -7 & -7 \\ 0 & -14 & -16 & -22 \end{array}\right) \sim \left(\begin{array}{cccc} 1 & 2 & 3 & 5 \\ 0 & 1 & 1 & 1 \\ 0 & -14 & -16 & -22 \end{array}\right) \sim \left(\begin{array}{ccc} 1 & 2 & 3 & 5 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & -2 & -8 \end{array}\right){−2c=−8,b+c=1,a+2b+3c=5;{c=4,b=−3,a=−1.\left\{ \begin{array}{l} -2c = -8, \\ b + c = 1, \\ a + 2b + 3c = 5; \end{array} \right. \Bigg\{ \begin{array}{l} c = 4, \\ b = -3, \\ a = -1. \end{array}

Answer: 4.

Question

2. Solve the set of linear equations by the matrix method. Solve for bb.


{a+3b+2c=3,2a−b−3c=−8,5a+2b+c=9.\left\{ \begin{array}{l} a + 3b + 2c = 3, \\ 2a - b - 3c = -8, \\ 5a + 2b + c = 9. \end{array} \right.

Solution

(1322−1−3521)(abc)=(3−89)\left(\begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array}\right) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = \left(\begin{array}{c} 3 \\ -8 \\ 9 \end{array}\right)(1322−1−3521)=A;det⁡A=−1−45+8+10−6+6=−28;\left(\begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array}\right) = A; \det A = -1 - 45 + 8 + 10 - 6 + 6 = -28;A11=−1+6=5;A12=−(2+15)=−17;A13=4+5=9;A21=−(3−4)=1;A22=1−10=−9;A23=−(2−15)=13;A31=−9+2=−7;A32=−(−3−4)=7;A33=−1−6=−7.A_{11} = -1 + 6 = 5; \quad A_{12} = -(2 + 15) = -17; \quad A_{13} = 4 + 5 = 9; \quad A_{21} = -(3 - 4) = 1; \quad A_{22} = 1 - 10 = -9; \quad A_{23} = -(2 - 15) = 13; \quad A_{31} = -9 + 2 = -7; \quad A_{32} = -(-3 - 4) = 7; \quad A_{33} = -1 - 6 = -7.A−1=−128(51−7−17−97913−7).A^{-1} = \frac{-1}{28} \left( \begin{array}{ccc} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{array} \right).−128(51−7−17−97913−7)(1322−1−3521)(abc)=−128(51−7−17−97913−7)(3−89);\frac{-1}{28} \left( \begin{array}{ccc} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{array} \right) \left(\begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = \frac{-1}{28} \left( \begin{array}{ccc} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{array} \right) \left(\begin{array}{c} 3 \\ -8 \\ 9 \end{array}\right);(abc)=−128(51−7−17−97913−7)(3−89);\left(\begin{array}{c} a \\ b \\ c \end{array}\right) = \frac{-1}{28} \left( \begin{array}{ccc} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{array} \right) \left(\begin{array}{c} 3 \\ -8 \\ 9 \end{array}\right);(abc)=−128(−5684−140);\left(\begin{array}{c} a \\ b \\ c \end{array}\right) = \frac{-1}{28} \left( \begin{array}{c} -56 \\ 84 \\ -140 \end{array}\right);(abc)=(2−35).\left(\begin{array}{c} a \\ b \\ c \end{array}\right) = \left(\begin{array}{c} 2 \\ -3 \\ 5 \end{array}\right).

Answer: -3.

Question

3. Solve the set of linear equations by Gaussian elimination method, find bb.


{a+2b+3c=5,3a−b+2c=8,4a−6b−4c=−2.\left\{ \begin{array}{l} a + 2b + 3c = 5, \\ 3a - b + 2c = 8, \\ 4a - 6b - 4c = -2. \end{array} \right.

Solution

(12353−1284−6−4−2)∼(12350−7−7−70−14−16−22)∼(123501110−14−16−22)∼(12301100−2).\left( \begin{array}{cccc} 1 & 2 & 3 & 5 \\ 3 & -1 & 2 & 8 \\ 4 & -6 & -4 & -2 \end{array} \right) \sim \left( \begin{array}{cccc} 1 & 2 & 3 & 5 \\ 0 & -7 & -7 & -7 \\ 0 & -14 & -16 & -22 \end{array} \right) \sim \left( \begin{array}{cccc} 1 & 2 & 3 & 5 \\ 0 & 1 & 1 & 1 \\ 0 & -14 & -16 & -22 \end{array} \right) \sim \left( \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \end{array} \right).


Answer: -3.

Question

4. Solve the set of linear equations by Gaussian elimination method, find aa.


{x+2y+3z=5,3x−y+2z=8,4x−6y−4z=−2.\left\{ \begin{array}{l} x + 2y + 3z = 5, \\ 3x - y + 2z = 8, \\ 4x - 6y - 4z = -2. \end{array} \right.

Solution

(12353−1284−6−4−2)∼(12350−7−7−70−14−16−22)∼(123501110−14−16−22)∼(12301100−2).\left( \begin{array}{cccc} 1 & 2 & 3 & 5 \\ 3 & -1 & 2 & 8 \\ 4 & -6 & -4 & -2 \end{array} \right) \sim \left( \begin{array}{cccc} 1 & 2 & 3 & 5 \\ 0 & -7 & -7 & -7 \\ 0 & -14 & -16 & -22 \end{array} \right) \sim \left( \begin{array}{cccc} 1 & 2 & 3 & 5 \\ 0 & 1 & 1 & 1 \\ 0 & -14 & -16 & -22 \end{array} \right) \sim \left( \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \end{array} \right).


Answer: -1.

Question

5. Solve the set of linear equations by the matrix method. Solve for aa.


{a+3b+2c=3,2a−b−3c=−8,5a+2b+c=9.\left\{ \begin{array}{l} a + 3b + 2c = 3, \\ 2a - b - 3c = -8, \\ 5a + 2b + c = 9. \end{array} \right.

Solution

(1322−1−3521)(abc)=(3−89)\left( \begin{array}{rrr} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right) \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right)(1322−1−3521)=A;det⁡A=−1−45+8+10−6+6=−28;\left( \begin{array}{rrr} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right) = A; \det A = -1 - 45 + 8 + 10 - 6 + 6 = -28;A11=−1+6=5;A12=−(2+15)=−17;A13=4+5=9;A21=−(3−4)=1;A22=1−10=−9;A23=−(2−15)=13;A31=−9+2=−7;A32=−(−3−4)=7;A33=−1−6=−7.A_{11} = -1 + 6 = 5; A_{12} = -(2 + 15) = -17; A_{13} = 4 + 5 = 9; A_{21} = -(3 - 4) = 1; A_{22} = 1 - 10 = -9; A_{23} = -(2 - 15) = 13; A_{31} = -9 + 2 = -7; A_{32} = -(-3 - 4) = 7; A_{33} = -1 - 6 = -7.A−1=−128(51−7−17−97913−7).A^{-1} = \frac{-1}{28} \begin{pmatrix} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{pmatrix}.−128(51−7−17−97913−7)(1322−1−3521)(abc)=−128(51−7−17−97913−7)(3−89);\frac{-1}{28} \begin{pmatrix} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{pmatrix} \begin{pmatrix} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \frac{-1}{28} \begin{pmatrix} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{pmatrix} \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix};(abc)=−128(51−7−17−97913−7)(3−89);\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \frac{-1}{28} \begin{pmatrix} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{pmatrix} \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix};(abc)=−128(−5684−140);\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \frac{-1}{28} \begin{pmatrix} -56 \\ 84 \\ -140 \end{pmatrix};(abc)=(2−35).\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}.


Answer: 2.

Question

6. For a relation R in A, if aa is a ______ relation, then:

- transitive

- reflexive

- symmetric

- associative

Answer: the statement of question is not complete.

Question

7. For a relation R in A, [Math Processing Error], it implies that a=ba = b, then R is an ______ relation.

- associative

- anti-symmetric

- complex

- transitive

Answer: anti-symmetric.

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