Answer on Question #58060 – Math – Linear Algebra
Question
1. Solve the set of linear equations by Gaussian elimination method, find c c c
{ a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2. \left\{
\begin{array}{l}
a + 2b + 3c = 5, \\
3a - b + 2c = 8, \\
4a - 6b - 4c = -2.
\end{array}
\right. ⎩ ⎨ ⎧ a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2. Solution
( 1 2 3 5 3 − 1 2 8 4 − 6 − 4 − 2 ) ∼ ( 1 2 3 5 0 − 7 − 7 − 7 0 − 14 − 16 − 22 ) ∼ ( 1 2 3 5 0 1 1 1 0 − 14 − 16 − 22 ) ∼ ( 1 2 3 5 0 1 1 1 0 0 − 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) ⎝ ⎛ 1 3 4 2 − 1 − 6 3 2 − 4 5 8 − 2 ⎠ ⎞ ∼ ⎝ ⎛ 1 0 0 2 − 7 − 14 3 − 7 − 16 5 − 7 − 22 ⎠ ⎞ ∼ ⎝ ⎛ 1 0 0 2 1 − 14 3 1 − 16 5 1 − 22 ⎠ ⎞ ∼ ⎝ ⎛ 1 0 0 2 1 0 3 1 − 2 5 1 − 8 ⎠ ⎞ { − 2 c = − 8 , b + c = 1 , a + 2 b + 3 c = 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} ⎩ ⎨ ⎧ − 2 c = − 8 , b + c = 1 , a + 2 b + 3 c = 5 ; { c = 4 , b = − 3 , a = − 1. Answer: 4.
Question
2. Solve the set of linear equations by the matrix method. Solve for b b b
{ a + 3 b + 2 c = 3 , 2 a − b − 3 c = − 8 , 5 a + 2 b + c = 9. \left\{
\begin{array}{l}
a + 3b + 2c = 3, \\
2a - b - 3c = -8, \\
5a + 2b + c = 9.
\end{array}
\right. ⎩ ⎨ ⎧ a + 3 b + 2 c = 3 , 2 a − b − 3 c = − 8 , 5 a + 2 b + c = 9. Solution
( 1 3 2 2 − 1 − 3 5 2 1 ) ( a b c ) = ( 3 − 8 9 ) \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) ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ ⎝ ⎛ a b c ⎠ ⎞ = ⎝ ⎛ 3 − 8 9 ⎠ ⎞ ( 1 3 2 2 − 1 − 3 5 2 1 ) = 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; ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ = A ; det A = − 1 − 45 + 8 + 10 − 6 + 6 = − 28 ; 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_{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 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 = − 1 28 ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) . A^{-1} = \frac{-1}{28} \left( \begin{array}{ccc}
5 & 1 & -7 \\
-17 & -9 & 7 \\
9 & 13 & -7
\end{array} \right). A − 1 = 28 − 1 ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ . − 1 28 ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) ( 1 3 2 2 − 1 − 3 5 2 1 ) ( a b c ) = − 1 28 ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) ( 3 − 8 9 ) ; \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); 28 − 1 ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ ⎝ ⎛ a b c ⎠ ⎞ = 28 − 1 ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ ⎝ ⎛ 3 − 8 9 ⎠ ⎞ ; ( a b c ) = − 1 28 ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) ( 3 − 8 9 ) ; \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); ⎝ ⎛ a b c ⎠ ⎞ = 28 − 1 ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ ⎝ ⎛ 3 − 8 9 ⎠ ⎞ ; ( a b c ) = − 1 28 ( − 56 84 − 140 ) ; \left(\begin{array}{c}
a \\
b \\
c
\end{array}\right)
=
\frac{-1}{28} \left( \begin{array}{c}
-56 \\
84 \\
-140
\end{array}\right); ⎝ ⎛ a b c ⎠ ⎞ = 28 − 1 ⎝ ⎛ − 56 84 − 140 ⎠ ⎞ ; ( a b c ) = ( 2 − 3 5 ) . \left(\begin{array}{c}
a \\
b \\
c
\end{array}\right)
=
\left(\begin{array}{c}
2 \\
-3 \\
5
\end{array}\right). ⎝ ⎛ a b c ⎠ ⎞ = ⎝ ⎛ 2 − 3 5 ⎠ ⎞ . Answer: -3.
Question
3. Solve the set of linear equations by Gaussian elimination method, find b b b
{ a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2. \left\{ \begin{array}{l} a + 2b + 3c = 5, \\ 3a - b + 2c = 8, \\ 4a - 6b - 4c = -2. \end{array} \right. ⎩ ⎨ ⎧ a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2. Solution
( 1 2 3 5 3 − 1 2 8 4 − 6 − 4 − 2 ) ∼ ( 1 2 3 5 0 − 7 − 7 − 7 0 − 14 − 16 − 22 ) ∼ ( 1 2 3 5 0 1 1 1 0 − 14 − 16 − 22 ) ∼ ( 1 2 3 0 1 1 0 0 − 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). ⎝ ⎛ 1 3 4 2 − 1 − 6 3 2 − 4 5 8 − 2 ⎠ ⎞ ∼ ⎝ ⎛ 1 0 0 2 − 7 − 14 3 − 7 − 16 5 − 7 − 22 ⎠ ⎞ ∼ ⎝ ⎛ 1 0 0 2 1 − 14 3 1 − 16 5 1 − 22 ⎠ ⎞ ∼ ⎝ ⎛ 1 0 0 2 1 0 3 1 − 2 ⎠ ⎞ .
Answer: -3.
Question
4. Solve the set of linear equations by Gaussian elimination method, find a a a
{ x + 2 y + 3 z = 5 , 3 x − y + 2 z = 8 , 4 x − 6 y − 4 z = − 2. \left\{ \begin{array}{l} x + 2y + 3z = 5, \\ 3x - y + 2z = 8, \\ 4x - 6y - 4z = -2. \end{array} \right. ⎩ ⎨ ⎧ x + 2 y + 3 z = 5 , 3 x − y + 2 z = 8 , 4 x − 6 y − 4 z = − 2. Solution
( 1 2 3 5 3 − 1 2 8 4 − 6 − 4 − 2 ) ∼ ( 1 2 3 5 0 − 7 − 7 − 7 0 − 14 − 16 − 22 ) ∼ ( 1 2 3 5 0 1 1 1 0 − 14 − 16 − 22 ) ∼ ( 1 2 3 0 1 1 0 0 − 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). ⎝ ⎛ 1 3 4 2 − 1 − 6 3 2 − 4 5 8 − 2 ⎠ ⎞ ∼ ⎝ ⎛ 1 0 0 2 − 7 − 14 3 − 7 − 16 5 − 7 − 22 ⎠ ⎞ ∼ ⎝ ⎛ 1 0 0 2 1 − 14 3 1 − 16 5 1 − 22 ⎠ ⎞ ∼ ⎝ ⎛ 1 0 0 2 1 0 3 1 − 2 ⎠ ⎞ .
Answer: -1.
Question
5. Solve the set of linear equations by the matrix method. Solve for a a a
{ a + 3 b + 2 c = 3 , 2 a − b − 3 c = − 8 , 5 a + 2 b + c = 9. \left\{ \begin{array}{l} a + 3b + 2c = 3, \\ 2a - b - 3c = -8, \\ 5a + 2b + c = 9. \end{array} \right. ⎩ ⎨ ⎧ a + 3 b + 2 c = 3 , 2 a − b − 3 c = − 8 , 5 a + 2 b + c = 9. Solution
( 1 3 2 2 − 1 − 3 5 2 1 ) ( a b c ) = ( 3 − 8 9 ) \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) ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ ⎝ ⎛ a b c ⎠ ⎞ = ⎝ ⎛ 3 − 8 9 ⎠ ⎞ ( 1 3 2 2 − 1 − 3 5 2 1 ) = 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; ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ = A ; det A = − 1 − 45 + 8 + 10 − 6 + 6 = − 28 ; 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_{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 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 = − 1 28 ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) . A^{-1} = \frac{-1}{28} \begin{pmatrix} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{pmatrix}. A − 1 = 28 − 1 ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ . − 1 28 ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) ( 1 3 2 2 − 1 − 3 5 2 1 ) ( a b c ) = − 1 28 ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) ( 3 − 8 9 ) ; \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}; 28 − 1 ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ ⎝ ⎛ a b c ⎠ ⎞ = 28 − 1 ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ ⎝ ⎛ 3 − 8 9 ⎠ ⎞ ; ( a b c ) = − 1 28 ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) ( 3 − 8 9 ) ; \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}; ⎝ ⎛ a b c ⎠ ⎞ = 28 − 1 ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ ⎝ ⎛ 3 − 8 9 ⎠ ⎞ ; ( a b c ) = − 1 28 ( − 56 84 − 140 ) ; \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \frac{-1}{28} \begin{pmatrix} -56 \\ 84 \\ -140 \end{pmatrix}; ⎝ ⎛ a b c ⎠ ⎞ = 28 − 1 ⎝ ⎛ − 56 84 − 140 ⎠ ⎞ ; ( a b c ) = ( 2 − 3 5 ) . \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}. ⎝ ⎛ a b c ⎠ ⎞ = ⎝ ⎛ 2 − 3 5 ⎠ ⎞ .
Answer: 2.
Question
6. For a relation R in A, if a a a
- 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 = b a = b a = b
- associative
- anti-symmetric
- complex
- transitive
Answer: anti-symmetric.
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#339914 on Dec 2023
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