Answer on Question #55751 – Math – Linear Algebra
6. Solve the set of linear equations by Gaussian elimination method: a + 2 b + 3 c = 5 a + 2b + 3c = 5 a + 2 b + 3 c = 5 3 a − b + 2 c = 8 3a - b + 2c = 8 3 a − b + 2 c = 8 4 a − 6 b − 4 c = − 2 4a - 6b - 4c = -2 4 a − 6 b − 4 c = − 2
4;5;9;10
Solution
a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2 ( 1 2 3 5 3 − 1 2 8 4 − 6 − 4 − 2 ) → 1 ( 1 2 3 5 0 7 3 7 3 7 3 0 7 2 4 22 4 ) → 2 ( 1 2 3 5 0 7 2 7 2 7 2 0 0 − 1 2 − 4 2 ) \begin{array}{l}
a + 2b + 3c = 5, 3a - b + 2c = 8, 4a - 6b - 4c = -2 \\
\left( \begin{array}{rrrr} 1 & 2 & 3 & 5 \\ 3 & -1 & 2 & 8 \\ 4 & -6 & -4 & -2 \end{array} \right) \xrightarrow{1} \left( \begin{array}{rrrr} 1 & 2 & 3 & 5 \\ 0 & \frac{7}{3} & \frac{7}{3} & \frac{7}{3} \\ 0 & \frac{7}{2} & 4 & \frac{22}{4} \end{array} \right) \xrightarrow{2} \left( \begin{array}{rrrr} 1 & 2 & 3 & 5 \\ 0 & \frac{7}{2} & \frac{7}{2} & \frac{7}{2} \\ 0 & 0 & -\frac{1}{2} & -\frac{4}{2} \end{array} \right)
\end{array} a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2 ⎝ ⎛ 1 3 4 2 − 1 − 6 3 2 − 4 5 8 − 2 ⎠ ⎞ 1 ⎝ ⎛ 1 0 0 2 3 7 2 7 3 3 7 4 5 3 7 4 22 ⎠ ⎞ 2 ⎝ ⎛ 1 0 0 2 2 7 0 3 2 7 − 2 1 5 2 7 − 2 4 ⎠ ⎞
1) The second and the third rows are multiplied by − 1 / 3 -1/3 − 1/3 − 1 / 4 -1/4 − 1/4
2) The second row is multiplied by 3 / 2 3/2 3/2
So, we get that: − 1 2 c = − 4 2 → c = 4 -\frac{1}{2}c = -\frac{4}{2} \rightarrow c = 4 − 2 1 c = − 2 4 → c = 4
Answer: c = 4 c = 4 c = 4
7. Solve the set of linear equations by the matrix method: a + 3 b + 2 c = 3 a + 3b + 2c = 3 a + 3 b + 2 c = 3 2 a − b − 3 c = − 8 2a - b - 3c = -8 2 a − b − 3 c = − 8 5 a + 2 b + c = 9 5a + 2b + c = 9 5 a + 2 b + c = 9
9; -3; 5; -4
Solution
Method 1 (matrix method)
Let
A = ( 1 3 2 2 − 1 − 3 5 2 1 ) , X = ( a b c ) , B = ( 3 − 8 9 ) A = \left( \begin{array}{rrr} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right), \quad X = \left( \begin{array}{c} a \\ b \\ c \end{array} \right), \quad B = \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right) A = ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ , X = ⎝ ⎛ a b c ⎠ ⎞ , B = ⎝ ⎛ 3 − 8 9 ⎠ ⎞
The solution of the system of equations is sought in the form
X = A − 1 ∗ B X = A^{-1} * B X = A − 1 ∗ B
Let's find A − 1 A^{-1} A − 1
The determinant of the system is
Δ = ∣ 1 3 2 2 − 1 − 3 5 2 1 ∣ = 1 ∣ − 1 − 3 2 1 ∣ − 3 ∣ 2 − 3 5 1 ∣ + 2 ∣ 2 − 1 5 2 ∣ = 1 ( − 1 + 6 ) − 3 ( 2 + 15 ) + 2 ( 4 + 5 ) \Delta = \left| \begin{array}{rrr} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right| = 1 \quad \left| \begin{array}{rrr} -1 & -3 \\ 2 & 1 \end{array} \right| - 3 \quad \left| \begin{array}{rr} 2 & -3 \\ 5 & 1 \end{array} \right| + 2 \quad \left| \begin{array}{rr} 2 & -1 \\ 5 & 2 \end{array} \right| = 1 (-1 + 6) - 3 (2 + 15) + 2 (4 + 5) Δ = ∣ ∣ 1 2 5 3 − 1 2 2 − 3 1 ∣ ∣ = 1 ∣ ∣ − 1 2 − 3 1 ∣ ∣ − 3 ∣ ∣ 2 5 − 3 1 ∣ ∣ + 2 ∣ ∣ 2 5 − 1 2 ∣ ∣ = 1 ( − 1 + 6 ) − 3 ( 2 + 15 ) + 2 ( 4 + 5 ) Δ = 5 − 51 + 18 = − 28 \Delta = 5 - 51 + 18 = -28 Δ = 5 − 51 + 18 = − 28 A 11 = ( − 1 ) 1 + 1 ∣ − 1 − 3 2 1 ∣ = − 1 + 6 = 5 , A 12 = ( − 1 ) 1 + 2 ∣ 2 − 3 5 1 ∣ = − ( 2 + 15 ) = − 17 , A_{11} = (-1)^{1+1} \left| \begin{array}{rr} -1 & -3 \\ 2 & 1 \end{array} \right| = -1 + 6 = 5, \quad A_{12} = (-1)^{1+2} \left| \begin{array}{rr} 2 & -3 \\ 5 & 1 \end{array} \right| = -(2 + 15) = -17, A 11 = ( − 1 ) 1 + 1 ∣ ∣ − 1 2 − 3 1 ∣ ∣ = − 1 + 6 = 5 , A 12 = ( − 1 ) 1 + 2 ∣ ∣ 2 5 − 3 1 ∣ ∣ = − ( 2 + 15 ) = − 17 , A 13 = ( − 1 ) 1 + 3 ∣ 2 − 1 5 2 ∣ = ( 4 + 5 ) = 9 , A_{13} = (-1)^{1+3} \left| \begin{array}{rr} 2 & -1 \\ 5 & 2 \end{array} \right| = (4 + 5) = 9, A 13 = ( − 1 ) 1 + 3 ∣ ∣ 2 5 − 1 2 ∣ ∣ = ( 4 + 5 ) = 9 , A 21 = ( − 1 ) 2 + 1 ∣ 3 2 2 1 ∣ = − ( 3 − 4 ) = − ( − 1 ) = 1 , A 22 = ( − 1 ) 2 + 2 ∣ 1 2 5 1 ∣ = ( 1 − 10 ) = − 9 , A_{21} = (-1)^{2+1} \left| \begin{array}{rr} 3 & 2 \\ 2 & 1 \end{array} \right| = -(3 - 4) = -(-1) = 1, \quad A_{22} = (-1)^{2+2} \left| \begin{array}{rr} 1 & 2 \\ 5 & 1 \end{array} \right| = (1 - 10) = -9, A 21 = ( − 1 ) 2 + 1 ∣ ∣ 3 2 2 1 ∣ ∣ = − ( 3 − 4 ) = − ( − 1 ) = 1 , A 22 = ( − 1 ) 2 + 2 ∣ ∣ 1 5 2 1 ∣ ∣ = ( 1 − 10 ) = − 9 , A 23 = ( − 1 ) 2 + 3 ∣ 1 3 5 2 ∣ = − ( 2 − 15 ) = − ( − 13 ) = 13 , A_{23} = (-1)^{2+3} \left| \begin{array}{rr} 1 & 3 \\ 5 & 2 \end{array} \right| = -(2 - 15) = -(-13) = 13, A 23 = ( − 1 ) 2 + 3 ∣ ∣ 1 5 3 2 ∣ ∣ = − ( 2 − 15 ) = − ( − 13 ) = 13 , A 31 = ( − 1 ) 3 + 1 ∣ 3 2 − 1 − 3 ∣ = − 9 + 2 = − 7 , A 32 = ( − 1 ) 3 + 2 ∣ 3 2 − 1 − 3 ∣ = − ( − 9 + 2 ) = 7 , A_{31} = (-1)^{3+1} \left| \begin{array}{rr} 3 & 2 \\ -1 & -3 \end{array} \right| = -9 + 2 = -7, \quad A_{32} = (-1)^{3+2} \left| \begin{array}{rr} 3 & 2 \\ -1 & -3 \end{array} \right| = -(-9 + 2) = 7, A 31 = ( − 1 ) 3 + 1 ∣ ∣ 3 − 1 2 − 3 ∣ ∣ = − 9 + 2 = − 7 , A 32 = ( − 1 ) 3 + 2 ∣ ∣ 3 − 1 2 − 3 ∣ ∣ = − ( − 9 + 2 ) = 7 , A 33 = ( − 1 ) 3 + 1 ∣ 1 3 2 − 1 ∣ = − 1 − 6 = − 7 A_{33} = (-1)^{3+1} \left| \begin{array}{rr} 1 & 3 \\ 2 & -1 \end{array} \right| = -1 - 6 = -7 A 33 = ( − 1 ) 3 + 1 ∣ ∣ 1 2 3 − 1 ∣ ∣ = − 1 − 6 = − 7 A ~ = ( A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ) T = ( 5 − 17 9 1 − 9 13 − 7 7 − 7 ) T = ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) \tilde{A} = \left( \begin{array}{rrrr} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array} \right)^T = \left( \begin{array}{rrr} 5 & -17 & 9 \\ 1 & -9 & 13 \\ -7 & 7 & -7 \end{array} \right)^T = \left( \begin{array}{rrr} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{array} \right) A ~ = ⎝ ⎛ A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ⎠ ⎞ T = ⎝ ⎛ 5 1 − 7 − 17 − 9 7 9 13 − 7 ⎠ ⎞ T = ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞
The inverse of A is
A − 1 = 1 Δ A ~ = − 1 28 ( 5 1 − 7 − 17 − 9 7 ) A^{-1} = \frac{1}{\Delta} \tilde{A} = -\frac{1}{28} \begin{pmatrix} 5 & 1 & -7 \\ -17 & -9 & 7 \end{pmatrix} A − 1 = Δ 1 A ~ = − 28 1 ( 5 − 17 1 − 9 − 7 7 ) ( 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 ⎠ ⎞ b = − 1 28 ⋅ ( − 17 ⋅ 3 − 9 ⋅ ( − 8 ) + 7 ⋅ 9 ) = − 84 28 = − 3 b = -\frac{1}{28} \cdot (-17 \cdot 3 - 9 \cdot (-8) + 7 \cdot 9) = -\frac{84}{28} = -3 b = − 28 1 ⋅ ( − 17 ⋅ 3 − 9 ⋅ ( − 8 ) + 7 ⋅ 9 ) = − 28 84 = − 3 Method 2 (Gaussian elimination method)
a + 3 b + 2 c = 3 , 2 a − b − 3 c = − 8 , 5 a + 2 b + c = 9 a + 3b + 2c = 3, \quad 2a - b - 3c = -8, \quad 5a + 2b + c = 9 a + 3 b + 2 c = 3 , 2 a − b − 3 c = − 8 , 5 a + 2 b + c = 9 ( 1 3 2 3 2 − 1 − 3 − 8 5 2 1 9 ) → 1 ( 1 3 2 0 7 7 0 2 2 0 13 5 9 5 ) → 2 ( 1 3 2 0 13 5 13 5 0 0 4 26 5 5 ) \begin{pmatrix} 1 & 3 & 2 & 3 \\ 2 & -1 & -3 & -8 \\ 5 & 2 & 1 & 9 \end{pmatrix} \xrightarrow{1} \begin{pmatrix} 1 & 3 & 2 \\ 0 & 7 & 7 \\ 0 & 2 & 2 \\ 0 & \frac{13}{5} & \frac{9}{5} \end{pmatrix} \xrightarrow{2} \begin{pmatrix} 1 & 3 & 2 \\ 0 & \frac{13}{5} & \frac{13}{5} \\ 0 & 0 & 4 \\ \frac{26}{5} & 5 \end{pmatrix} ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 3 − 8 9 ⎠ ⎞ 1 ⎝ ⎛ 1 0 0 0 3 7 2 5 13 2 7 2 5 9 ⎠ ⎞ 2 ⎝ ⎛ 1 0 0 5 26 3 5 13 0 5 2 5 13 4 ⎠ ⎞
1) The second and the third rows are multiplied by − 1 / 2 -1/2 − 1/2 − 1 / 5 -1/5 − 1/5
2) The second row is multiplied by 26 / 35 26/35 26/35
So, we get that: 4 5 c = 20 5 → c = 5 \frac{4}{5}c = \frac{20}{5} \rightarrow c = 5 5 4 c = 5 20 → c = 5
From the second row we get that: 13 5 b + 13 5 c = 26 5 → 13 5 b + 13 5 5 = 26 5 → 13 5 b + 13 = 26 5 → b = − 3 \frac{13}{5}b + \frac{13}{5}c = \frac{26}{5} \rightarrow \frac{13}{5}b + \frac{13}{5}5 = \frac{26}{5} \rightarrow \frac{13}{5}b + 13 = \frac{26}{5} \rightarrow b = -3 5 13 b + 5 13 c = 5 26 → 5 13 b + 5 13 5 = 5 26 → 5 13 b + 13 = 5 26 → b = − 3
Answer: b = − 3 b = -3 b = − 3
8. Solve the set of linear equations by Gaussian elimination method: a + 2 b + 3 c = 5 a + 2b + 3c = 5 a + 2 b + 3 c = 5 3 a − b + 2 c = 8 3a - b + 2c = 8 3 a − b + 2 c = 8 4 a − 6 b − 4 c = − 2 4a - 6b - 4c = -2 4 a − 6 b − 4 c = − 2
Find b
4; -5; -3; 5
Solution
a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2 a + 2b + 3c = 5, \quad 3a - b + 2c = 8, \quad 4a - 6b - 4c = -2 a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2
From part 6 we got that c = 4 c = 4 c = 4
From the second row we get that 7 2 b + 7 2 c = 7 2 → 7 2 b + 7 2 ⋅ 4 = 7 2 → 7 2 b + 14 = 7 2 → b = − 3 \frac{7}{2}b + \frac{7}{2}c = \frac{7}{2} \rightarrow \frac{7}{2}b + \frac{7}{2} \cdot 4 = \frac{7}{2} \rightarrow \frac{7}{2}b + 14 = \frac{7}{2} \rightarrow b = -3 2 7 b + 2 7 c = 2 7 → 2 7 b + 2 7 ⋅ 4 = 2 7 → 2 7 b + 14 = 2 7 → b = − 3
Answer: b = − 3 b = -3 b = − 3
9. Solve the set of linear equations by Gaussian elimination method: a + 2 b + 3 c = 5 a + 2b + 3c = 5 a + 2 b + 3 c = 5 3 a − b + 2 c = 8 3a - b + 2c = 8 3 a − b + 2 c = 8 4 a − 6 b − 4 c = − 2 4a - 6b - 4c = -2 4 a − 6 b − 4 c = − 2
Find a
-1; 4; 5; -11
Solution
a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2 a + 2b + 3c = 5, \quad 3a - b + 2c = 8, \quad 4a - 6b - 4c = -2 a + 2 b + 3 c = 5 , 3 a − b + 2 c = 8 , 4 a − 6 b − 4 c = − 2
From part 8 we get that b = − 3 b = -3 b = − 3 c = 4 c = 4 c = 4
From the first row of ( 1 2 3 0 7 2 7 2 0 0 − 1 2 ) ∣ 5 7 2 − 4 2 ⟩ \left( \begin{array}{ccc}1 & 2 & 3\\ 0 & \frac{7}{2} & \frac{7}{2}\\ 0 & 0 & -\frac{1}{2} \end{array} \right)\left| \begin{array}{c}5\\ \frac{7}{2}\\ -\frac{4}{2} \end{array} \right\rangle ⎝ ⎛ 1 0 0 2 2 7 0 3 2 7 − 2 1 ⎠ ⎞ ∣ ∣ 5 2 7 − 2 4 ⟩
1 a + 2 b + 3 c = 5 → a + 2 ∗ ( − 3 ) + 3 ∗ 4 = 5 → a − 6 + 12 = 5 → a = − 1. 1a + 2b + 3c = 5 \rightarrow a + 2 * (-3) + 3 * 4 = 5 \rightarrow a - 6 + 12 = 5 \rightarrow a = -1. 1 a + 2 b + 3 c = 5 → a + 2 ∗ ( − 3 ) + 3 ∗ 4 = 5 → a − 6 + 12 = 5 → a = − 1.
Answer: a = − 1 a = -1 a = − 1
10. Solve the set of linear equations by the matrix method: a + 3 b + 2 c = 3 a + 3b + 2c = 3 a + 3 b + 2 c = 3 2 a − b − 3 c = − 8 2a - b - 3c = -8 2 a − b − 3 c = − 8 5 a + 2 b + c = 9 5a + 2b + c = 9 5 a + 2 b + c = 9
Solution
a + 3 b + 2 c = 3 , 2 a − b − 3 c = − 8 , 5 a + 2 b + c = 9 a + 3b + 2c = 3,2a - b - 3c = -8,5a + 2b + c = 9 a + 3 b + 2 c = 3 , 2 a − b − 3 c = − 8 , 5 a + 2 b + c = 9
Let
A = ( 1 3 2 2 − 1 − 3 5 2 1 ) , X = ( a b c ) , B = ( 3 − 8 9 ) A = \left( \begin{array}{c c c} 1 & 3 & 2 \\ 2 & - 1 & - 3 \\ 5 & 2 & 1 \end{array} \right), X = \left( \begin{array}{c} a \\ b \\ c \end{array} \right), B = \left( \begin{array}{c} 3 \\ - 8 \\ 9 \end{array} \right) A = ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ , X = ⎝ ⎛ a b c ⎠ ⎞ , B = ⎝ ⎛ 3 − 8 9 ⎠ ⎞
The solution of the system of equations is sought in the form
X = A − 1 ∗ B X = A ^ {- 1} * B X = A − 1 ∗ B
Let's find A − 1 A^{-1} A − 1
The determinant of the system is
Δ = ∣ 1 3 2 2 − 1 − 3 5 2 1 ∣ = 1 ∣ − 1 − 3 2 1 ∣ − 3 ∣ 2 − 3 5 1 ∣ + 2 ∣ 2 − 1 5 2 ∣ = 1 ( − 1 + 6 ) − 3 ( 2 + 15 ) + 2 ( 4 + 5 ) \Delta = \left| \begin{array}{c c c} 1 & 3 & 2 \\ 2 & - 1 & - 3 \\ 5 & 2 & 1 \end{array} \right| = 1 \left| \begin{array}{c c} - 1 & - 3 \\ 2 & 1 \end{array} \right| - 3 \left| \begin{array}{c c} 2 & - 3 \\ 5 & 1 \end{array} \right| + 2 \left| \begin{array}{c c} 2 & - 1 \\ 5 & 2 \end{array} \right| = 1 (- 1 + 6) - 3 (2 + 15) + 2 (4 + 5) Δ = ∣ ∣ 1 2 5 3 − 1 2 2 − 3 1 ∣ ∣ = 1 ∣ ∣ − 1 2 − 3 1 ∣ ∣ − 3 ∣ ∣ 2 5 − 3 1 ∣ ∣ + 2 ∣ ∣ 2 5 − 1 2 ∣ ∣ = 1 ( − 1 + 6 ) − 3 ( 2 + 15 ) + 2 ( 4 + 5 ) Δ = 5 − 51 + 18 = − 28 \Delta = 5 - 5 1 + 1 8 = - 2 8 Δ = 5 − 51 + 18 = − 28 A 11 = ( − 1 ) 1 + 1 ∣ − 1 − 3 2 1 ∣ = − 1 + 6 = 5 , A 12 = ( − 1 ) 1 + 2 ∣ 2 − 3 5 1 ∣ = − ( 2 + 15 ) = − 17 , A _ {1 1} = (- 1) ^ {1 + 1} \left| \begin{array}{c c} - 1 & - 3 \\ 2 & 1 \end{array} \right| = - 1 + 6 = 5, \quad A _ {1 2} = (- 1) ^ {1 + 2} \left| \begin{array}{c c} 2 & - 3 \\ 5 & 1 \end{array} \right| = - (2 + 1 5) = - 1 7, A 11 = ( − 1 ) 1 + 1 ∣ ∣ − 1 2 − 3 1 ∣ ∣ = − 1 + 6 = 5 , A 12 = ( − 1 ) 1 + 2 ∣ ∣ 2 5 − 3 1 ∣ ∣ = − ( 2 + 15 ) = − 17 , A 13 = ( − 1 ) 1 + 3 ∣ 2 − 1 5 2 ∣ = ( 4 + 5 ) = 9 , A _ {1 3} = (- 1) ^ {1 + 3} \left| \begin{array}{c c} 2 & - 1 \\ 5 & 2 \end{array} \right| = (4 + 5) = 9, A 13 = ( − 1 ) 1 + 3 ∣ ∣ 2 5 − 1 2 ∣ ∣ = ( 4 + 5 ) = 9 , A 21 = ( − 1 ) 2 + 1 ∣ 3 2 2 1 ∣ = − ( 3 − 4 ) = − ( − 1 ) = 1 , A 22 = ( − 1 ) 2 + 2 ∣ 1 2 5 1 ∣ = ( 1 − 10 ) = − 9 , A _ {2 1} = (- 1) ^ {2 + 1} \left| \begin{array}{c c} 3 & 2 \\ 2 & 1 \end{array} \right| = - (3 - 4) = - (- 1) = 1, A _ {2 2} = (- 1) ^ {2 + 2} \left| \begin{array}{c c} 1 & 2 \\ 5 & 1 \end{array} \right| = (1 - 1 0) = - 9, A 21 = ( − 1 ) 2 + 1 ∣ ∣ 3 2 2 1 ∣ ∣ = − ( 3 − 4 ) = − ( − 1 ) = 1 , A 22 = ( − 1 ) 2 + 2 ∣ ∣ 1 5 2 1 ∣ ∣ = ( 1 − 10 ) = − 9 , A 23 = ( − 1 ) 2 + 3 ∣ 1 3 5 2 ∣ = − ( 2 − 15 ) = − ( − 13 ) = 13 , A _ {2 3} = (- 1) ^ {2 + 3} \left| \begin{array}{c c} 1 & 3 \\ 5 & 2 \end{array} \right| = - (2 - 1 5) = - (- 1 3) = 1 3, A 23 = ( − 1 ) 2 + 3 ∣ ∣ 1 5 3 2 ∣ ∣ = − ( 2 − 15 ) = − ( − 13 ) = 13 , A 31 = ( − 1 ) 3 + 1 ∣ 3 2 − 1 − 3 ∣ = − 9 + 2 = − 7 , A 32 = ( − 1 ) 3 + 2 ∣ 3 2 − 1 − 3 ∣ = − ( − 9 + 2 ) = 7 , A _ {3 1} = (- 1) ^ {3 + 1} \left| \begin{array}{c c} 3 & 2 \\ - 1 & - 3 \end{array} \right| = - 9 + 2 = - 7, A _ {3 2} = (- 1) ^ {3 + 2} \left| \begin{array}{c c} 3 & 2 \\ - 1 & - 3 \end{array} \right| = - (- 9 + 2) = 7, A 31 = ( − 1 ) 3 + 1 ∣ ∣ 3 − 1 2 − 3 ∣ ∣ = − 9 + 2 = − 7 , A 32 = ( − 1 ) 3 + 2 ∣ ∣ 3 − 1 2 − 3 ∣ ∣ = − ( − 9 + 2 ) = 7 , A 33 = ( − 1 ) 3 + 1 ∣ 1 3 2 − 1 ∣ = − 1 − 6 = − 7 A _ {3 3} = (- 1) ^ {3 + 1} \left| \begin{array}{c c} 1 & 3 \\ 2 & - 1 \end{array} \right| = - 1 - 6 = - 7 A 33 = ( − 1 ) 3 + 1 ∣ ∣ 1 2 3 − 1 ∣ ∣ = − 1 − 6 = − 7 A ~ = ( A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ) T = ( 5 − 17 9 1 − 9 13 − 7 7 − 7 ) T = ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) \tilde {A} = \left( \begin{array}{c c c} A _ {1 1} & A _ {1 2} & A _ {1 3} \\ A _ {2 1} & A _ {2 2} & A _ {2 3} \\ A _ {3 1} & A _ {3 2} & A _ {3 3} \end{array} \right) ^ {T} = \left( \begin{array}{c c c} 5 & - 1 7 & 9 \\ 1 & - 9 & 1 3 \\ - 7 & 7 & - 7 \end{array} \right) ^ {T} = \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 7 \end{array} \right) A ~ = ⎝ ⎛ A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ⎠ ⎞ T = ⎝ ⎛ 5 1 − 7 − 17 − 9 7 9 13 − 7 ⎠ ⎞ T = ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ A ~ = ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) \tilde {A} = \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 7 \end{array} \right) A ~ = ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞
The inverse of A is
A − 1 = 1 Δ A ~ = − 1 28 ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) A ^ {- 1} = \frac {1}{\Delta} \tilde {A} = - \frac {1}{2 8} \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 7 \end{array} \right) A − 1 = Δ 1 A ~ = − 28 1 ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ ( 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}{2 8} \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 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 = − 1 28 ∗ ( 5 ∗ 3 − 8 − 7 ∗ 9 ) = 56 28 = 2. a = - \frac {1}{28} * (5 * 3 - 8 - 7 * 9) = \frac {56}{28} = 2. a = − 28 1 ∗ ( 5 ∗ 3 − 8 − 7 ∗ 9 ) = 28 56 = 2.
Answer: a = 2 a = 2 a = 2
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#339914 on Dec 2023