Answer on Question #59824 – Math – Linear Algebra
Question
Find the solution to the following linear system by using
(I) Cramer rules
(II) Inverse of matrix
(III) Math Lab with two different methods
{ X + 2 Y − Z = 2 , 3 X + Y + Z = 4 , X − Y + Z = 6. \left\{ \begin{array}{l} X + 2Y - Z = 2, \\ 3X + Y + Z = 4, \\ X - Y + Z = 6. \end{array} \right. ⎩ ⎨ ⎧ X + 2 Y − Z = 2 , 3 X + Y + Z = 4 , X − Y + Z = 6. Solution
(I) Using the Cramer rule
Δ = ∣ 1 2 − 1 3 1 1 1 − 1 1 ∣ = 1 ⋅ ∣ 1 1 − 1 1 ∣ − 2 ∣ 3 1 1 1 ∣ + ( − 1 ) ⋅ ∣ 3 1 1 − 1 ∣ = 1 + 1 − 2 ( 3 − 1 ) − ( − 3 − 1 ) = 2 − 4 + 4 = 2 , \Delta = \left| \begin{array}{ccc} 1 & 2 & -1 \\ 3 & 1 & 1 \\ 1 & -1 & 1 \end{array} \right| = 1 \cdot \left| \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right| - 2 \left| \begin{array}{cc} 3 & 1 \\ 1 & 1 \end{array} \right| + (-1) \cdot \left| \begin{array}{cc} 3 & 1 \\ 1 & -1 \end{array} \right| = 1 + 1 - 2(3 - 1) - (-3 - 1) = 2 - 4 + 4 = 2, Δ = ∣ ∣ 1 3 1 2 1 − 1 − 1 1 1 ∣ ∣ = 1 ⋅ ∣ ∣ 1 − 1 1 1 ∣ ∣ − 2 ∣ ∣ 3 1 1 1 ∣ ∣ + ( − 1 ) ⋅ ∣ ∣ 3 1 1 − 1 ∣ ∣ = 1 + 1 − 2 ( 3 − 1 ) − ( − 3 − 1 ) = 2 − 4 + 4 = 2 , Δ X = ∣ 2 2 − 1 4 1 1 6 − 1 1 ∣ = 2 ⋅ 1 ⋅ 1 + 2 ⋅ 1 ⋅ 6 + ( − 1 ) ⋅ 4 ⋅ ( − 1 ) − 6 ⋅ 1 ⋅ ( − 1 ) − 1 ⋅ 4 ⋅ 2 − ( − 1 ) ⋅ 1 ⋅ 2 = 2 + 12 + 4 + 6 − 8 + 2 = 18 , \Delta_X = \left| \begin{array}{ccc} 2 & 2 & -1 \\ 4 & 1 & 1 \\ 6 & -1 & 1 \end{array} \right| = 2 \cdot 1 \cdot 1 + 2 \cdot 1 \cdot 6 + (-1) \cdot 4 \cdot (-1) - 6 \cdot 1 \cdot (-1) - 1 \cdot 4 \cdot 2 - (-1) \cdot 1 \cdot 2 = 2 + 12 + 4 + 6 - 8 + 2 = 18, Δ X = ∣ ∣ 2 4 6 2 1 − 1 − 1 1 1 ∣ ∣ = 2 ⋅ 1 ⋅ 1 + 2 ⋅ 1 ⋅ 6 + ( − 1 ) ⋅ 4 ⋅ ( − 1 ) − 6 ⋅ 1 ⋅ ( − 1 ) − 1 ⋅ 4 ⋅ 2 − ( − 1 ) ⋅ 1 ⋅ 2 = 2 + 12 + 4 + 6 − 8 + 2 = 18 , Δ Y = ∣ 1 2 − 1 3 4 1 1 6 1 ∣ = 1 ⋅ 4 ⋅ 1 + 2 ⋅ 1 ⋅ 1 + ( − 1 ) ⋅ 3 ⋅ 6 − 1 ⋅ 4 ⋅ ( − 1 ) − 6 ⋅ 1 ⋅ 1 − 1 ⋅ 3 ⋅ 2 = 4 + 2 − 18 + 4 − 6 − 6 = − 20 , \Delta_Y = \left| \begin{array}{ccc} 1 & 2 & -1 \\ 3 & 4 & 1 \\ 1 & 6 & 1 \end{array} \right| = 1 \cdot 4 \cdot 1 + 2 \cdot 1 \cdot 1 + (-1) \cdot 3 \cdot 6 - 1 \cdot 4 \cdot (-1) - 6 \cdot 1 \cdot 1 - 1 \cdot 3 \cdot 2 = 4 + 2 - 18 + 4 - 6 - 6 = -20, Δ Y = ∣ ∣ 1 3 1 2 4 6 − 1 1 1 ∣ ∣ = 1 ⋅ 4 ⋅ 1 + 2 ⋅ 1 ⋅ 1 + ( − 1 ) ⋅ 3 ⋅ 6 − 1 ⋅ 4 ⋅ ( − 1 ) − 6 ⋅ 1 ⋅ 1 − 1 ⋅ 3 ⋅ 2 = 4 + 2 − 18 + 4 − 6 − 6 = − 20 , Δ Z = ∣ 1 2 2 3 1 4 1 − 1 6 ∣ = 1 ⋅ 1 ⋅ 6 + 2 ⋅ 4 ⋅ 1 + 2 ⋅ 3 ⋅ ( − 1 ) − 1 ⋅ 1 ⋅ 2 − ( − 1 ) ⋅ 4 ⋅ 1 − 6 ⋅ 2 ⋅ 3 = 6 + + 8 − 6 − 2 + 4 − 36 = − 26. \Delta_Z = \left| \begin{array}{ccc} 1 & 2 & 2 \\ 3 & 1 & 4 \\ 1 & -1 & 6 \end{array} \right| = 1 \cdot 1 \cdot 6 + 2 \cdot 4 \cdot 1 + 2 \cdot 3 \cdot (-1) - 1 \cdot 1 \cdot 2 - (-1) \cdot 4 \cdot 1 - 6 \cdot 2 \cdot 3 = 6 + +8 - 6 - 2 + 4 - 36 = -26. Δ Z = ∣ ∣ 1 3 1 2 1 − 1 2 4 6 ∣ ∣ = 1 ⋅ 1 ⋅ 6 + 2 ⋅ 4 ⋅ 1 + 2 ⋅ 3 ⋅ ( − 1 ) − 1 ⋅ 1 ⋅ 2 − ( − 1 ) ⋅ 4 ⋅ 1 − 6 ⋅ 2 ⋅ 3 = 6 + + 8 − 6 − 2 + 4 − 36 = − 26.
The solution is
X = Δ X Δ = 18 2 = 9 , Y = Δ Y Δ = − 20 2 = − 10 , Z = Δ Z Δ = − 26 2 = − 13. X = \frac{\Delta_X}{\Delta} = \frac{18}{2} = 9, \quad Y = \frac{\Delta_Y}{\Delta} = \frac{-20}{2} = -10, \quad Z = \frac{\Delta_Z}{\Delta} = \frac{-26}{2} = -13. X = Δ Δ X = 2 18 = 9 , Y = Δ Δ Y = 2 − 20 = − 10 , Z = Δ Δ Z = 2 − 26 = − 13.
(II) We write the system of equations in the form
A X ˉ = B , A \bar{X} = B, A X ˉ = B ,
where A = ( 1 2 − 1 3 1 1 1 − 1 1 ) A = \begin{pmatrix} 1 & 2 & -1 \\ 3 & 1 & 1 \\ 1 & -1 & 1 \end{pmatrix} A = ⎝ ⎛ 1 3 1 2 1 − 1 − 1 1 1 ⎠ ⎞ X ˉ = ( X Y Z ) \bar{X} = \begin{pmatrix} X \\ Y \\ Z \end{pmatrix} X ˉ = ⎝ ⎛ X Y Z ⎠ ⎞ B = ( 2 4 6 ) B = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} B = ⎝ ⎛ 2 4 6 ⎠ ⎞
Multiplying the equation A X ˉ = B A\bar{X} = B A X ˉ = B A − 1 A^{-1} A − 1
A − 1 A X ˉ = A − 1 B → X ˉ = A − 1 B A^{-1}A\bar{X} = A^{-1}B \rightarrow \bar{X} = A^{-1}B A − 1 A X ˉ = A − 1 B → X ˉ = A − 1 B
The inverse matrix A − 1 A^{-1} A − 1 A A A
A − 1 = 1 Δ ( A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ) , A^{-1} = \frac{1}{\Delta} \begin{pmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{pmatrix}, A − 1 = Δ 1 ⎝ ⎛ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ⎠ ⎞ ,
where
Δ \Delta Δ A A A
A i j A_{ij} A ij a i j a_{ij} a ij
Δ = ∣ 1 2 − 1 3 1 1 1 − 1 1 ∣ = 1 ⋅ 1 ⋅ 1 + 2 ⋅ 1 ⋅ 1 + ( − 1 ) ⋅ 3 ⋅ ( − 1 ) − 1 ⋅ 1 ⋅ ( − 1 ) − 1 ⋅ 1 ⋅ ( − 1 ) − \Delta = \left| \begin{array}{ccc} 1 & 2 & -1 \\ 3 & 1 & 1 \\ 1 & -1 & 1 \end{array} \right| = 1 \cdot 1 \cdot 1 + 2 \cdot 1 \cdot 1 + (-1) \cdot 3 \cdot (-1) - 1 \cdot 1 \cdot (-1) - 1 \cdot 1 \cdot (-1) - Δ = ∣ ∣ 1 3 1 2 1 − 1 − 1 1 1 ∣ ∣ = 1 ⋅ 1 ⋅ 1 + 2 ⋅ 1 ⋅ 1 + ( − 1 ) ⋅ 3 ⋅ ( − 1 ) − 1 ⋅ 1 ⋅ ( − 1 ) − 1 ⋅ 1 ⋅ ( − 1 ) − − 1 ⋅ 2 ⋅ 3 = 1 + 2 + 3 + 1 + 1 − 6 = 2 ; A 11 = ( − 1 ) 1 + 1 ∣ 1 1 − 1 1 ∣ = 1 + 1 = 2 ; A 12 = ( − 1 ) 1 + 2 ∣ 3 1 1 1 ∣ = − ( 3 − 1 ) = − 2 ; A 13 = ( − 1 ) 1 + 3 ∣ 3 1 1 − 1 ∣ = − 3 − 1 = − 4 ; A 21 = ( − 1 ) 2 + 1 ∣ 2 − 1 − 1 1 ∣ = − ( 2 − 1 ) = − 1 ; A 22 = ( − 1 ) 2 + 2 ∣ 1 − 1 1 1 ∣ = 1 + 1 = 2 ; A 23 = ( − 1 ) 2 + 3 ∣ 1 2 1 − 1 ∣ = − ( − 1 − 2 ) = 3 ; A 31 = ( − 1 ) 3 + 1 ∣ 2 − 1 1 1 ∣ = 2 + 1 = 3 ; A 32 = ( − 1 ) 3 + 2 ∣ 1 − 1 3 1 ∣ = − ( 1 + 3 ) = − 4 ; A 33 = ( − 1 ) 3 + 3 ∣ 1 2 3 1 ∣ = 1 − 6 = − 5. \begin{array}{l}
-1 \cdot 2 \cdot 3 = 1 + 2 + 3 + 1 + 1 - 6 = 2; \\
A_{11} = (-1)^{1+1} \left| \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right| = 1 + 1 = 2; \quad A_{12} = (-1)^{1+2} \left| \begin{array}{cc} 3 & 1 \\ 1 & 1 \end{array} \right| = -(3 - 1) = -2; \\
A_{13} = (-1)^{1+3} \left| \begin{array}{cc} 3 & 1 \\ 1 & -1 \end{array} \right| = -3 - 1 = -4; \quad A_{21} = (-1)^{2+1} \left| \begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array} \right| = -(2 - 1) = -1; \\
A_{22} = (-1)^{2+2} \left| \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right| = 1 + 1 = 2; \quad A_{23} = (-1)^{2+3} \left| \begin{array}{cc} 1 & 2 \\ 1 & -1 \end{array} \right| = -(-1 - 2) = 3; \\
A_{31} = (-1)^{3+1} \left| \begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array} \right| = 2 + 1 = 3; \quad A_{32} = (-1)^{3+2} \left| \begin{array}{cc} 1 & -1 \\ 3 & 1 \end{array} \right| = -(1 + 3) = -4; \\
A_{33} = (-1)^{3+3} \left| \begin{array}{cc} 1 & 2 \\ 3 & 1 \end{array} \right| = 1 - 6 = -5.
\end{array} − 1 ⋅ 2 ⋅ 3 = 1 + 2 + 3 + 1 + 1 − 6 = 2 ; A 11 = ( − 1 ) 1 + 1 ∣ ∣ 1 − 1 1 1 ∣ ∣ = 1 + 1 = 2 ; A 12 = ( − 1 ) 1 + 2 ∣ ∣ 3 1 1 1 ∣ ∣ = − ( 3 − 1 ) = − 2 ; A 13 = ( − 1 ) 1 + 3 ∣ ∣ 3 1 1 − 1 ∣ ∣ = − 3 − 1 = − 4 ; A 21 = ( − 1 ) 2 + 1 ∣ ∣ 2 − 1 − 1 1 ∣ ∣ = − ( 2 − 1 ) = − 1 ; A 22 = ( − 1 ) 2 + 2 ∣ ∣ 1 1 − 1 1 ∣ ∣ = 1 + 1 = 2 ; A 23 = ( − 1 ) 2 + 3 ∣ ∣ 1 1 2 − 1 ∣ ∣ = − ( − 1 − 2 ) = 3 ; A 31 = ( − 1 ) 3 + 1 ∣ ∣ 2 1 − 1 1 ∣ ∣ = 2 + 1 = 3 ; A 32 = ( − 1 ) 3 + 2 ∣ ∣ 1 3 − 1 1 ∣ ∣ = − ( 1 + 3 ) = − 4 ; A 33 = ( − 1 ) 3 + 3 ∣ ∣ 1 3 2 1 ∣ ∣ = 1 − 6 = − 5.
Hence
A − 1 = 1 2 ( 2 − 1 3 − 2 2 − 4 − 4 3 − 5 ) = ( 1 − 1 2 3 2 − 1 1 − 2 − 2 3 2 − 5 2 ) . A^{-1} = \frac{1}{2} \left( \begin{array}{ccc} 2 & -1 & 3 \\ -2 & 2 & -4 \\ -4 & 3 & -5 \end{array} \right) = \left( \begin{array}{ccc} 1 & -\frac{1}{2} & \frac{3}{2} \\ -1 & 1 & -2 \\ -2 & \frac{3}{2} & -\frac{5}{2} \end{array} \right). A − 1 = 2 1 ⎝ ⎛ 2 − 2 − 4 − 1 2 3 3 − 4 − 5 ⎠ ⎞ = ⎝ ⎛ 1 − 1 − 2 − 2 1 1 2 3 2 3 − 2 − 2 5 ⎠ ⎞ .
The solution is given by
( X Y Z ) = A − 1 B = ( 1 − 1 2 3 2 − 1 1 − 2 − 2 3 2 − 5 2 ) ( 2 4 6 ) = ( 1 ⋅ 2 − 1 2 ⋅ 4 + 3 2 ⋅ 6 − 1 ⋅ 2 + 1 ⋅ 4 + ( − 2 ) ⋅ 6 − 2 ⋅ 2 + 3 2 ⋅ 4 − 5 2 ⋅ 6 ) = ( 2 − 2 + 9 − 2 + 4 − 12 − 4 + 6 − 15 ) = = ( 9 − 10 − 13 ) . X = 9 , Y = − 10 , Z = − 13. \begin{array}{l}
\left( \begin{array}{c} X \\ Y \\ Z \end{array} \right) = A^{-1} B = \left( \begin{array}{ccc} 1 & -\frac{1}{2} & \frac{3}{2} \\ -1 & 1 & -2 \\ -2 & \frac{3}{2} & -\frac{5}{2} \end{array} \right) \left( \begin{array}{c} 2 \\ 4 \\ 6 \end{array} \right) = \left( \begin{array}{c} 1 \cdot 2 - \frac{1}{2} \cdot 4 + \frac{3}{2} \cdot 6 \\ -1 \cdot 2 + 1 \cdot 4 + (-2) \cdot 6 \\ -2 \cdot 2 + \frac{3}{2} \cdot 4 - \frac{5}{2} \cdot 6 \end{array} \right) = \left( \begin{array}{c} 2 - 2 + 9 \\ -2 + 4 - 12 \\ -4 + 6 - 15 \end{array} \right) = \\
= \left( \begin{array}{c} 9 \\ -10 \\ -13 \end{array} \right). \\
X = 9, Y = -10, Z = -13.
\end{array} ⎝ ⎛ X Y Z ⎠ ⎞ = A − 1 B = ⎝ ⎛ 1 − 1 − 2 − 2 1 1 2 3 2 3 − 2 − 2 5 ⎠ ⎞ ⎝ ⎛ 2 4 6 ⎠ ⎞ = ⎝ ⎛ 1 ⋅ 2 − 2 1 ⋅ 4 + 2 3 ⋅ 6 − 1 ⋅ 2 + 1 ⋅ 4 + ( − 2 ) ⋅ 6 − 2 ⋅ 2 + 2 3 ⋅ 4 − 2 5 ⋅ 6 ⎠ ⎞ = ⎝ ⎛ 2 − 2 + 9 − 2 + 4 − 12 − 4 + 6 − 15 ⎠ ⎞ = = ⎝ ⎛ 9 − 10 − 13 ⎠ ⎞ . X = 9 , Y = − 10 , Z = − 13.
(III) Using MatLab solution of the system can be obtained by means of the inverse matrix:
To get started, select "MATLAB Help" from the Help menu.
>> A=[1 2 -1 ;3 1 1 ;1 -1 1 ]
A =
1 2 -1
3 1 1
1 -1 1
>> B=[2 ;4 ;6 ]
B =
2
4
6
>> inv(A)*B
ans =
9.0000
-10.0000
-13.0000
>>
We solve the system of equations by Gauss
>> A=[1 2 -1;3 1 1;1 -1 1]; B=[2;4;6]; C=[A B];
>> D=rref(C)
D =
1 0 0 9
0 1 0 -10
0 0 1 -13
>>
The solution of the system of the last column of the matrix D
Answer: X = 9 X = 9 X = 9 Y = − 10 Y = -10 Y = − 10 Z = − 13 Z = -13 Z = − 13
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#339914 on Dec 2023