Solve the following vector equation: 2 5 +4-4-6-7 x1-2+x2 3 +. -3 24 x1 = x2 = x3= How to enter the solution: If the value of a variable is a number, just enter the number. •If the value of a variable is a formula involving free variables, enter the formula. For example, if you obtain that x₁ = 1-2x2 +3x3 where x2 and x3 are free, then as the value of x1 you should enter 1 - 2*x_2 + 3*x_3 . Use the underscore to indicate subscripts of variables, and to indicate multiplication. - . If a variable is a free variable, enter the variable name as its value. For example, if x2 is a free variable, then you should enter X-2 as the value of x2. If the system has no solutions enter None as the value of one of the variables. In such case you can leave values of the remaining variables blank.
Solve the following vector equation: 2 5 +4-4-6-7 x1-2+x2 3 +. -3 24 x1 = x2 = x3= How to enter the solution: If the value of a variable is a number, just enter the number. •If the value of a variable is a formula involving free variables, enter the formula. For example, if you obtain that x₁ = 1-2x2 +3x3 where x2 and x3 are free, then as the value of x1 you should enter 1 - 2*x_2 + 3*x_3 . Use the underscore to indicate subscripts of variables, and to indicate multiplication. - . If a variable is a free variable, enter the variable name as its value. For example, if x2 is a free variable, then you should enter X-2 as the value of x2. If the system has no solutions enter None as the value of one of the variables. In such case you can leave values of the remaining variables blank.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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