Question:Let A be the n x n matrix with 0's on the main diagonal, and 1's everywhere else. For an arbitrary vector bin n, solve the linear system Ax=b, expressing the components x1,.......,xnof xin terms of the components of b. See Exercise 69 for the case n=3 .

Short Answer

Expert verified

The solution of the linear system Ax=b isxi=b1+......+bnn-1-bi,i=1,2,....,n .

Step by step solution

01

Consider the system.

IfA is an n x m matrix with row vectorsω1,..........ωnandxis a vector in Rm then, .

Ax=[-ω1-...-ωn-]x=[-ω1.x-...-ωn.x-]

Consider the linear system.

y+z=ay+z=by+y=c

The matrix form of the system is,

011|1101|b110|c

The solution is, x=b+c-a2,y=a+c-b2,z=a+b-c2.

02

Compute the system

Consider the linear system, x1,.......,xnof xxin terms of the components ofb.

x1+x2+.......+xn=b1x1+x2+.......+xn=b1.;x1+x2+.......+xn=bn-1x1+x2+.......+xn=bn

The solution, xi=b1+......+bnn-1-bi.

Where, i=1,2,....,n

Hence, xi=b1+......+bnn-1-bi,i=1,2,....,n is the solution of the linear systemAx=b.

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Explain why a set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) in \({\mathbb{R}^5}\) must be linearly independent when \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent and \({{\mathop{\rm v}\nolimits} _4}\) is not in Span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\).

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