Describe all solutions of Ax=0 in parametric vector form, where A is row equivalent to the given matrix. [130−4260−8]

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Describe all solutions of Ax=0 in parametric vector form, where A is row equivalent to the given matrix.  [130−4260−8]

Nichole Watt · Accepted Answer

Step 1  The argument matrix is   [130−40260−80]  Reduce matrix to row echelon form   [130−40260−80] R1↔R2 [260−8000000]  [260−8000000]12 R1↔R2 [130−4000000]  Thus we get  x1+3x2−4x4=0⇒x1=−3x2+4x4  Hence, all solution of Ax=0 parametric vector form is  x=[x1x2x3x4]=[−3x2+4x4x2x3x4]=[−3x2x200]+[00x30]+[4x400x4]=x2[−3100]+x3[0010]+x4[4001]  Result:  x=x2[−3100]+x3[0010]+x4[4001]

Don Sumner · Accepted Answer

Row operations:1. Replace R2 with R2 - 2R1.The resulting row-echelon form is [130−40000].Let&#039;s denote the variables as x1, x2, x3, and x4 for the corresponding columns.Since the second row consists of all zeros, it represents the equation 0x1+0x2+0x3+0x4=0, which simplifies to 0=0. This equation is always true, indicating that we have a free variable.Therefore, the system of equations has infinitely many solutions, and we can express the solutions in parametric vector form as follows:x1&amp;amp;=−3x3+4x4x2&amp;amp;=x3x3&amp;amp;=x3x4&amp;amp;=x4Using the parametric variable t for x3 and x4, the solutions can be written as:x1&amp;amp;=−3t+4sx2&amp;amp;=tx3&amp;amp;=tx4&amp;amp;=s where s and t are arbitrary real numbers.

RizerMix · Accepted Answer

To find all the solutions of the equation Ax=0, where A is row equivalent to the given matrix [130−4260−8], we can perform row operations on A to obtain its row echelon form. The row echelon form of A can then be used to describe the solutions in parametric vector form.Performing row operations on A:[130−4260−8]→R2−2R1[130−40000]The resulting row echelon form shows that the second row consists of all zeros. This means that the second equation in the system is redundant and can be ignored. We are left with a single equation x1+3x2−4x4=0.To express the solutions in parametric vector form, we introduce parameters for the free variables. In this case, x2 and x4 are free variables. We can choose them to be x2=t and x4=s, where t and s are arbitrary real numbers.Now, we can express the solution vector x in terms of the parameters t and s:x1=−3t+4sx2=tx3=0x4=sTherefore, the solutions to the equation Ax=0 in parametric vector form are:[x1x2x3x4]=[−3t+4st0s], where t and s are arbitrary real numbers.

Vasquez · Accepted Answer

Step 1: Swap the first and second rows.[260−8130−4]Step 2: Subtract twice the first row from the second row.[260−80000]Step 3: Divide the first row by 2.[130−40000]Now, we can write the system of equations corresponding to the row-echelon form of the matrix:x+3y−4w=00=0Notice that the second equation 0=0 doesn&#039;t provide any additional information since it is always satisfied. Thus, we can disregard it.Now, we express the variables in terms of the free variable y:x=−3y+4wy=yw=wWe can write the solutions in parametric vector form using the free variable y as a parameter:[xyw]=[−3y+4wyw]=y[−310]+w[401]Therefore, the set of all solutions of Ax=0 can be expressed as:x=−3y+4wy=yw=w

Describe all solutions of Ax=0 in parametric vector form, where A is row equival

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