Find all x in R4 that are mapped into the zero vector by the transformation X↦AX for the given matrix A. A=[1392103−40123−2305]

Question

Find all x in R4 that are mapped into the zero vector by the transformation X↦AX for the given matrix A.  A=[1392103−40123−2305]

May Dunn · Accepted Answer

Step 1 Solve Ax=0 to find all x∈R4.The audmented matrix is [13920103−4001230−23050]. Reduce matrix to reduced row echelon from [1392|0103−4|00123|0−2305|0]R1↔R4R2+1/2R1→R2∼R4+1/2R1→R4[−2305|00323−32|00123|0092992|0] [−2305|00323−32|00123|0092992|0]R2↔R4R3−2/9R2→R3∼R4+1/3R2→R4[−2305|0092992|00002|0000−3|0] [−2305|0092992|00002|0000−3|0]−1/3R4→R4R3−2R4→R3R2−9/2R4→R2[−2305|009290|00000|000010]

memomzungup4 · Accepted Answer

Step 2  [−2305|009290|00000|00001|0]R1−5R4→R1R3↔R42/9R2→R2[−2300|00120|00001|000000]  [−2300|00120|00001|00000|0]R1−3R2→R1−1/2R1→R1[1030|00120|00001|000000].  Terefore, we get  x1+3x3=0⇒x1=−3x3  x1+2x3=0⇒x1=−2x3  x3=x3  x4=0.  That is  x=[x1x2x3x4]=[−3x3−2x3x30]=x3[−3−210]  Hense, the required vectors are  x=x3[−3−210]  where x3∈R4.  Result  x=x3[−3−210]

Vasquez · Accepted Answer

Answer:X={[4x4+9t−2x4−3ttx4]|x4,t∈ℝ}Explanation:Let X=[x1x2x3x4] be the vector in ℝ4. Multiplying A with X, we have:[1392103−40123−2305][x1x2x3x4]=[0000].Performing the matrix multiplication, we get the following system of equations:x1+3x2+9x3+2x4=0x1+3x3−4x4=0x2+2x3+3x4=0−2x1+3x2+5x4=0To solve this system, we can use various methods such as Gaussian elimination or matrix inversion. Let&#039;s use Gaussian elimination here to find the solutions.Converting the system of equations into an augmented matrix, we have:[1392|0103−4|00123|0−2305|0]Performing row operations to obtain the row-echelon form:[1392|00−3−6−6|00123|003189|0]⟹[1392|00123|00−3−6−6|003189|0]⟹[1392|00123|00000|000018|0]From the row-echelon form, we can see that the third and fourth rows correspond to the sameequation. This means there are two free variables in the system. We can choose x2 and x4 as the free variables.Expressing x1 and x3 in terms of the free variables, we have:x1=−3x2−9x3−2x4x3=t (where t is a parameter representing x3)Substituting these expressions back into the equation for x2, we get:x2=−2x4−3tFinally, the general solution for x is:x=[−3x2−9x3−2x4x2x3x4]=[−3(−2x4−3t)−9t−2x4−2x4−3ttx4]=[6x4+9t−2x4−2x4−3ttx4]=[4x4+9t−2x4−3ttx4]where x4 and t can take any real values.Thus, the set of all solutions x is given by:X={[4x4+9t−2x4−3ttx4]|x4,t∈ℝ}

RizerMix · Accepted Answer

To find all the vectors in ℝ⁴ that are mapped to the zero vector by the transformation X → AX, where A is the given matrix:A=[1392103−40123−2305]We need to solve the equation AX = 0.First, let&#039;s write the augmented matrix [A|0] and row reduce it to its echelon form:[13920103−4001230−23050]Row 2 = Row 2 - Row 1:[139200−3−6−6001230−23050]Row 4 = Row 4 + 2 * Row 1:[139200−3−6−6001230091890]Row 4 = Row 4 + 3 * Row 2:[139200−3−6−6001230000−90]Row 3 = Row 3 + (1/3) * Row 2:[139200−3−6−6000010000−90]Row 1 = Row 1 - 9 * Row 3:[130−1600−3−6−6000010000−90]Row 2 = Row 2 + 2 * Row 3:[130−1600−3−60000010000−90]Row 1 = Row 1 + 16 * Row 4:[130000−3−60000010000−90]Row 2 = - (1/3) * Row 2:[130000120000010000−90]Row 1 = Row 1 - 3 * Row 2:[10−6000120000010000−90]Row 1 = Row 1 + 6 * Row 3:[100000120000010000−90]Row 4 = Row 4 + 9 * Row 3:[10000012000001000000]From the row echelon form, we can see that the system has infinitely many solutions. We can write the solution set as:x=[−2t2tt0],t∈ℝTherefore, all the vectors in ℝ⁴ that are mapped to the zero vector by the transformation X → AX are of the form [−2t2tt0], where t is a real number.

Find all x in \mathbb{R}^{4} that are mapped into the zero vector by the t

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