I need to find a unique description of Nul A, namely by listing the vectors that measure the null space.A=[15−4−3101−21000000]

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Benedict · Accepted Answer

First find the general solution of&amp;nbsp;Ax=0&amp;nbsp;in terms of the free variables. Since[A&amp;nbsp;0]=[15−4−31001−2100000000]R1→R1−5R2[106−81001−2100000000]the general solution isx1=−6x3+8x4−x5x2=2x3−x4with&amp;nbsp;x3,x4&amp;nbsp;and&amp;nbsp;x5&amp;nbsp;free. Sox=[x1x2x3x4x5]=x3[−62100]+x4[8−1010]+x5[−10001]and a spanning set for Nul A is{[−62100],[8−1010],[−10001]}Result:{[−62100],[8−1010],[−10001]}&amp;nbsp;

Jazz Frenia · Accepted Answer

To find a unique description of the null space of matrix A, we need to list the vectors that measure the null space. The matrix A is given by:A=[15−4−3101−21000000]To determine the null space, we solve the equation A𝐱=0, where 𝐱 is a vector in the null space and 0 is the zero vector. Writing this in augmented form, we have:[15−4−3101−21000000][x1x2x3x4x5]=[000]Simplifying the matrix equation, we get the following system of equations:x1+5x2−4x3−3x4+x5=0x2−2x3+x4=00=0To find the null space, we solve this system of equations. First, we can express x1, x2, and x5 in terms of the remaining variables:x1=4x3+3x4−x5x2=2x3−x4Now, we can write the general solution for the null space vector 𝐱 as follows:𝐱=[4x3+3x4−x52x3−x4x3x4x5]where x3, x4, and x5 are arbitrary constants.Thus, the unique description of the null space of matrix A is given by the set of vectors 𝐱:𝐱=[4x3+3x4−x52x3−x4x3x4x5]where x3, x4, and x5 can be any real numbers.

Nick Camelot · Accepted Answer

The null space of a matrix A is the set of all solutions 𝐱 to the equation A𝐱=0, where 0 is the zero vector.For the given matrix A=[15−4−3101−21000000], we need to find the null space. Writing the equation in augmented form and reducing the matrix to echelon form, we get:[15−4−3101−21000000]⟶[106−2101−21000000]From this, we can see that the first and fourth columns of the matrix contain the leading variables x1 and x4, respectively. The second and third columns contain the free variables x2 and x3, respectively.Using the leading and free variables, we can express the general solution for the null space vector 𝐱 as:𝐱=[−6x3+2x4−x52x3−x4x3x4x5]=x3[−62100]+x4[2−1010]+x5[−10001]where x3, x4, and x5 are arbitrary constants.Therefore, the null space of matrix A can be described as the span of the set of vectors:{[−62100],[2−1010],[−10001]}which consists of three linearly independent vectors, and thus forms a basis for the null space.

Mr Solver · Accepted Answer

Answer:[−32100] and [2−1010]Explanation:To find a unique description of the null space of matrix A, we need to determine the vectors that measure the null space. Given the matrix:A=[15−4−3101−21000000]We can find the null space of A by solving the equation A𝐱=0, where 𝐱 is a vector in the null space. Let&#039;s write the augmented matrix and perform row reduction to solve for 𝐱:[15−4−31001−2100000000]Performing row operations, we can obtain:[103−21001−2100000000]Now, we can express the solution in terms of free variables. Let x3=t and x4=s, where t and s are arbitrary parameters. Then, the solution vector 𝐱 is given by:𝐱=[−3t+2s2t−sts0]Thus, the null space of matrix A can be expressed as:Null(A)={[−3t+2s2t−sts0]∣t,s∈ℝ}This means that any vector in the null space of A can be written as a linear combination of the vectors [−32100] and [2−1010].

Find an explicit description of Nul A by listing vectors that span the null spac

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