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The linear transformation T: R3 → R2 is defined by T(x) = AX, where - A= -=[ 0 - 2 6 ]

[10 0 13 ] Find Ker(T). {(-13, 30, 10)} {(-13t, 30t, 10t), t ∈R} {(-13t, 30t, -10t), t∈R} {(13t, 30t, -10t), t∈R}

User KBog
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1 Answer

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To find the kernel (null space) of the linear transformation T:
\displaystyle\sf R^(3)\rightarrow R^(2), defined by
\displaystyle\sf T(x)=Ax, where
\displaystyle\sf A=\begin{bmatrix} 0 & -2 & 6 \\ 10 & 0 & 13 \end{bmatrix}, we need to solve the equation
\displaystyle\sf T(x)=0.

Using
\displaystyle\sf tags for formatting, the matrix equation can be represented as:


\displaystyle\sf A\begin{bmatrix} x_(1) \\ x_(2) \\ x_(3) \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}

Substituting the values of
\displaystyle\sf A, we have:


\displaystyle\sf \begin{bmatrix} 0 & -2 & 6 \\ 10 & 0 & 13 \end{bmatrix}\begin{bmatrix} x_(1) \\ x_(2) \\ x_(3) \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}

To find the kernel, we need to solve the augmented matrix
\displaystyle\sf [A|0] using row reduction techniques:


\displaystyle\sf \begin{bmatrix} 0 & -2 & 6 & 0 \\ 10 & 0 & 13 & 0 \end{bmatrix}

Performing row reduction, we get:


\displaystyle\sf \begin{bmatrix} 10 & 0 & 13 & 0 \\ 0 & -2 & 6 & 0 \end{bmatrix}

Dividing the first row by 10 and the second row by -2, we have:


\displaystyle\sf \begin{bmatrix} 1 & 0 & (13)/(10) & 0 \\ 0 & 1 & -3 & 0 \end{bmatrix}

From the row-reduced form, we can see that
\displaystyle\sf x_(1) and
\displaystyle\sf x_(2) are leading variables, while
\displaystyle\sf x_(3) is a free variable.

Therefore, the kernel (null space) of the transformation T is given by:


\displaystyle\sf \begin{bmatrix} x_(1) \\ x_(2) \\ x_(3) \end{bmatrix}= \begin{bmatrix} -(13)/(10)x_(3) \\ 3x_(3) \\ x_(3) \end{bmatrix}

In set notation, we can represent the kernel as:


\displaystyle\sf \text{Ker}(T)=\left\{ \left( -(13)/(10) t, 3t, t \right) \ \middle|\ t\in \mathbb{R} \right\}

Therefore, the correct option is:
\displaystyle\sf \left( -13t, 30t, 10t \right),\ t\in \mathbb{R}.

User Niraj Chauhan
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