Consider the linear transformation T: ℝ^2 →ℝ^3. If T

Consider the linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$. If $T\begin{bmatrix} -3 \ -5 \end{bmatrix} = \begin{bmatrix} 4 \ -3 \ 5 \end{bmatrix}$ and $T\begin{bmatrix} 2 \ 3 \end{bmatrix} = \begin{bmatrix} 4 \ 4 \ -1 \end{bmatrix}$, find the matrix $A$ such that $T(\vec{x}) = A\vec{x}$. A =

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00:01 In this question they given that that is the linear transformation.

00:08 So linear transformation by using this transformation here we have to find out the first one matrix a and egan values and the third one egan vectors.

00:32 So let come to the solution here.

00:37 So from the given information they given that that is the linear transformation here so which is t map r square extends to r square so let us assume that here that is t of xy is equal to x minus y so let us see the graph for this linear transformation so this is the graph for linear transformation so this one we can write as a here so t of one zero is equal to one zero here so this can be we can write like so 1 multiply by 1 0 here so that therefore here we have to take that is lambda is equal to 1 is egan value and x is equal to 1 which is egan vector at lambda is equal to 1 here so that let us assume that the first one equal equal to 1 1 1 minus that is lambda is equal to 1 comma minus 1 and the second one egan vectors are x is equal to 1 0 comma y is equal to here 0 1 so 0 1 so these are the again values and egan vectors so from this here we have to assume that that is let so beta is equal to set of 1 0 comma 0 1 be the pair of which is paid for which is paid for for r square so for this reason t of one zero is equal to one zero so which is equal to one x x it will change it to zero one so that therefore the matrices of a here so the matrix for t is is equal to have to consider that that is equal to here 1 0 minus 1.

03:16 So this is the matrices and this is the first question answer.

03:21 That means the matrix of a here.

03:25 So now let come to the eigen values...

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