(adsbygoogle = window.adsbygoogle || []).push({}); [tex]A = \left[\begin{array}{ccccc} 1 & -1 \\ 2 & 5 \\ 3 & 4 \end{array}\right][/tex]

Let [tex]T_{A}: R^2 \rightarrow R^3[/tex] be the matrix transformation that maps a [tex]2 \times 1[/tex] column vectorxin R^{2}into the [tex]3 \times 1[/tex] column vector Axin R^{3}.

The relationship can be expressed as T_{A}(x) = Ax

Find a vector x in R^{2}whose image under T_{A}is [tex]\left(\begin{array}{ccc}7\\0\\7\end{ar ray}\right)[/tex]

3. The attempt at a solution

If we write this in component form as:

[tex]\left[\begin{array}{ccccc} 1 & -1 \\ 2 & 5 \\ 3 & 4 \end{array}\right][/tex] [tex]\left[\begin{array}{ccccc} x_{1} \\ x_{2} \end{array}\right][/tex] [tex]=\left[\begin{array}{ccccc} 7 \\ 0 \\ 7 \end{array}\right][/tex]

I could solve for [tex]\left[\begin{array}{ccccc} x_{1} \\ x_{2} \end{array}\right][/tex]

However a linear equation in the matrix form "AX = B" can be solved easily for X if "A" is invertible and its inverse is known, you get: X = A^{-1}B

But [tex]A = \left[\begin{array}{ccccc} 1 & -1 \\ 2 & 5 \\ 3 & 4 \end{array}\right][/tex] is NOT square and therefore it has no inverse!

Does anyone know how to find the solution?

Thank you

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# Homework Help: Linear Transformations

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