- #1
roam
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[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 vector x in R2 into the [tex]3 \times 1[/tex] column vector Ax in R3.
The relationship can be expressed as TA(x) = Ax
Find a vector x in R2 whose image under TA 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-1B
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
Let [tex]T_{A}: R^2 \rightarrow R^3[/tex] be the matrix transformation that maps a [tex]2 \times 1[/tex] column vector x in R2 into the [tex]3 \times 1[/tex] column vector Ax in R3.
The relationship can be expressed as TA(x) = Ax
Find a vector x in R2 whose image under TA 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-1B
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