How Do You Solve a Non-Square Linear Transformation Equation?

[roam]

A = \left[\begin{array}{ccccc} 1 & -1 \\ 2 & 5 \\ 3 & 4 \end{array}\right]

Let T_{A}: R^2 \rightarrow R^3 be the matrix transformation that maps a 2 \times 1 column vector x in R² into the 3 \times 1 column vector Ax in R³.

The relationship can be expressed as T_A(x) = Ax


Find a vector x in R² whose image under T_A is \left(\begin{array}{ccc}7\\0\\7\end{ar ray}\right)



3. The Attempt at a Solution

If we write this in component form as:

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


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

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

Does anyone know how to find the solution?

Thank you

 

[HallsofIvy]

A = \left[\begin{array}{ccccc} 1 & -1 \\ 2 & 5 \\ 3 & 4 \end{array}\right]

Let T_{A}: R^2 \rightarrow R^3 be the matrix transformation that maps a 2 \times 1 column vector x in R² into the 3 \times 1 column vector Ax in R³.

The relationship can be expressed as T_A(x) = Ax


Find a vector x in R² whose image under T_A is \left(\begin{array}{ccc}7\\0\\7\end{ar ray}\right)



3. The Attempt at a Solution

If we write this in component form as:

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


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

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

Does anyone know how to find the solution?

Thank you

n equations in m unknowns with n> m MAY not have a solution at all since a linear transformation from an m dimensional vector space to another vector space maps it into an at most m dimensional subset of the second vector space. That's why there is no inverse matrix. There will be a solution only if the given vector, here <7 0 7> happens to lie in that subspace. In any case, finding an inverse matrix is seldom the best method of solving a matrix equation anyway.

I think the simplest thing to do for as simple a system as this is just write out the equations themselve rather that use matrices:
x_1- x_2= 7, 2x_1+ 5x_2= 0, and 3x_2+ 4x_2= 7
The first equation says that x_1=7+ x_2. Putting that into the second equation 2(7+ x_2)+ 5x_2= 14+ 7x_2= 0 so x_2= -2 and then x_1= 7- 2= 5. Now check to see whether they satisfy the third equation: 3x_1+ 4x_2= 3(5)+ 4(-2)= 15- 8= 7.

If you must use matrices, write the augmented matrix:
\begin{bmatrix} 1 &amp; -1 \\ 2 &amp; 5 \\ 3 &amp; 4\end{bmatrix}\begin{bmatrix} 7 \\ 0 \\ 7\end{bmatrix}
and row reduce. Subtract 2 times the first row from the second and 3 times the first row from the third to get
\begin{bmatrix} 1 &amp; -1 \\ 0 &amp; 7 \\ 0 &amp; 7\end{bmatrix}\begin{bmatrix} 7 \\ -14 \\ -14\end{bmatrix}
Now divide the second row by 1, add the second row to the first row and subtract the second row from the third row to get
\begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1 \\ 0 &amp; 0\end{bmatrix}\begin{bmatrix} 5 \\ -2 \\ 0\end{bmatrix}
The fact that the third row is now completely zeros tells us that is a correct solution.