# Homework Help: Linear Transformations

1. Jan 9, 2009

### 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 TA(x) = Ax

Find a vector x in R2 whose image under TA 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

2. Jan 9, 2009

### HallsofIvy

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:
$$\begin{bmatrix} 1 & -1 \\ 2 & 5 \\ 3 & 4\end{bmatrix}\begin{bmatrix} 7 \\ 0 \\ 7\end{bmatrix}$$
and row reduce. Subtract 2 times the first row from the second and 3 times the first row from the third to get
$$\begin{bmatrix} 1 & -1 \\ 0 & 7 \\ 0 & 7\end{bmatrix}\begin{bmatrix} 7 \\ -14 \\ -14\end{bmatrix}$$
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
$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix} 5 \\ -2 \\ 0\end{bmatrix}$$
The fact that the third row is now completely zeros tells us that is a correct solution.