Find a Matrix that Maps Given Vectors to Given Vectors

Problems and solutions in Linear Algebra

Problem 44

Suppose that a real matrix $A$ maps each of the following vectors
\[\mathbf{x}_1=\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, \mathbf{x}_2=\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}, \mathbf{x}_3=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \] into the vectors
\[\mathbf{y}_1=\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \mathbf{y}_2=\begin{bmatrix}
-1 \\
0 \\
3
\end{bmatrix}, \mathbf{y}_3=\begin{bmatrix}
3 \\
1 \\
1
\end{bmatrix},\] respectively.
That is, $A\mathbf{x}_i=\mathbf{y}_i$ for $i=1,2,3$.
Find the matrix $A$.

(Kyoto University Exam)
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We will give two solutions.

Solution 1

By the condition that $A\mathbf{x}_i=\mathbf{y}_i$, we have a matrix equality $A[\mathbf{x}_1 \mathbf{x}_2 \mathbf{x}_3]=[\mathbf{y}_1 \mathbf{y}_2 \mathbf{y}_3]$.
Explicitly we have
\[ A\begin{bmatrix}
1 & 0 & 0 \\
1 &1 &0 \\
1 & 1 & 1
\end{bmatrix}
=\begin{bmatrix}
1 & -1 & 3 \\
2 &0 &1 \\
0 & 3 & 1
\end{bmatrix}.\] To find the matrix $A$, we find the inverse matrix of $\begin{bmatrix}
1 & 0 & 0 \\
1 &1 &0 \\
1 & 1 & 1
\end{bmatrix}$ and multiply on the right by it.

We use Gauss-Jordan elimination to transform the augmented matrix $[A|I]$ into $[I|A^{-1}]$.
We have
\begin{align*}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 1 &0 & 0 \\
1 & 1 & 0 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 & 1 \\
\end{array}\right] \xrightarrow[R_3-R_1]{R_2-R_1}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 1 &0 & 0 \\
0 & 1 & 0 & -1 & 1 & 0 \\
0 & 1 & 1 & -1 & 0 & 1 \\
\end{array}\right]\\
\xrightarrow{R_3-R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 1 &0 & 0 \\
0 & 1 & 0 & -1 & 1 & 0 \\
0 & 0 & 1 & 0 & -1 & 1 \\
\end{array}\right].
\end{align*}

Thus the inverse matrix is
\[\begin{bmatrix}
1 &0 & 0 \\
-1 & 1 & 0 \\
0 & -1 & 1 \\
\end{bmatrix}.\] Hence we have
\begin{align*}
A &=\begin{bmatrix}
1 & -1 & 3 \\
2 &0 &1 \\
0 & 3 & 1
\end{bmatrix}
\begin{bmatrix}
1 &0 & 0 \\
-1 & 1 & 0 \\
0 & -1 & 1 \\
\end{bmatrix}
=
\begin{bmatrix}
2 & -4 & 3 \\
2 &-1 &1 \\
-3 & 2 & 1
\end{bmatrix}.
\end{align*}

Solution 2

Note that in general $A$ is a matrix whose $i$-th column is the image of the $i$-th unit vector $\mathbf{e}_i$.
For given vectors $\mathbf{x}_i$, it is easy to see that
\begin{align*}
\mathbf{e}_1 &=\mathbf{x}_1-\mathbf{x}_2 \\
\mathbf{e}_2 &=\mathbf{x}_2-\mathbf{x}_3 \\
\mathbf{e}_3 &=\mathbf{x}_3.
\end{align*}

Thus we have
\begin{align*}
A\mathbf{e}_1=A\mathbf{x}_1-A\mathbf{x}_2=\mathbf{y}_1-\mathbf{y}_2=\begin{bmatrix}
2 \\
2 \\
-3
\end{bmatrix}.
\end{align*}
Similarly, we obtain
\begin{align*}
A\mathbf{e}_2=\begin{bmatrix}
-4 \\
-1 \\
2
\end{bmatrix},
\text{ and }
A\mathbf{e}_3=\begin{bmatrix}
3 \\
1 \\
1
\end{bmatrix}.
\end{align*}

Therefore we have
\[A=[A\mathbf{e}_1 \, A\mathbf{e}_2 \, A\mathbf{e}_3]= \begin{bmatrix}
2 & -4 & 3 \\
2 &-1 &1 \\
-3 & 2 & 1
\end{bmatrix}.\] Of course, this agrees with the answer of Solution 1.

Comment.

This is a typical problem in linear algebra, but if you noticed the second approach then you could save time in the exam.

Once you obtained the matrix $A$, you can check whether you computed right by calculating $A\mathbf{x}_i$ with your $A$ and see if it is equal to the vector $\mathbf{y}_i$.


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