Section 4.3 Solving Matrix Equations \(\tta\ttx=\ttb\)
We concluded the last chapter with a discussion about solving numerical equations like \(ax=b\) for \(x\text{.}\) We have seen how to solve equations of the form \(\ttaxb\) by identifying them as systems of linear equations. In this section we will learn how to solve the general matrix equation \(\tta\ttx=\ttb\) for \(\ttx\text{.}\)
We will start by considering the best case scenario when solving \(\ttaxb\text{;}\) that is, when \(\tta\) is square and we have exactly one solution. For instance, suppose we want to solve \(\ttaxb\) where
\begin{equation*}
\tta = \bbm 1 \amp 1\\2 \amp 1\ebm \quad \text{and} \quad \vb = \bbm 0\\1\ebm\text{.}
\end{equation*}
We know how to solve this; put the appropriate matrix into reduced row echelon form and interpret the result.
\begin{equation*}
\bbm 1 \amp 1 \amp 0\\2\amp 1\amp 1\ebm \quad \arref \quad \bbm 1\amp 0\amp 1\\0\amp 1\amp -1\ebm
\end{equation*}
We read from this that
\begin{equation*}
\vx = \bbm 1\\-1\ebm\text{.}
\end{equation*}
Written in a more general form, we found our solution by forming the augmented matrix
\begin{equation*}
\bbm\tta \amp \vb\ebm
\end{equation*}
and interpreting its reduced row echelon form:
\begin{equation*}
\bbm\tta \amp \vb\ebm \quad \arref \quad \bbm\tti \amp \vx \ebm\text{.}
\end{equation*}
Notice that when the reduced row echelon form of \(\tta\) is the identity matrix \(\tti\) we have exactly one solution. This, again, is the best case scenario.
We apply the same general technique to solving the matrix equation \(\tta\ttx=\ttb\) for \(\ttx\text{.}\) We’ll assume that \(\tta\) is a square matrix (\(\ttb\) need not be) and we’ll form the augmented matrix
\begin{equation*}
\bbm\tta \amp \ttb\ebm\text{.}
\end{equation*}
Putting this matrix into reduced row echelon form will give us \(\ttx\text{,}\) much like we found \(\vx\) before.
\begin{equation*}
\bbm\tta \amp \ttb\ebm \quad \arref \quad \bbm\tti \amp \ttx\ebm\text{.}
\end{equation*}
As long as the reduced row echelon form of \(\tta\) is the identity matrix, this technique works great. After a few examples, we’ll discuss why this technique works, and we’ll also talk just a little bit about what happens when the reduced row echelon form of \(\tta\) is not the identity matrix.
First, some examples.
Example 4.3.1. Solving a matrix equation.
Solve the matrix equation \(\tta\ttx=\ttb\) where
\begin{equation*}
\tta = \bbm 1 \amp -1\\ 5 \amp 3\ebm\quad \text{and} \quad \ttb = \bbm -8\amp -13\amp 1\\32\amp -17\amp 21\ebm\text{.}
\end{equation*}
Solution.
To solve \(\tta\ttx=\ttb\) for \(\ttx\text{,}\) we form the proper augmented matrix, put it into reduced row echelon form, and interpret the result.
\begin{equation*}
\bbm 1\amp -1\amp -8\amp -13\amp 1\\5\amp 3\amp 32\amp -17\amp 21\ebm\quad \arref \quad
\bbm 1\amp 0\amp 1\amp -7\amp 3\\0\amp 1\amp 9\amp 6\amp 2\ebm
\end{equation*}
We read from the reduced row echelon form of the matrix that
\begin{equation*}
\ttx = \bbm 1 \amp -7 \amp 3\\9\amp 6\amp 2\ebm\text{.}
\end{equation*}
We can easily check to see if our answer is correct by multiplying \(\tta\ttx\text{.}\)
Example 4.3.2. Another matrix equation.
Solve the matrix equation \(\tta\ttx = \ttb\) where
\begin{equation*}
\tta = \bbm 1 \amp 0 \amp 2\\0\amp -1\amp -2\\2\amp -1\amp 0\ebm \quad\text{and}\quad \ttb = \bbm-1\amp 2\\2\amp -6\\2\amp -4\ebm\text{.}
\end{equation*}
Solution.
To solve, let’s again form the augmented matrix
\begin{equation*}
\bbm \tta \amp \ttb\ebm\text{,}
\end{equation*}
put it into reduced row echelon form, and interpret the result.
\begin{equation*}
\bbm 1 \amp 0 \amp 2\amp -1\amp 2\\0\amp -1\amp -2\amp 2\amp -6\\2\amp -1\amp 0\amp 2\amp -4\ebm\quad\arref\quad \bbm 1\amp 0\amp 0\amp 1\amp 0\\0\amp 1\amp 0\amp 0\amp 4\\0\amp 0\amp 1\amp -1\amp 1\ebm
\end{equation*}
We see from this that
\begin{equation*}
\ttx = \bbm 1 \amp 0\\0 \amp 4\\-1\amp 1\ebm\text{.}
\end{equation*}
Why does this work? To see the answer, let’s define five matrices.
\begin{equation*}
\tta = \bbm 1 \amp 2\\3 \amp 4\ebm, \ \ \vu =\bbm 1\\1\ebm, \ \ \vvv =\bbm-1\\1\ebm, \ \ \vw = \bbm 5\\6\ebm \ \ \text{and} \ \ \ttx = \bbm 1\amp -1\amp 5\\1\amp 1\amp 6\ebm\text{.}
\end{equation*}
Notice that \(\vu\text{,}\) \(\vvv\) and \(\vw\) are the first, second and third columns of \(\ttx\text{,}\) respectively. Now consider this list of matrix products: \(\tta\vu\text{,}\) \(\tta\vvv\text{,}\) \(\tta\vw\) and \(\tta\ttx\text{.}\)
\begin{align*}
\tta\vu \amp = \bbm 1\amp 2\\3\amp 4\ebm\bbm 1\\1\ebm\\
\amp = \bbm 3\\7\ebm\\
\tta\vw \amp = \bbm 1\amp 2\\3\amp 4\ebm\bbm 5\\6\ebm \\
\amp = \bbm 17\\39\ebm\\
\tta\vvv \amp = \bbm 1\amp 2\\3\amp 4\ebm\bbm-1\\1\ebm\\
\amp = \bbm 1\\1\ebm\\
\tta\ttx \amp = \bbm 1\amp 2\\3\amp 4\ebm\bbm 1\amp -1\amp 5\\1\amp 1\amp 6\ebm\\
\amp = \bbm 3\amp 1\amp 17\\7\amp 1\amp 39\ebm \text{.}
\end{align*}
So again note that the columns of \(\ttx\) are \(\vu\text{,}\) \(\vvv\text{,}\) and \(\vw\text{;}\) that is, we can write
\begin{equation*}
\ttx = \bbm\vu \amp \vvv\amp \vw \ebm\text{.}
\end{equation*}
Notice also that the columns of \(\tta\ttx\) are \(\tta\vu\text{,}\) \(\tta\vvv\) and \(\tta\vw\text{,}\) respectively. Thus we can write
\begin{align*}
\tta\ttx \amp = \tta\bbm\vu \amp \vvv\amp \vw \ebm\\
\amp =\bbm \tta\vu \amp \tta\vvv \amp \tta\vw \ebm\\
\amp =\bbm \bbm 3\\7\ebm \amp \bbm 1\\1\ebm \amp \bbm 17\\39\ebm \ebm \\
\amp =\bbm 3\amp 1\amp 17\\7\amp 1\amp 39\ebm\text{.}
\end{align*}
This is exactly the same sort of thing we did in
Section 3.5 when we had several vectors and we wanted to determine whether or not each of them belonged to a given span. Rather than perform the same set of row operations for each vector separately, we can do them all together. (See the discussion following
Example 3.5.8.)
Thus, we are once again making use of the following fact:
The columns of a matrix product \(\tta\ttx\) are \(\tta\) times the columns of \(\ttx\text{.}\)
How does this help us solve the matrix equation \(\tta\ttx=\ttb\) for \(\ttx\text{?}\) Assume that \(\tta\) is a square matrix (that forces \(\ttx\) and \(\ttb\) to be the same size). We’ll let \(\vec{x}_{1},\vec{x}_{2}, \cdots \vec{x}_{n}\) denote the columns of the (unknown) matrix \(\ttx\text{,}\) and we’ll let \(\vec{b}_{1},\vec{b}_{2}, \cdots \vec{b}_{n}\) denote the columns of \(\ttb\text{.}\) We want to solve \(\tta\ttx = \ttb\) for \(\ttx\text{.}\) That is, we want \(\ttx\) where
\begin{align}
\tta\ttx \amp = \ttb\tag{4.3.1}\\
\tta\bbm\vec{x}_{1}\amp \vec{x}_{2}\amp \cdots \amp \vec{x}_{n} \ebm \amp = \bbm\vec{b}_{1}\amp \vec{b}_{2}\amp \cdots \amp \vec{b}_{n} \ebm\tag{4.3.2}\\
\bbm\tta\vec{x}_{1}\amp \tta\vec{x}_{2}\amp \cdots \amp \tta\vec{x}_{n} \ebm \amp = \bbm\vec{b}_{1}\amp \vec{b}_{2}\amp \cdots \amp \vec{b}_{n} \ebm\text{.}\tag{4.3.3}
\end{align}
If the matrix on the left hand side is equal to the matrix on the right, then their respective columns must be equal. This means we need to solve \(n\) equations:
\begin{align*}
\tta\vec{x}_{1} \amp = \vec{b}_{1}\\
\tta\vec{x}_{2} \amp = \vec{b}_{2}\\
\vdots\, \, \amp = \, \vdots \\
\tta\vec{x}_{n} \amp = \vec{b}_{n}\text{.}
\end{align*}
We already know how to do this; this is what we learned in the previous section. Let’s do this in a concrete example. In our above work we defined matrices \(\tta\) and \(\ttx\text{,}\) and looked at the product \(\tta\ttx\text{.}\) Let’s call the product \(\ttb\text{;}\) that is, set \(\ttb = \tta\ttx\text{.}\) Now, let’s pretend that we don’t know what \(\ttx\) is, and let’s try to find the matrix \(\ttx\) that satisfies the equation \(\tta\ttx=\ttb\text{.}\) As a refresher, recall that
\begin{equation*}
\tta = \bbm 1 \amp 2\\3 \amp 4\ebm \quad \text{and} \quad \ttb = \bbm 3\amp 1\amp 17\\7\amp 1\amp 39\ebm\text{.}
\end{equation*}
Since \(\tta\) is a \(2\times 2\) matrix and \(\ttb\) is a \(2\times 3\) matrix, what dimensions must \(\ttx\) be in the equation \(\tta\ttx=\ttb\text{?}\) The number of rows of \(\ttx\) must match the number of columns of \(\tta\text{;}\) the number of columns of \(\ttx\) must match the number of columns of \(\ttb\text{.}\) Therefore we know that \(\ttx\) must be a \(2\times 3\) matrix.
We’ll call the three columns of \(\ttx\) \(\vec{x}_{1}\text{,}\) \(\vec{x}_{2}\) and \(\vec{x}_{3}\text{.}\) Our previous explanation tells us that if \(\tta\ttx = \ttb\text{,}\) then:
\begin{align*}
\tta\ttx \amp = \ttb\\
\tta\bbm\vec{x}_{1} \amp \vec{x}_{2} \amp \vec{x}_{3}\ebm \amp = \bbm 3\amp 1\amp 17\\7\amp 1\amp 39\ebm \\
\bbm\tta\vec{x}_{1} \amp \tta\vec{x}_{2} \amp \tta\vec{x}_{3}\ebm \amp =\bbm 3\amp 1\amp 17\\7\amp 1\amp 39\ebm\text{.}
\end{align*}
Hence
\begin{align*}
\tta\vec{x}_{1} \amp = \bbm 3\\7\ebm\\
\tta\vec{x}_{2} \amp = \bbm 1\\1\ebm\\
\tta\vec{x}_{3} \amp = \bbm 17\\39\ebm\text{.}
\end{align*}
To find \(\vec{x}_{1}\text{,}\) we form the proper augmented matrix and put it into reduced row echelon form and interpret the results.
\begin{equation*}
\bbm 1 \amp 2 \amp 3\\3\amp 4\amp 7\ebm \quad \arref \quad \bbm 1\amp 0\amp 1\\0\amp 1\amp 1\ebm
\end{equation*}
This shows us that
\begin{equation*}
\vec{x}_{1} = \bbm 1\\1\ebm\text{.}
\end{equation*}
To find \(\vec{x}_{2}\text{,}\) we again form an augmented matrix and interpret its reduced row echelon form.
\begin{equation*}
\bbm 1 \amp 2 \amp 1\\3\amp 4\amp 1\ebm \quad \arref \quad \bbm 1\amp 0\amp -1\\0\amp 1\amp 1\ebm\text{.}
\end{equation*}
Thus
\begin{equation*}
\vec{x}_{2} = \bbm-1\\1\ebm\text{,}
\end{equation*}
which matches with what we already knew from above.
Before continuing on in this manner to find \(\vec{x}_{3}\text{,}\) we should stop and think. If the matrix vector equation \(\ttaxb\) is consistent, then the steps involved in putting
\begin{equation*}
\bbm\tta \amp \vb\ebm
\end{equation*}
into reduced row echelon form depend only on \(\tta\text{;}\) it does not matter what is. So when we put the two matrices
\begin{equation*}
\bbm 1 \amp 2 \amp 3\\3\amp 4\amp 7\ebm \quad \text{and} \quad \bbm 1\amp 2\amp 1\\3\amp 4\amp 1\ebm
\end{equation*}
from above into reduced row echelon form, we performed exactly the same steps! (In fact, those steps are: \(-3R_1+R_2\rightarrow R_2\text{;}\) \(-\frac12R_2\rightarrow R_2\text{;}\) \(-2R_2+R_1\rightarrow R_1\text{.}\))
This is just as we noted after
Example 3.5.8. Instead of solving for each column of
\(\ttx\) separately, performing the same steps to put the necessary matrices into reduced row echelon form three different times, why don’t we just do it all at once? (Unless you enjoy doing unnecessary work.) Instead of individually putting
\begin{equation*}
\bbm 1 \amp 2 \amp 3\\3\amp 4\amp 7\ebm, \quad \bbm 1\amp 2\amp 1\\3\amp 4\amp 1\ebm \quad \text{and} \quad \bbm 1\amp 2\amp 17\\3\amp 4\amp 39\ebm
\end{equation*}
into reduced row echelon form, let’s just put
\begin{equation*}
\bbm 1 \amp 2 \amp 3\amp 1\amp 17\\3\amp 4\amp 7\amp 1\amp 39\ebm
\end{equation*}
into reduced row echelon form.
\begin{equation*}
\bbm 1 \amp 2 \amp 3\amp 1\amp 17\\3\amp 4\amp 7\amp 1\amp 39\ebm \quad \arref \quad \bbm 1\amp 0\amp 1\amp -1\amp 5\\0\amp 1\amp 1\amp 1\amp 6\ebm\text{.}
\end{equation*}
By looking at the last three columns, we see \(\ttx\text{:}\)
\begin{equation*}
\ttx = \bbm 1 \amp -1 \amp 5\\1\amp 1\amp 6\ebm\text{.}
\end{equation*}
In each of the examples we’ve considered so far, the reduced row echelon form
\(R\) of the matrix
\(A\) was equal to the
\(n\times n\) identity matrix:
\(R=I\text{.}\) It follows from
Definition 3.6.14 in
Section 3.6 that for an
\(n\times n\) matrix
\(A\text{,}\) we have
\(R=I\) if and only if the rank of
\(A\) is equal to
\(n\text{.}\) At this point we should recall
Theorem 3.6.15 from
Section 3.6, and the discussion that followed. One of the things
Theorem 3.6.15 tells us is that if
\(A\) is an
\(n\times n\) matrix and
\(\operatorname{rank}(A)=n\text{,}\) then the equation
\(A\vec{x}=\vec{b}\) is
guaranteed to have a unique solution, no matter what the vector
\(\vec{b}\) is.
But what if \(\operatorname{rank}(A)\lt n\text{?}\) In this case, the reduced row echelon form of \(A\) is an \(n\times n\) matrix \(R\) with at least row of zeros on the bottom. Our experience with solving systems of the form \(\ttaxb\) tells us that in this case, the matrix equation \(AX=B\) may have infinitely many solutions, or no solution at all. Let us consider an example.
Example 4.3.3. Solving \(AX=B\) when \(\operatorname{rank}(A)\lt n\).
Solve the matrix equations \(AX=B\) and \(AX=C\text{,}\) where
\begin{equation*}
A = \bbm 1 \amp 0 \amp -3\\-2\amp 3\amp 4\\0\amp 6\amp -4\ebm, B = \bbm 1\amp 2\amp 0\\1\amp 1\amp 3\\6\amp 3\amp 0\ebm, \text{ and } C = \bbm 4\amp -2\amp -3\\-6\amp 5\amp 7\\4\amp 2\amp 2\ebm\text{.}
\end{equation*}
Solution.
We proceed as in the previous examples. For the equation \(AX=B\text{,}\) we have
\begin{align*}
\bbm A\amp B\ebm \amp = \left[\begin{array}{ccc|ccc}1\amp 0\amp -3\amp 1\amp 2\amp 0\\-2\amp 3\amp 4\amp 1\amp 1\amp 3\\0\amp 6\amp -4\amp 6\amp 3\amp 0\end{array}\right] \\
\amp \to\left[\begin{array}{ccc|ccc}1\amp 0\amp -3\amp 1\amp 2\amp 0\\0\amp 3\amp -2\amp 3\amp 5\amp 3\\0\amp 0\amp 0\amp 0\amp -7\amp -6\end{array}\right]\text{.}
\end{align*}
We stopped before reaching the reduced row echelon form, but there’s no reason to continue: we already have a row of zeros on the left-hand side of the augmented matrix, and two non-zero entries in that same row, on the right. What this tells us is that it will be impossible to solve for the second and third columns of \(X\text{;}\) thus, there is no solution in this case.
For the equation \(AX=C\text{,}\) we have
\begin{align*}
\bbm A\amp C\ebm \amp = \left[\begin{array}{ccc|ccc}1\amp 0\amp -3\amp 4\amp -2\amp -3\\-2\amp 3\amp 4\amp -6\amp 5\amp 7\\0\amp 6\amp -4\amp 4\amp 2\amp 2\end{array}\right]\\
\amp \arref\left[\begin{array}{ccc|ccc}1\amp 0\amp -3\amp 4\amp -2\amp 3\\0\amp 1\amp -2/3\amp 2/3\amp 1/3\amp 1/3\\0\amp 0\amp 0\amp 0\amp 0\amp 0\end{array}\right]\text{.}
\end{align*}
In this case, we are able to solve for each column of \(X\text{,}\) but in each case there are infinitely many possibilities: we find \(X = \bbm \vec{x}_1\amp \vec{x}_2\amp \vec{x}_3\ebm\text{,}\) where
\begin{equation*}
\vec{x}_1 = \bbm 4+3r\\ \frac{2}{3}+\frac{2}{3}r\\r\ebm, \vec{x}_2 = \bbm -2+3s\\ \frac{1}{3}+\frac{2}{3}s\\s\ebm, \vec{x}_3 = \bbm -3+3t\\ \frac{1}{3}+\frac{2}{3}t\\ t\ebm,
\end{equation*}
for parameters \(r, s, t\text{.}\) Any choice of values for each of these parameters provides us with a solution. For simple example, we can set all three parameters equal to zero, giving us
\begin{equation*}
X = \bbm 4 \amp -2 \amp -3\\2/3\amp 1/3\amp 1/3\\0\amp 0\amp 0\ebm\text{.}
\end{equation*}
It’s easy to check that indeed, \(AX=C\) in this case.
In the previous example, we saw that the equation
\(AX=B\) had no solution, while we were able to solve
\(AX=C\text{.}\) How do we know which of these will be the case? Let’s go back to
Equation (4.3.2) above. From this equation, we can see that each column of
\(B\) is of the form
\(\vec{b}_i = A\vec{x}_i\) for some vector
\(\vec{x}_i\text{.}\) Now recall from
Section 2.7 that the set of vectors of the form
\(A\vx\) is precisely the column space of
\(A\text{.}\) In the case of the matrix
\(C\text{,}\) we can check that if we write
\begin{equation*}
A = \bbm \vec{a}_1 \amp \vec{a}_2\amp \vec{a}_3\ebm \text{ and } C = \bbm \vec{c}_1 \amp \vec{c}_2 \amp \vec{c}_3\ebm\text{,}
\end{equation*}
then \(\vec{c}_1 = \vec{a}_1-\vec{a}_2\text{,}\) \(\vec{c}_2 = \vec{a}_1+\vec{a}_2+\vec{a}_3\text{,}\) and \(\vec{c}_3 = \vec{a}_2+\vec{a}_3\text{.}\) Thus, all three columns of \(C\) are linear combinations of the columns of \(A\text{,}\) which is what allowed us to find a solution to \(AX=C\) even though the rank of \(A\) was less than \(n\text{.}\)
Now that we’ve justified the technique we’ve been using in this section to solve \(\tta\ttx=\ttb\) for \(\ttx\text{,}\) we reinforce its importance by restating it as a Key Idea.
These simple steps cause us to ask certain questions. One of these we asked (and answered) above: What if
\(\tta\) does not have maximum rank, so that the reduced row echelon form of
\(A\) is not equal to
\(I\text{?}\) Second, we specify above that
\(\tta\) should be a square matrix. What happens if
\(\tta\) isn’t square? Is a solution still possible? If you study what happens in
Example 4.3.3 carefully, you can probably guess that a similar argument applies, by applying the ideas of
Section 3.6 to each column of
\(B\) individually.
These questions are good to ask, and we leave it to the reader to discover their answers. Instead of tackling these questions, we instead tackle the problem of “Why do we care about solving \(\tta\ttx=\ttb\text{?}\)” The simple answer is that, for now, we only care about the special case when \(\ttb=\tti\text{.}\) By solving \(\tta\ttx = \tti\) for \(\ttx\text{,}\) we find a matrix \(\ttx\) that, when multiplied by \(\tta\text{,}\) gives the identity \(\tti\text{.}\) That will be very useful.
Exercises Exercises
Exercise Group.
Matrices \(\tta\) and \(\ttb\) are given. Solve the matrix equation \(\tta\ttx=\ttb\text{.}\)
1.
\(\tta = \bbm -3 \amp -6 \ 4 \amp 0 \ebm \text{,}\) \(\ttb=\bbm 48\amp -30\\ 0\amp -8 \ebm \)
2.
\(\tta = \bbm 0 \amp -2 \amp 1 \ 0\amp 2\amp 2 \\ 1\amp 2\amp -3\ebm \text{,}\) \(\ttb=\tti_3 \)
3.
\(\tta = \bbm -4 \amp 1 \ -1 \amp -2 \ebm \text{,}\) \(\ttb=\bbm -2\amp -10\amp 19\\ 13\amp 2\amp -2 \ebm \)
4.
\(\tta = \bbm 1 \amp -3 \ -3 \amp 6 \ebm \text{,}\) \(\ttb=\bbm 12\amp -10\\ -27 \amp 27 \ebm \)
5.
\(\tta = \bbm -1 \amp -2 \ -2 \amp -3 \ebm \text{,}\) \(\ttb=\bbm 13\amp 4\amp 7\\ 22\amp 5\amp 12 \ebm \)
6.
\(\tta = \bbm 3 \amp 3 \ 6 \amp 4 \ebm \text{,}\) \(\ttb=\bbm 15\amp -39\\ 16\amp -66 \ebm \)
7.
\(\tta = \bbm 4 \amp -1 \ -7 \amp 5 \ebm \text{,}\) \(\ttb=\bbm 8 \amp -31\\ -27 \amp 38 \ebm \)
8.
\(\tta = \bbm 2 \amp 2 \ 3 \amp 1 \ebm \text{,}\) \(\ttb=\tti_2 \)
9.
\(\tta = \bbm -2 \amp 0 \amp 4 \ -5\amp -4\amp 5 \\ -3\amp 5\amp -3\ebm \text{,}\) \(\ttb=\bbm -18\amp 2\amp -14\\ -38\amp 18\amp -13 \\ 10\amp 2\amp -18 \ebm \)
10.
\(\tta = \bbm -5 \amp -4 \amp -1 \ 8\amp -2\amp -3 \\ 6\amp 1\amp -8\ebm \text{,}\) \(\ttb=\bbm -21\amp -8\amp -19\\ 65\amp -11\amp -10 \\ 75 \amp -51\amp 33 \ebm \)
11.
\(\tta = \bbm -3 \amp 3 \amp -2 \ 1\amp -3\amp 2 \\ -1\amp -1\amp 2\ebm \text{,}\) \(\ttb=\tti_3 \)
12.
\(\tta = \bbm 1 \amp 0 \ 3 \amp -1 \ebm \text{,}\) \(\ttb=\tti_2 \)
