Skip to main content

Matrix Calculations A subspace of linear algebra

Chrissy Safranski

Section 2.5 Elementary matrices and invertibility

We have seen how the matrix inverse can be used to solve systems of equations and how systems of equations can arise in many different scenarios and contexts. The examples we’ve used had only a few unknowns, at most 4 or 5.
🔗
However, many of the most useful linear systems, in areas like machine learning and data analysis, have tens or hundreds of variables. It becomes very important to maximize efficiency, and the material we are learning now lays a foundation for that which will continue to the next section also.
🔗

Subsection Prepare

We’ll start by seeing how performing an elementary row operation can be done via matrix multiplication.
🔗

Example 2.5.1. Row operation and matrix multiplication.

Consider the matrix \(\begin{bmatrix} 1 \amp 2 \\ 3 \amp 5 \end{bmatrix}\text{.}\)
🔗
The first step in Gauss-Jordan elimination is \(-3R_1+R_2 \rightarrow R_2\text{,}\)
\begin{align*} \amp \amp \amp \begin{bmatrix} 1 \amp 2 \\ 3 \amp 5 \end{bmatrix}\\ -3R_1+R_2 \amp\rightarrow R_2 \amp \amp \begin{bmatrix} 1 \amp 2 \\ 0 \amp -1 \end{bmatrix}\text{.} \end{align*}
🔗
However, we want to perform this row operation by finding a matrix to multiply by \(A\text{.}\) If we instead perform that same row operation on the \(2\times 2\) identity matrix, we get
\begin{align*} \amp \amp \amp \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}\\ -3R_1+R_2 \amp\rightarrow R_2 \amp \amp \begin{bmatrix} 1 \amp 0 \\ -3 \amp 1 \end{bmatrix}\text{.} \end{align*}
Call this matrix, \(E_1\text{,}\) so \(E_1=\begin{bmatrix} 1 \amp 0 \\ -3 \amp 1 \end{bmatrix}\text{.}\)
🔗
Now check that multiplying \(E_1\) by \(A\) gives us
\begin{align*} E_1A\amp =\begin{bmatrix} 1 \amp 0 \\ -3 \amp 1 \end{bmatrix}\begin{bmatrix} 1 \amp 2 \\ 3 \amp 5 \end{bmatrix}\\ \amp = \begin{bmatrix} 1 \amp 2 \\ 0 \amp -1 \end{bmatrix} \text{,} \end{align*}
which is the exact same effect as performing \(-3R_1+R_2\rightarrow R_2\) on \(A\) directly.
🔗
🔗
We just turned an operation on rows of a matrix into a matrix multiplication. You might be wondering, “Does this work for other elementary row operations?” The answer is, “Yes!” We can turn row operations into matrix multiplication by performing the row operation on the appropriately sized identity matrix.
🔗

Activity 2.5.2. Row operations and matrix multiplication.

🔗
There is a name for the kinds of matrices we have been considering.
🔗

Definition 2.5.3. Elementary Matrices.

Matrices which are obtained by performing one elementary row operation to the identity matrix are called elementary matrices.
🔗
🔗

Example 2.5.4. Continuing Gauss-Jordan with matrix multiplication.

Let’s return to the matrix in Example 2.5.1 and continue our row reducing process. We already defined \(E_1\) and calculated
\begin{equation*} E_1A = \left[\begin{array}{rrr} 1 \amp 2 \\ 0 \amp -1 \end{array}\right] \end{equation*}
The next step in Gauss-Jordan elimination is \(-R_2\rightarrow R_2\text{.}\) Doing this operation to the \(2 \times 2\) identity matrix yields
\begin{align*} \amp \amp \amp \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}\\ -R_2 \amp\rightarrow R_2 \amp \amp \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix}\text{.} \end{align*}
If we let \(E_2=\begin{bmatrix}1 \amp 0 \\ 0 \amp -1 \end{bmatrix}\text{,}\) and multiply \(E_2\) by the matrix we already had after multiplying by \(E_1\text{,}\) we get
\begin{align*} E_2 (E_1A) \amp = \begin{bmatrix}1 \amp 0 \\ 0 \amp -1 \end{bmatrix} \begin{bmatrix} 1 \amp 2 \\ 0 \amp -1 \end{bmatrix} \\ \amp = \begin{bmatrix} 1 \amp 2 \\ 0 \amp 1 \end{bmatrix} \text{.} \end{align*}
🔗
There’s one more row operation we need to do to finish row reducing \(A\text{.}\) Let’s create \(E_3\) by performing \(-2R_2+R_1 \rightarrow R_1\) on the \(2\times 2\) identity matrix
\begin{align*} \amp \amp \amp \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}\\ -2R_2+R_1 \amp\rightarrow R_1 \amp \amp \begin{bmatrix} 1 \amp -2 \\ 0 \amp 1 \end{bmatrix}\text{,} \end{align*}
and multiplying \(E_3\) by the matrix we had gives us
\begin{align*} E_3 (E_2E_1A) \amp = \begin{bmatrix}1 \amp -2 \\ 0 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \amp 2 \\ 0 \amp 1 \end{bmatrix} \\ \amp = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix} \text{.} \end{align*}
🔗
🔗
To recap what we just did: we row reduced the matrix \(A\) by multiplying on the left by three matrices in turn. In this case, the reduced row echelon form was the identity matrix, so we also found matrices whose product is the identity. Associativity means that
\begin{equation*} (E_3E_2E_1)A=I\text{,} \end{equation*}
which means that \(A\) is invertible and \(A^{-1}=E_3E_2E_1\text{.}\)
🔗
This is kind of a big deal. We said in Calculating inverses of any size that in order to calculate inverses we augment by the identity matrix and then row reduce. If a matrix row reduces to \(I\) on the left then the inverse appears on the right. Now we see another connection between inverses and row reducing: if a matrix row reduces to the identity, the inverse is the product of the elementary matrices which perform the row reduction.
🔗
It is possible to show that since the product \(E_3E_2E_1\) is invertible (and it is since its inverse is \(A\)) it is the case that \(E_3\text{,}\) \(E_2\text{,}\) and \(E_1\) are all invertible, too. Instead, let’s try to see why that has to be the case based on properties of elementary matrices.
🔗

Activity 2.5.5. Inverses of elementary matrices.

There are 3 kinds of elementary row operations and so there are 3 kinds of elementary matrices: ones which result from 1) scaling a row, 2) from interchanging two rows, and 3) from adding a multiple of a row to another row.
🔗
(a) Scaling operation inverses.
🔗
(b) Scaling inverse generally.
Say in your own words what the inverse (or opposite) is of scaling a row by a number \(c\text{.}\)
🔗
🔗
(c) Interchanging operation inverses.
🔗
(d) Interchanging inverse generally.
Say in your own words what the inverse is of interchanging two rows.
🔗
🔗
(e) Multiply-and-add operation inverses.
🔗
(f) Multiply-and-add inverse generally.
Say in your own words what the inverse is of multiplying a row by \(c\) and adding to another row.
🔗
🔗
🔗
Let’s return to our running example of this section. We started with
\begin{equation*} A=\begin{bmatrix} 1 \amp 2 \\ 3 \amp 5 \end{bmatrix} \end{equation*}
and through row reducing by multiplying by elementary matrices we found that
\begin{equation*} E_3E_2E_1A=I\text{.} \end{equation*}
We’ve seen that elementary row operations can be undone by another elementary row operation, which means that every elementary matrix is invertible, and the inverse is another elementary matrix. Thus we can multiply both sides of the above equation by \(E_3^{-1}\) and get that
\begin{align*} E_3^{-1}(E_3E_2E_1A)\amp = E_3^{-1}I \\ (E_3^{-1}E_3)E_2E_1A\amp = E_3^{-1} \\ E_2E_1A\amp = E_3^{-1} \text{.} \end{align*}
Next we undo \(E_2\) by multiplying both sides by \(E_2^{-1}\) to get
\begin{align*} E_2^{-1}(E_2E_1A)\amp = E_2^{-1}E_3^{-1} \\ E_1A\amp = E_2^{-1}E_3^{-1} \text{.} \end{align*}
Finally, we undo \(E_1\) in the same way and get
\begin{align*} E_1^{-1}E_1A\amp = E_1^{-1}E_2^{-1}E_3^{-1} \\ A\amp = E_1^{-1}E_2^{-1}E_3^{-1} \text{.} \end{align*}
🔗
Since the inverse of an elementary matrix is also an elementary matrix, we just wrote \(A\) as a product of elementary matrices. Moreover, there wasn’t anything special about this example; the fact that the reduced row echelon form of \(A\) is \(I\) meant that we could multiply \(A\) by elementary matrices to get \(I\text{,}\) which meant that \(A\) is invertible and also that \(A\) can be expressed as the product of elementary matrices.
🔗
We have now seen a lot of different things that are equivalent to a matrix having an inverse (and there will be still more in the future). Let’s collect what we know so far.
🔗
A theorem in math is a statement which has a proof that it’s true. We don’t get into full proofs, but we can still explore and understand some of the reasons for each of the above equivalences. We did a little of that just above the theorem, when we were seeing how row reducing to the identity led to invertibility and being able to be written as a product of elementary matrices, and we’ll explore more during the Participate activities.
🔗

Reading Questions Reading Questions

1. Order of multiplication.
If you have found elementary matrices \(E_1\) and \(E_2\) which perform row operations, and you want to perform \(E_1\) first on a matrix \(M\) and then to the result perform \(E_2\text{,}\) which product do you calculate?
🔗
  • \(E_1E_2M\)
  • Just like with functions \(f(g(x))\text{,}\) the \(g(x)\) comes second as we read left-to-right, but that means perform \(g(x)\) first and plug the result into \(f(x)\text{.}\)
    So too with matrix multiplication. By associativity, \(E_1E_2M\) is the same as \(E_1(E_2M)\text{,}\) which means multiply \(E_2M\) and then multiply \(E_1\) by the result.
  • \(E_2E_1M\)
  • Correct!
🔗
2. Elementary matrices and invertibility.
    Every elementary matrix is invertible, and the inverse of an elementary matrix is also an elementary matrix.
    🔗

  • True.

  • The inverse of scaling is scaling by the reciprocal; the inverse of swapping two rows is swapping the same two rows again; the inverse of multiplying a row by \(c\) and adding to another row is multiplying the row by \(-c\) and adding.

  • False.

  • The inverse of scaling is scaling by the reciprocal; the inverse of swapping two rows is swapping the same two rows again; the inverse of multiplying a row by \(c\) and adding to another row is multiplying the row by \(-c\) and adding.
🔗
3. Reflection.
(a)
Reflect on your confidence level.
🔗
    How confident do you feel with the material you just read about?
    🔗
  • 1.
    Not at all confident or didn’t do the reading.
    🔗
  • 2.
    Not very confident.
    🔗
  • 3.
    Somewhat confident.
    🔗
  • 4.
    Mostly confident.
    🔗
  • 5.
    Confident so far and ready to engage more deeply.
    🔗
🔗
(b)
Ask a question about the material. What additional information do you think someone would need to become more confident in their understanding?
🔗
🔗
🔗
🔗
🔗

Worksheet Participate

1.

Consider the matrix \(A=\left[\begin{array}{rrr} 1 \amp 0 \amp 2 \\ 0 \amp 1 \amp -1 \\ -3 \amp 0 \amp -7 \end{array}\right]\text{.}\) Note that performing Gauss-Jordan elimination requires performing in order
🔗
\begin{align*} 3R_1+R_3 \amp \rightarrow R_3 \\ -R_3 \amp \rightarrow R_3 \\ R_3+R_2 \amp \rightarrow R_2 \\ -2R_3+R_1 \amp \rightarrow R_1 \text{.} \end{align*}
🔗
(a)
Calculate \(E_1\text{,}\) \(E_2\text{,}\) \(E_3\text{,}\) and \(E_4\) without using technology. What is the inverse of each matrix you found?
🔗
🔗
(b)
Use the sage cell below and modify it to use what you got for \(E_1\) and verify that \(E_1A\) performs \(3R_1+R_3 \rightarrow R_3\) on \(A\text{.}\)
🔗
🔗
(c)
Now similarly modify the sage cell below with what you found for \(E_2\text{,}\) \(E_3\text{,}\) and \(E_4\) and verify that \(E_4E_3E_2E_1A=I\text{.}\)
🔗
🔗
(d)
What do the calculations in the cell below tell us about \(A^{-1}\) and elementary matrices?
🔗
🔗
(e)
Use sage to verify your calculations of the inverses in the first part, and to verify that
\begin{equation*} A=E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}\text{.} \end{equation*}
🔗
🔗
🔗

2.

The theorem containing various things equivalent to the property of being invertible is sometimes called “The Invertible Matrix Theorem”. But the star of the show is really the property of the reduced row echelon form of \(A\) being \(I\text{.}\) Let’s explore a bit how some of the various properties are related.
🔗
(a)
Suppose you know that \(A\) is an \(n\times n\) matrix whose reduced row echelon form is \(I\text{.}\) Set up the augmented matrix \(\big[A | I \big]\text{.}\) What happens after you row reduce this matrix? What does mean for the invertibility of \(A\text{?}\)
🔗
🔗
(b)
Suppose you know that \(A\) is an \(n\times n\) matrix whose reduced row echelon form is \(I\text{.}\) Now we know that performing a row operation is the same as multiplying by an elementary matrix. Performing a row operation on \(\big[A | I \big]\) has the same result as \(\big[E_1A | E_1I \big]\text{.}\) The reduced row echelon form of \(A\) being \(I\) means that there is a sequence of elementary matrices \(E_1, E_2, \ldots E_m\) such that \(E_m\cdots E_2 E_1A=I\text{.}\) What does this mean about \(E_m\cdots E_2 E_1\text{?}\)
🔗
🔗
(c)
Suppose you know that \(A\) is an invertible \(n\times n\) matrix and you have a linear system with \(A\) as its coefficient matrix. Then you can write \(AX=B\) where \(X\) is a column matrix of variables and \(B\) is your column matrix of constants. How many solutions are there for \(X\text{?}\) How do you know?
🔗
🔗
(d)
Suppose you know that \(A\) is an invertible \(n\times n\) matrix and that \(X\) is a matrix with \(n\) rows such that \(AX=\mathbf{0}\text{.}\) What does \(X\) have to equal? How do you know?
🔗
🔗
(e)
Suppose that \(A\) is an \(n\times n\) matrix and \(X\) is a column matrix with \(n\) rows. Now suppose that the only solution to \(AX=\mathbf{0}\) is that \(X=\mathbf{0}\text{.}\) (Why isn’t it possible that there are no solutions?) Set up \(\big[A | \mathbf{0} \big]\text{.}\) After row reducing to solve the system, how many free variables can there be? What does the reduced row echelon form have to look like?
🔗
🔗
🔗
🔗

Summary.

  • Elementary matrices are the result of performing one elementary row operation on the identity matrix. Multiplying a matrix on the left by an elementary matrix has the same result as performing that row operation on the matrix.
    🔗
    🔗
  • We have a list of conditions that are all equivalent to a square matrix being invertible, including row reducing to the identity matrix and having exactly one solution to a linear system (not 0, not infinitely many solutions).
    🔗
    🔗
  • This material is setting us up to learn more about how computers are programmed to calculate with matrices, to be efficient and fast when there are hundreds, or even thousands, of numbers to add and multiply.
    🔗
    🔗
🔗

Subsection Practice

Exercise 2.5.1. Scaling row operation with inverse.

Assume that A is a matrix with four rows. Find the elementary matrix E such that E A gives the matrix resulting from A after the given row operation is performed. Then find \(E^{-1}\) and give the elementary row operation corresponding to \(E^{-1}\text{.}\)
🔗
\begin{equation*} -5 R_{4} \rightarrow R_{4} \end{equation*}
🔗
\(E =\) (4 × 4 array)
🔗
\(E^{-1} =\) (4 × 4 array)
🔗
The elementary row operation corresponding to \(E^{-1}\) is:
🔗
  • \(\displaystyle \frac{1}{-5} R_{1}\rightarrow R_{4}\)
    🔗
    🔗
  • \(\displaystyle \frac{1}{-5} R_{4}\rightarrow R_{4}\)
    🔗
    🔗
  • \(\displaystyle -5 R_{4}\rightarrow R_{4}\)
    🔗
    🔗
  • \(\displaystyle \frac{1}{-5} R_{1}\rightarrow R_{1}\)
    🔗
    🔗
  • \(\displaystyle \frac{1}{-5} R_{4}+R_{1}\rightarrow R_{4}\)
    🔗
    🔗
🔗
🔗

Exercise 2.5.2. Interchanging row operation with inverse.

Assume that A is a matrix with four rows. Find the elementary matrix E such that E A gives the matrix resulting from A after the given row operation is performed. Then find \(E^{-1}\) and give the elementary row operation corresponding to \(E^{-1}\text{.}\)
🔗
\(R_{4}\leftrightarrow R_{3}\)
🔗
\(E =\) (4 × 4 array)
🔗
\(E^{-1} =\) (4 × 4 array)
🔗
The elementary row operation corresponding to \(E^{-1}\) is:
🔗
  • \(\displaystyle R_{3}\leftrightarrow R_{2}\)
    🔗
    🔗
  • \(\displaystyle R_{4}\leftrightarrow R_{2}\)
    🔗
    🔗
  • \(\displaystyle R_{4}\leftrightarrow R_{3}\)
    🔗
    🔗
🔗
🔗

Exercise 2.5.3. Lots of row operation with inverses.

a. Suppose that \(E_1 \left[\begin{array}{cc} -3 \amp 3\cr -3 \amp -1 \end{array}\right] = \left[\begin{array}{cc} -18 \amp 18\cr -3 \amp -1 \end{array}\right]\text{.}\) Find \(E_1\) and \(E_1^{-1}\text{.}\)
🔗
\(E_1 =\) (2 × 2 array), \(E_1^{-1} =\) (2 × 2 array).
🔗
b. Suppose that \(E_2 \left[\begin{array}{cc} -3 \amp 3\cr -3 \amp -1 \end{array}\right] = \left[\begin{array}{cc} -3 \amp -1\cr -3 \amp 3 \end{array}\right]\text{.}\) Find \(E_2\) and \(E_2^{-1}\text{.}\)
🔗
\(E_2 =\) (2 × 2 array), \(E_2^{-1} =\) (2 × 2 array).
🔗
c. Suppose that \(E_3 \left[\begin{array}{cc} -3 \amp 3\cr -3 \amp -1 \end{array}\right] = \left[\begin{array}{cc} -15 \amp -1\cr -3 \amp -1 \end{array}\right]\text{.}\) Find \(E_3\) and \(E_3^{-1}\text{.}\)
🔗
\(E_3 =\) (2 × 2 array), \(E_3^{-1} =\) (2 × 2 array).
🔗
d. Suppose that \(E_4 \left[\begin{array}{ccc} -2 \amp 2 \amp 2\cr -5 \amp -2 \amp 2\cr 2 \amp -4 \amp -4 \end{array}\right] = \left[\begin{array}{ccc} -2 \amp 2 \amp 2\cr -15 \amp -6 \amp 6\cr 2 \amp -4 \amp -4 \end{array}\right]\text{.}\)
🔗
Find \(E_4\) and \(E_4^{-1}\text{.}\)
🔗
\(E_4 =\) (3 × 3 array),
🔗
\(E_4^{-1} =\) (3 × 3 array).
🔗
e. Suppose that \(E_5 \left[\begin{array}{ccc} -2 \amp 2 \amp 2\cr -5 \amp -2 \amp 2\cr 2 \amp -4 \amp -4 \end{array}\right] = \left[\begin{array}{ccc} -2 \amp 2 \amp 2\cr 2 \amp -4 \amp -4\cr -5 \amp -2 \amp 2 \end{array}\right]\text{.}\)
🔗
Find \(E_5\) and \(E_5^{-1}\text{.}\)
🔗
\(E_5 =\) (3 × 3 array),
🔗
\(E_5^{-1} =\) (3 × 3 array).
🔗
f. Suppose that \(E_6 \left[\begin{array}{ccc} -2 \amp 2 \amp 2\cr -5 \amp -2 \amp 2\cr 2 \amp -4 \amp -4 \end{array}\right] = \left[\begin{array}{ccc} -2 \amp 2 \amp 2\cr -5 \amp -2 \amp 2\cr -6 \amp 4 \amp 4 \end{array}\right]\text{.}\)
🔗
Find \(E_6\) and \(E_6^{-1}\text{.}\)
🔗
\(E_6 =\) (3 × 3 array),
🔗
\(E_6^{-1} =\) (3 × 3 array).
🔗
🔗

Exercise 2.5.4. Product of elementary matrices.

Consider the following Gauss-Jordan reduction:
🔗
\begin{equation*} \underbrace{\left[\begin{array}{ccc} 1 \amp 0 \amp 9\cr 0 \amp 0 \amp -8\cr 4 \amp 1 \amp 0 \end{array}\right]}_A \to \underbrace{\left[\begin{array}{ccc} 1 \amp 0 \amp 9\cr 0 \amp 0 \amp 1\cr 4 \amp 1 \amp 0 \end{array}\right]}_{E_1 A} \to \underbrace{\left[\begin{array}{ccc} 1 \amp 0 \amp 9\cr 4 \amp 1 \amp 0\cr 0 \amp 0 \amp 1 \end{array}\right]}_{E_2E_1 A} \to \underbrace{\left[\begin{array}{ccc} 1 \amp 0 \amp 0\cr 4 \amp 1 \amp 0\cr 0 \amp 0 \amp 1 \end{array}\right]}_{E_3E_2E_1 A} \to \underbrace{\left[\begin{array}{ccc} 1 \amp 0 \amp 0\cr 0 \amp 1 \amp 0\cr 0 \amp 0 \amp 1 \end{array}\right]}_{E_4E_3E_2E_1 A}=I \end{equation*}
🔗
Find
🔗
\(E_1 =\) (3 × 3 array),
🔗
\(E_2 =\) (3 × 3 array),
🔗
\(E_3 =\) (3 × 3 array).
🔗
\(E_4 =\) (3 × 3 array).
🔗
Write \(A\) as a product \(A=E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}\) of elementary matrices:
🔗
\(\left[\begin{array}{ccc} 1 \amp 0 \amp 9\cr 0 \amp 0 \amp -8\cr 4 \amp 1 \amp 0 \end{array}\right] =\) (3 × 3 array) (3 × 3 array) (3 × 3 array) (3 × 3 array)
🔗
🔗

Exercise 2.5.5. True/False invertibility.

Determine which of the following statements are true and which are false.
🔗
  1. Every invertible matrix can be written as a product of elementary matrices.
    🔗
    🔗
  2. The reduced row echelon form of a square matrix is the identity matrix.
    🔗
    🔗
  3. The inverse of an elementary matrix is an elementary matrix.
    🔗
    🔗
  4. If \(A\) is an invertible matrix and \(AX=\mathbf{0}\) then \(X=\mathbf{0}\)
    🔗
    🔗
🔗
🔗
🔗

Exercises Additional Practice

1.

(a)
Calculate \(B^{-1}\) for \(B=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -4 \end{array}\right]\text{.}\)
🔗
Answer.
\(B^{-1}=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -\frac{1}{4}\end{array}\right]\)
🔗
🔗
🔗
(b)
Calculate \(C^{-1}\) for \(C=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 3 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right]\text{.}\)
🔗
Answer.
\(C^{-1}=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ -3 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right]\)
🔗
🔗
🔗
(c)
Calculate \(D^{-1}\) for \(D=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{array}\right]\text{.}\)
🔗
Answer.
\(D^{-1}=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{array}\right]\)
🔗
🔗
🔗
(d)
True or False? \((BCD)^{-1}=B^{-1}C^{-1}D^{-1}\)
🔗
Answer.
False. \((BCD)^{-1}BCD=I\text{,}\) which means \((BCD)^{-1}=D^{-1}C^{-1}B^{-1}\)
🔗
🔗
🔗
(e)
Calculate \((BCD)^{-1}\text{.}\)
🔗
Answer.
\(=\)
\begin{align*} (BCD)^{-1}\amp = D^{-1}C^{-1}B^{-1} \\ \amp = \left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{array}\right] \left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ -3 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right] \left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -\frac{1}{4}\end{array}\right]\\ \amp = \left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp -\frac{1}{4} \\ -3 \amp 1 \amp 0\end{array}\right] \end{align*}
🔗
🔗
🔗
🔗

2.

Consider the matrix
\begin{equation*} A = \left[\begin{array}{rrr} 1 \amp 3 \amp -2 \\ -3 \amp -6 \amp 3 \\ 2 \amp 0 \amp -2 \\ \end{array}\right]\text{.} \end{equation*}
🔗
(a)
Calculate the elementary matrix \(E_1\) which performs \(3R_1+R_2\rightarrow R_2\text{.}\)
🔗
Answer.
\(E_1=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 3 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right]\)
🔗
🔗
🔗
(b)
Calculate the elementary matrix \(E_2\) which performs \(-2R_1+R_3\rightarrow R_3\text{.}\)
🔗
Answer.
\(E_2=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ -2 \amp 0 \amp 1 \end{array}\right]\)
🔗
🔗
🔗
(c)
Calculate the elementary matrix \(E_3\) which performs \(2R_2+R_3\rightarrow R_3\text{.}\)
🔗
Answer.
\(E_3=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 2 \amp 1 \end{array}\right]\)
🔗
🔗
🔗
(d)
Calculate the elementary matrix \(E_4\) which performs \(-\frac{1}{4}R_3\rightarrow R_3\text{.}\)
🔗
Answer.
\(E_4=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -\frac{1}{4} \end{array}\right]\)
🔗
🔗
🔗
(e)
Calculate the elementary matrix \(E_5\) which performs \(3R_3+R_2\rightarrow R_2\text{.}\)
🔗
Answer.
\(E_5=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 3 \\ 0 \amp 0 \amp 1 \end{array}\right]\)
🔗
🔗
🔗
(f)
Calculate the elementary matrix \(E_6\) which performs \(2R_3+R_1\rightarrow R_1\text{.}\)
🔗
Answer.
\(E_6=\left[\begin{array}{rrr} 1 \amp 0 \amp 2 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right]\)
🔗
🔗
🔗
(g)
Calculate the elementary matrix \(E_7\) which performs \(\frac{1}{3}R_2\rightarrow R_2\text{.}\)
🔗
Answer.
\(E_7=\left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp \frac{1}{3} \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right]\)
🔗
🔗
🔗
(h)
Calculate the elementary matrix \(E_8\) which performs \(-3R_2+R_1\rightarrow R_1\text{.}\)
🔗
Answer.
\(E_8=\left[\begin{array}{rrr} 1 \amp -3 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right]\)
🔗
🔗
🔗
🔗

3.

Suppose you know that \(E_8E_7E_6E_5E_4E_3E_2E_1A=I\) where each \(E_i\) is an elementary matrix.
🔗
(a)
What is \(A^{-1}\text{?}\)
🔗
Hint.
\(A^{-1}\) is the matrix you multiply by \(A\) to get \(I\text{.}\)
🔗
🔗
Answer.
\(A^{-1}=E_8E_7E_6E_5E_4E_3E_2E_1\text{.}\)
🔗
🔗
🔗
(b)
Write \(A\) as a product of elementary matrices.
🔗
Hint.
Start with the equation above and multiply both sides on the left by \(E_8^{-1}\text{.}\) What do you get?
🔗
🔗
Answer.
\(A=E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}E_5^{-1}E_6^{-1}E_7^{-1}E_8^{-1}\)
🔗
🔗
🔗
🔗
🔗
🔗
PrevTopNext
Feedback