Title: Default for 7.1 Introduction to Matrices
Course: HELM Workbooks 2023-2024

Applications of Matrices

The solution of simultaneous linear equations is a task which frequently occurs in science and engineering.  In electrical engineering the analysis of circuits provides a ready example.

However the simultaneous equations arise, we need to study two things:

1. how we can conveniently represent large systems of linear equations;
2. how we might find the solution of such equations.

We shall discover that knowledge of the theory of matrices is an essential mathematical tool in this area.

Representing simultaneous linear equations

Suppose that we wish to solve the following three equations in three unknowns \(x_1,x_2\) and \(x_3\):

\[\begin{alignedat}{4} 3x_1 & {}+{} & 2x_2 & {}-{} & x_3 &= 3 \\ x_1 & {}-{} & x_2 & {}+{} & x_3 &= 4 \\ 2x_1 & {}+{} & 3x_2 & {}+{} & 4x_3 &= 5 \end{alignedat}\]

We can isolate three facets of this system: the coefficients of \(x_1,x_2,x_3\); the unknowns \(x_1,x_2,x_3\); and the numbers on the right-hand sides.

Notice that in the system

\[\begin{alignedat}{4} 3x & {}+{} & 2y & {}-{} & z &= 3 \\ x & {}-{} & y & {}+{} & z &= 4 \\ 2x & {}+{} & 3y & {}+{} & 4z &= 5 \end{alignedat}\]

the only difference from the first system is the names given to the unknowns.  It can be checked that the first system has the solution \(x_1=2,\ x_2=-1,\ x_3=1.\)

The second system therefore has the solution \(x=2, y=-1, z=1.\)

We can isolate the three facets of the first system by using arrays of numbers and of unknowns:

\[\left[\begin{array}{rrr}3&2&-1\\1&-1&1\\2&3&4\end{array}\right]\left[\begin{array}{c}x_1\\ x_2\\x_3\end{array}\right]=\left[\begin{array}{c}3\\ 4\\ 5\end{array}\right].\]

Even more conveniently, we can represent the arrays with letters (usually capital letters)

\[AX\ \ \ \ =\ \ \ \ B.\]

Here, to be explicit, we write

\[ A=\left[\begin{array}{rrr}3&2&-1\\ 1&-1&1\\2&3&4\end{array}\right],\qquad X=\left[\begin{array}{c}x_1\\ x_2\\x_3\end{array}\right],\qquad B=\left[\begin{array}{c}3\\ 4\\ 5\end{array}\right].\]

Here \(A\) is called the matrix of coefficients, \(X\) is called the matrix of unknowns and \(B\) is called the matrix of constants.

If we now append  to \(A\) the column of right-hand sides we obtain the augmented matrix for the system:

\[ \left[\left.\begin{array}{rrr}3&2&-1\ \\ 1&-1&1\ \\ 2&3&4\ \end{array}\right|\right.\left.\begin{array}{c}3\\ 4\\ 5\end{array}\right].\]

The order of the entries, or elements, is crucial.  For example, all the entries in the second row relate to the second equation, the entries in column 1 are the coefficients of the unknown \(x_1\), and those in the last column are the constants on the right-hand sides of the equations.

In particular, the entry in row 2 column 3 is the coefficient of \(x_3\) in equation 2.

Representing networks

Shortest-distance problems are important in communications study.  The figure below illustrates schematically a system of four towns connected by a set of roads.

[Image: 7.1 A system of roads connecting towns]

The system can be represented by the matrix

\[ \ \ \ \ a\ \ \ \quad b\ \ \ \quad  c\ \ \quad d\]

\[\begin{array}{c}a\\ b\\ c\\ d\end{array}\left[\begin{array}{cccccccc}0&&1&&0&&0\\ 1&&0&&1&&1\\ 0&&1&&0&&1\\ 0&&1&&1&&0\end{array}\right].\]

The row refers to the town from which the road starts and the column refers to the town where the road ends.

An entry of 1 indicates that two towns are directly connected by a road (for example \(b\) and \(d\)) and an entry of zero indicates that there is no direct road (for example \(a\) and \(c\)). Of course, if there is a road from \(b\) to \(d\) (say) it is also a road from \(d\) to \(b\).

In this quiz we shall develop some basic ideas about matrices.

Definitions

An array of numbers, rectangular in shape, is called a matrix.  The first matrix below has \(3\) rows and \(2\) columns and is said to be a '\(3\) by \(2\)' matrix (written \(3\times 2\)). The second matrix is a '\(2\) by \(4\)' matrix (written \(2\times 4\)).

\[\left[\begin{array}{rr}1&4\\ -2&3\\ 2&1\end{array}\right]\ \qquad\qquad\left[\begin{array}{cccc}1&2&3&4\\ 5&6&7&9\end{array}\right]\]

The general \(3\times 3\) matrix can be written

\[A=\left[\begin{array}{lll}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right]\]

where \(a_{ij}\) denotes the element in row \(i\), column \(j\).

For example in the matrix:

\[A=\left[\begin{array}{lll}\hfill 0&\hfill -1&\hfill -3\\ \hfill0&\hfill 6&\hfill -12\\ \hfill 5&\hfill 7&\hfill 123\end{array}\right]\]

\[a_{11}= 0,\qquad a_{12}=-1,\qquad a_{13}=-3,\quad\dots \quad a_{22}=6,\quad \dots \quad a_{32}=7,\quad a_{33}=123\]

Key Point

The General Matrix

A general \(m\times n\) matrix \(A\)  has \(m\) rows and \(n\) columns. The entries in the matrix \(A\) are called the elements of \(A\). In matrix \(A\) the element in row \(i\) and column \(j\) is denoted by \(a_{ij}\).

A matrix with only one column is called a column vector (or column matrix). For example, \(\left[\begin{array}{c}x_1\\ x_2\\ x_3\end{array}\right]\) and \(\left[\begin{array}{c}3\\ 4\\ 5\end{array}\right]\) are both \(3\times 1\) column vectors. A matrix with only one row is called a row vector (or row matrix). For example,  \([2, -3, 8, 9]\) is a \(1\times 4\) row vector. Often the entries in a row vector are separated by commas for clarity.

Square matrices

When the number of rows is the same as the number of columns, i.e. \(m=n\), the matrix is said to be square and of order \(n\) (or \(m\)).

- In an \(n\times n\) square matrix \(A\), the leading diagonal (or principal diagonal) is the `north-west to south-east' collection of elements \(a_{11},a_{22},\ldots , a_{nn}.\)  The sum of the elements in the leading diagonal of \(A\) is called the trace of the matrix, denoted by \(\text{tr}(A)\).

\[A=\left[\begin{matrix} a_{11}&a_{12}&\dots&a_{1n}\\a_{21}&a_{22}&\dots&a_{2n}\\\vdots&\vdots&\vdots&\vdots\\a_{n1}&a_{n2}&\dots&a_{nn}\end{matrix}\right] \qquad\qquad \text{tr}(A)=a_{11}+a_{22}+\dots +a_{nn}\]

- A square matrix in which all the elements below the leading diagonal are zero is called an upper triangular matrix, often denoted by \(U\).

\[U=\left[\begin{matrix}u_{11}&u_{12}&\dots&\dots&u_{1n}\\0&u_{22}&\dots&\dots&u_{2n}\\0&0&\dots&\vdots&\vdots\\0&0&\dots&0&u_{nn}\end{matrix}\right] \qquad\qquad u_{ij}=0\quad \text{when}\ \ i>j\]
- A square matrix in which all the elements above the leading diagonal are zero is called a lower triangular matrix, often denoted by \(L\).

\[L=\left[\begin{matrix}l_{11}&0&0&\dots&0\\l_{21}&l_{22}&0&\dots&0\\\vdots&\vdots&\dots&\dots&0\\l_{n1}&l_{n2}&\vdots&\dots&l_{nn}\end{matrix}\right] \qquad\qquad l_{ij}=0\quad \text{when}\ \ i<j\]

- A square matrix where all the non-zero elements are along the leading diagonal is called a diagonal matrix, often denoted by \(D\).

\[ D=\left[\begin{matrix}d_{11}&0&0&\dots&0\\0&d_{22}&0&\dots&0\\0&0&\dots&\dots&0\\0&0&0&\dots&d_{nn}\end{matrix}\right] \qquad\qquad d_{ij}=0\quad \text{when}\ \ i\ne j\]

Some examples of matrices and their classification

\(\qquad A=\left[\begin{array}{ccc} 1&2&3\\ 4&5&6\end{array}\right]\) is \(2\times 3\).  It is not square.

\(\qquad B=\left[\begin{array}{cc}1&2\\ 3&4\end{array}\right]\) is \(2\times 2\). It is square.

Also, \(\text{tr}(A)\) does not exist as it is not a square matrix, but \(\text{tr}(B)=1+4=5\).

\(\qquad C=\left[\begin{array}{rrr}1&2&3\\ 0&-2&-5\\ 0&0&1\end{array}\right]\) and \(D=\left[\begin{array}{rrr} 4&0&3\\ 0&-2&5\\ 0&0&1\end{array}\right]\) are both \(3\times 3\), square and upper triangular.

Also, \(\text{tr}(C)=0\) and \(\text{tr}(D)=3\).

\(\qquad E=\left[\begin{array}{rrr}1&0&0\\ 2&-2&0\\ 3&-5&1\end{array}\right]\) and \(F=\left[\begin{array}{rrr}-1&0&0\\ 1&4&0\\ 0&1&1\end{array}\right]\) are both \(3\times 3\), square and lower triangular.

Also, \(\text{tr}(E)=0\) and \(\text{tr}(F)=4\).

\(\qquad G=\left[\begin{array}{rrr}1&0&0\\ 0&2&0\\ 0&0&-3\end{array}\right]\) and \(H=\left[\begin{array}{ccc}4&0&0\\ 0&2&0\\ 0&0&0\end{array}\right]\) are both \(3\times 3\), square and diagonal.

Also, \(\text{tr}(G)=0\) and \(\text{tr}(H)=6\).

Exercise

Give the size of the following matrices:

1. \([value]\) \(\times\)
2. \([value]\) \(\times\)

Find the trace of the following matrices:

1. \(A=[value]\), \(\text{tr}(A) = \)
2. \(B=[value]\), \(\text{tr}(B) = \)
Feedback: First, we find the size of the first two matrices. 1. The matrix \[[value]\] has [value] rows and [value] columns. Therefore, it is a \([value] \times [value]\) matrix. 2. The matrix \[[value]\] has [value] rows and [value] columns. Therefore, it is a \([value] \times [value]\) matrix. Next, we find the trace of the given matrices. Recall that the trace of a square matrix is the sum of its diagonal entries. 1. The matrix \[A = [value]\] has diagonal entries [value] and [value]. Therefore, its trace is \(\mathrm{tr}(A) = [value] + [value] = [value]\). 2. Similarly, the matrix \[B = [value]\] has trace \(\mathrm{tr}(B) = [value]\).

Q: Exercise

Classify the following \(3\times 3\) matrices. If multiple options are valid then choose the most restrictive option.

\(A=\left[\begin{array}{lll}\hfill 1&\hfill 1&\hfill 2\\ \hfill 1&\hfill 1&\hfill 1\\ \hfill 0&\hfill 1&\hfill 3 \end{array}\right]\), \(B=\left[\begin{array}{lll}\hfill 1&\hfill 0&\hfill 0\\ \hfill 1&\hfill 1&\hfill 0\\ \hfill 0&\hfill 1&\hfill 3 \end{array}\right]\), \(C=\left[\begin{array}{lll}\hfill 1&\hfill 2&\hfill 1\\ \hfill 0&\hfill 1&\hfill 4\\ \hfill 0&\hfill 0&\hfill 1 \end{array}\right]\), \(D=\left[\begin{array}{lll}\hfill 1&\hfill 0&\hfill 0\\ \hfill 0&\hfill 3&\hfill 0\\ \hfill 0&\hfill 0&\hfill 2 \end{array}\right]\)

Matrix \(A\) is: [...]

Matrix \(B\) is: [...]

Matrix \(C\) is: [...]

Matrix \(D\) is: [...]
Also offered: None of the above apply
A: A Square Matrix; Lower triangular; Upper triangular; Diagonal
If right: Correct answer, well done.
If wrong: Incorrect answer.

Equality of matrices

As we noted earlier, the terms in a matrix are called the elements or entries of the matrix.

\[\text{The elements of the matrix}\quad A=\left[\begin{array}{ll}\hfill1&\hfill 2\\ \hfill -1 &\hfill -4\end{array}\right]\quad \text{are}\quad1,2,-1,-4.\]

We say two matrices \(A\), \(B\) are equal to each other if and only if \(A\) and \(B\) have the same number of rows and the same number of columns and if each element of \(A\) is equal to the corresponding element of \(B\). When this is the case we write \(A=B\). For example if the following two matrices are equal:

\[A=\left[\begin{array}{ll}\hfill 1&\hfill \alpha \\ \hfill -1&\hfill -\beta\end{array}\right]\qquad  B=\left[\begin{array}{ll}\hfill 1&\hfill 2\\ \hfill -1&\hfill -4\end{array}\right]\]

then we can conclude that \(\alpha=2\) and \(\beta =4\).

The unit matrix

The unit matrix or the identity matrix, denoted by \(I_n\) (or, often, simply \(I\)), is the diagonal matrix of order \(n\) in which all diagonal elements are \(1\). Hence, for example, \(I_2=\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\) and \(I_3=\left[\begin{array}{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right]\).

The zero matrix

The zero matrix or null matrix is the matrix all of whose elements are zero.  There is a zero matrix for every size.  For example the \(2\times 3\) and \(2\times 2\) cases are:

\[\left[\begin{array}{ccc}0&0&0\\ 0&0&0\end{array}\right],\ \left[\begin{array}{cc}0&0\\ 0&0\end{array}\right].\]

 Zero matrices, of whatever size, are denoted by \({\underline 0}\) with the size often inferred from the context.

The transpose of a matrix

The transpose of a matrix \(A\) is a matrix where the rows of \(A\) become the columns of the new matrix and the columns of \(A\) become its rows.  For example

\[ A=\left[\begin{array}{ccc}1&2&3\\ 4&5&6\end{array}\right]\ \ \text{becomes}\ \left[\begin{array}{cc}1&4\\ 2&5\\ 3&6\end{array}\right]\]

The resulting matrix is called the transposed matrix of \(A\) and denoted \(A^T\).  In the previous example it is clear that \(A^T\) is not equal to \(A\) since the matrices are of different sizes.  If \(A\) is square \(n\times n\) then \(A^T\) will also be \(n\times n\).

Example

Find the transpose of the matrix  \(B=\left[\begin{array}{ccc}1&2&3\\ 4&5&6\\ 7&8&9\end{array}\right].\)

Solution

Interchanging rows with columns we find \[\qquad B^T=\left[\begin{array}{ccc}1&4&7\\ 2&5&8\\ 3&6&9\end{array}\right].\]

Both matrices are \(3\times 3\) but \(B\) and \(B^T\) are clearly different.

When the transpose of a matrix is equal to the original matrix i.e. \(A^T=A\), then we say that the matrix \(A\) is symmetric. (This is because it has symmetry about the leading diagonal.)

In the previous example \(B\) is not symmetric.

Example

Show that the matrix \(C=\left[\begin{array}{rrr}1&-2&3\\ -2&4&-5\\ 3&-5&6\end{array}\right]\) is symmetric.

Solution

Taking the transpose of \(C\):

\[C^T=\left[\begin{array}{rrr}1&-2&3\\ -2&4&-5\\ 3&-5&6\end{array}\right].\]

Clearly \(C^T=C\) and so \(C\) is a symmetric matrix.  Notice how the leading diagonal acts as a "mirror'' with the entries on the diagonal remaining fixed in place and, for example, \(c_{12}=-2\) and \(c_{21}=-2\) reflecting through the diagonal.

In general \(c_{ij}=c_{ji}\) for a symmetric matrix.

Addition and subtraction of matrices

Under what circumstances can we add two matrices i.e. define \(A+B\) for given matrices \(A\),\(B\)?

Consider

\[A=\left[\begin{array}{cc}1&2\\ 3&4\end{array}\right]\qquad\text{and}\qquad B=\left[\begin{array}{rrr}5&6&9\\ 7&8&10\end{array}\right].\]

There is no sensible way to define \(A+B\) in this case since \(A\) and \(B\) are different sizes. However, if we consider matrices of the same size then addition can be defined in a very natural way.

Consider \(A=\left[\begin{array}{cc}1&2\\ 3&4\end{array}\right]\) and \(B=\left[\begin{array}{cc}5&6\\ 7&8\end{array}\right]\).  The 'natural' way to add \(A\) and \(B\) is to add corresponding elements together:

\[A+B=\left[\begin{array}{cc}1+{ 5}&2+{ 6}\\3+{ 7}&4+{ 8}\end{array}\right]=\left[\begin{array}{rr}6&8\\10&12\end{array}\right].\]

In general if \(A\) and \(B\) are both  \(m\times n\) matrices, with elements \(a_{ij}\) and \(b_{ij}\) respectively, then their sum is a matrix \(C\), also \(m\times n\), such that the elements of \(C\) are

\[c_{ij}=a_{ij}+b_{ij}\qquad i=1,2,\dots ,m\quad j=1,2,\dots ,n.\]

In the above example

\[c_{11}=a_{11}+b_{11}= 1+5=6\qquad c_{21}=a_{21}+b_{21}=3+7=10\quad \text{and so on.}\]

Subtraction of matrices follows along similar lines:

\[D=A-B=\left[\begin{array}{cc}1-{ 5}&2-{ 6}\\3-{ 7}&4-{ 8}\end{array}\right]=\left[\begin{array}{rr}-4&-4\\-4&-4\end{array}\right].\]

Multiplication of a matrix by a number

There is also a natural way of defining the product of a matrix with a number. Using the matrix \(A\) above, we note that

\[ A+A=\left[\begin{array}{cc}1&2\\3&4\end{array}\right]+\left[\begin{array}{cc}1&2\\3&4\end{array}\right]=\left[\begin{array}{cc}2&4\\6&8\end{array}\right].\]

What we see is that \(2A\) (which is the shorthand notation for \(A+A\)) is obtained by multiplying every element of \(A\) by \(2\).

In general if \(A\) is an \(m\times n\) matrix with typical element \(a_{ij}\) then the product of a number \(k\) with \(A\) is written \(kA\) and has the corresponding elements \(ka_{ij}\).

Hence, again using the matrix \(A\) above,

\[7A=7\left[\begin{array}{cc}1&2\\3&4\end{array}\right]=\left[\begin{array}{rr}7&14\\ 21&28\end{array}\right].\]

Similarly:

\[ -3A=\left[\begin{array}{rr}-3&-6\\ -9&-12\end{array}\right].\]

Some simple matrix properties

Using the definition of matrix addition described above we can easily verify the following properties of matrix addition:

Key Point

Basic Properties of Matrices

Let \(A,B\) and \(C\) be matrices of the same size and let \(k\) be a scalar (i.e., \(k\) is a real number).

Matrix addition is commutative: \(A+B=B+A\)

Matrix addition is associative: \(A+(B+C)=(A+B)+C\)

The distributive law holds: \(k(A+B)=k\, A+k\, B\)

These Key Point results follow from the fact that \(a_{ij}+b_{ij}=b_{ij}+a_{ij}\) etc.

We can also show that the transpose of a matrix satisfies the following simple properties:

Key Point

Properties of Transposed Matrices

Let \(A\) and \(B\) be matrices of the same size.

\[(A+B)^T=A^T+B^T\]

\[(A-B)^T=A^T-B^T\]

\[(A^T)^T=A\]

Example

Show that \((A^T)^T=A\) for the matrix

\(A=\left[\begin{array}{ccc}1&2&3\\ 4&5&6\end{array}\right]\).

Solution

\(A^T=\left[\begin{array}{cc}1&4\\ 2&5\\ 3&6\end{array}\right]\) so that

\((A^T)^T=\left[\begin{array}{ccc}1&2&3\\ 4&5&6\end{array}\right]=A\).

Exercise

Find the transpose of each of the following matrices.

1. \(A = [value]\)
\(A^T = \)
Is \(A\) symmetric?
2. \(B = [value]\)
\(B^T = \)
Is \(B\) symmetric?
3. \(C = [value]\)
\(C^T = \)
Is \(C\) symmetric?
Feedback: In order to find the transpose of a matrix we swap the roles of rows and columns. If the resulting transpose is equal to the original matrix we say it is symmetric. 1. The transpose of the matrix \[A = [value]\] is given by \[A^T = [value].\] We see that \(A^T \neq A\), so \(A\) is not symmetric. 2. The transpose of the matrix \[B = [value]\] is given by \[B^T = [value].\] We see that \(B^T = B\), so \(B\) is symmetric. 3. The transpose of the matrix \[C = [value]\] is given by \[C^T = [value].\] We see that \(C^T = C\), so \(C\) is symmetric.

Exercise

For the following matrices find \(A+B,\ A-B,\ B-A,\ 2A\):

\[A = [value] \text{ and } B = [value].\]

1. \(A+B = \)
2. \(A-B = \)
3. \(B-A = \)
4. \(2A = \)
Feedback: We are give the matrices \[A = [value] \quad \text{and} \quad B = [value],\] and are asked to find the following. Recall that addition, subtraction and scalar multiplication of matrices is performed element-wise. 1. \[A + B = [value] + [value] = [value] = [value]\] 2. \[A - B = [value] - [value] = [value] = [value]\] 3. \[B - A = [value] - [value] = [value] = [value]\] 4. \[2A = 2 [value] = [value] = [value]\]

Exercise

For the matrices \(A = [value]\) and \(B = [value]\), calculate:

1. \(3(A+B) = \)
2. \(3A+3B = \)
3. \((A+B)^T = \)
4. \(A^T+B^T = \)

What do you notice about your answers to (a) and (b)? And (c) and (d)?
Feedback: We are given the matrices \(A = [value]\) and \(B = [value]\), and are asked to find the following: 1. \[\begin{aligned} 3(A + B) &= 3\cdot \left([value] + [value]\right) \\ &= 3 \cdot [value] \\ &= 3 \cdot [value] \\ &= [value] \\ &= [value] \end{aligned}\] 2. \[\begin{aligned} 3A + 3B &= 3 \cdot [value] + 3 \cdot [value] \\ &= [value] + [value] \\ &= [value] + [value] \\ &= [value] \\ &= [value] \end{aligned}\] 3. \[\begin{aligned} (A + B)^T &= \left([value] + [value]\right)^T \\ &= [value]^T \\ &= [value]^T \\ &= [value] \end{aligned}\] 4. \[\begin{aligned} A^T + B^T &= [value]^T + [value]^T \\ &= [value] + [value] \\ &= [value] \\ &= [value] \end{aligned}\] As you probably have noticed, the answers to (a) and (b) are equal, as are the answers to (c) and (d). This is no coincidence, since \(3(A + B) = 3A + 3B\) and \((A+B)^T = A^T + B^T\) by distributivity of scalar multiplication and transpose over addition.

Exercise

Find the coefficient matrix \(A\) of the system:

\[\begin{alignedat}{4} [value]x_1 & {}+{} & [value]x_2 & {}+{} & [value]x_3 &=[value] \\ [value]x_1 & {}+{} & [value]x_2 & {}+{} & [value]x_3 &=[value] \\ [value]x_1 & {}+{} & [value]x_2 & {}+{} & [value]x_3 &=&[value] \end{alignedat}\]

\(A = \)

If \(B=[value]\) determine \((3A^T-B)^T\).

\((3A^T-B)^T = \)
Feedback: We are asked to write down the coefficient matrix of the system: \[\begin{alignedat}{4} [value]x_1 & {}+{} & [value]x_2 & {}+{} & [value]x_3 &=[value] \\ [value]x_1 & {}+{} & [value]x_2 & {}+{} & [value]x_3 &=[value] \\ [value]x_1 & {}+{} & [value]x_2 & {}+{} & [value]x_3 &=&[value] \end{alignedat}\] In order to do this, we can just read off the coefficients of \(x_1\) into the first column, those of \(x_2\) into the second column and those of \(x_3\) into the third column. \[A = [value]\] Next, we are given the matrix \[B = [value]\] and are asked to compute \((3A^T - B)^T\). We can do this directly, but we can simplify the calculation by using the distributivity property of the transpose. \[(3A^T - B)^T = 3(A^T)^T - B^T = 3A - B^T\] We compute the latter expression, which is simpler. \[\begin{aligned} 3A - B^T &= 3 \cdot [value] - [value]^T \\ &= [value] - [value] \\ &= [value] - [value] \\ &= [value] \end{aligned}\]

Exercise

For the matrices \(A = [value]\) and \(B = [value]\), calculate:

1. \(3(A^T-B) = \)
2. \((3A-3B^T)^T = \)

What do you notice about your answers to (a) and (b)?
Feedback: We are given the matrices \(A = [value]\) and \(B = [value]\), and are asked to compute the following. 1. \[\begin{aligned} 3(A^T - B) &= 3\cdot \left([value]^T - [value]\right) \\ &= 3 \cdot \left([value] - [value]\right) \\ &= 3 \cdot [value] \\ &= [value] \end{aligned}\] 2. \[\begin{aligned} (3A - B^T)^T &= \left(3 \cdot [value] - 3 \cdot [value]^T\right)^T \\ &= \left(3 \cdot [value] - 3 \cdot [value]\right)^T \\ &= \left([value] - [value]\right)^T \\ &= \left([value]\right)^T \\ &= [value] \end{aligned}\] Notice how both part (a) and (b) gave the same answer. This is because \[3(A^T - B) = 3A^T - 3B = \left((3A^T - 3B)^T\right)^T = (3A - B^T)^T,\] by the distributivity properties.