2.1: Matrix Addition, Scalar Multiplication, and Transposition

Learning Objectives

• Add two matrices of the same size entry by entry.

• Multiply a matrix by a scalar and describe the effect on its entries.

• Recognize when matrix addition and scalar multiplication are not defined.

• Apply the properties of matrix addition, such as commutativity and associativity.

• Use the distributive properties of scalar multiplication with respect to matrix addition.

• Represent systems of equations and other data using matrix addition and scalar multiplication

Definition: A Matrix

A matrix is a rectangular array of numbers arranged in rows and columns. Matrices are usually denoted by uppercase letters: \(A, B, C, \dots \). If a matrix has $m$ rows and $n$ columns, its size is said to be \(m \times n\). The individual numbers inside a matrix are called entries. Each entry is written using a lowercase letter with subscripts that show its position in the matrix. The entry in row \(i\) and column \(j\) is called the \((i,j)\)-entry and is denoted by \(a_{ij}\).  
\[
A = [a_{ij}]_{m \times n} =
\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{bmatrix}
\]

Example

Given the following matrixes:

\[A = \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 0 & 5 & 6 \end{array} \right] \quad B = \left[ \begin{array}{rr} 1 & -1 \\ 0 & 2 \end{array} \right] \quad C = \left[ \begin{array}{r} 1 \\ 3 \\ 2 \end{array} \right] \nonumber \]

1- Find the sizes of the following matrices.

2- Find values for the following entries in the above matrices if exist:

• \(a_{12}\)
• \(a_{21}\)
• \(b_{22}\)
• \(c_{31}\).

Solution

1- Since the matrix \(A\) shown has \(2\) rows and \(3\) columns. In general, a matrix with \(m\) rows and \(n\) columns is referred to as an \(\boldsymbol{m} \times \boldsymbol{n}\) matrix or as having size \(\boldsymbol{m} \times \boldsymbol{n}\). Thus matrices \(A\), \(B\), and \(C\) above have sizes \(2 \times 3\), \(2 \times 2\), and \(3 \times 1\), respectively.

2- Here are the values of the given entries:

•  \(a_{12}\): The first index \(1\) means row 1, and the second index \(2\) means column 2. In matrix \(A\), this is \(2\). So \(a_{12} = 2\).  
•  \(a_{21}\): The first index \(2\) means row 2, and the second index \(1\) means column 1. In matrix \(A\), this is \(0\). So \(a_{21} = 0\).  
•   \(b_{22}\): The first index \(2\) means row 2, and the second index \(2\) means column 2. In matrix \(B\), this is \(2\). So \(b_{22} = 2\).  
•   \(c_{31}\): The first index \(3\) refers to row 3 and the second index \(1\) to column 1. In matrix \(C\), this is \(2\). So \(c_{31} = 2\).  
• The entry \(c_{13}\) does not exist. Matrix \(C\) has only one column of size \(3 \times 1\), which means it has 3 rows and only 1 column. The index \(c_{13}\) refers to the entry in row 1, column 3. Since there is no third column in \(C\), this entry cannot be found.

 

 

Definition: Equal Matrices

Two matrices \(A\) and \(B\) are called equal (written \(A = B\)) if and only if:

1. They have the same size.
2. Corresponding entries are equal.

Two points \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) in the plane are equal if and only if they have the same coordinates; that is, \((x_{1} = x_{2})\) and \((y_{1} = y_{2})\).  

Similarly, two matrices are equal if and only if they have the same size and all their corresponding entries are equal.

Example \(\PageIndex{1}\)

Given \(A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\), \(B = \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 3 & 0 & 1 \end{array} \right]\) and \(C = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right]\) discuss the possibility that \(A = B\), \(B = C\), \(A = C\).

Solution

\(A = B\) is impossible because \(A\) and \(B\) are of different sizes: \(A\) is \(2 \times 2\) whereas \(B\) is \(2 \times 3\). Similarly, \(B = C\) is impossible. But \(A = C\) is possible provided that corresponding entries are equal: \(\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right]\) means \(a = 1\), \(b = 0\), \(c = -1\), and \(d = 2\).

After understanding the general definition of a matrix, it is useful to classify matrices into special categories based on their shape or the arrangement of their entries. These classifications help us recognize important structures that appear frequently in applications and make certain computations easier. Let us look at some common \textbf{types of matrices}.

Type	Definition	Example
Row Matrix	A matrix with only one row (1 × n).	\[ \begin{bmatrix} 2 & -1 & 5 & 0 \end{bmatrix} \]
Column Matrix	A matrix with only one column (m × 1).	\[ \begin{bmatrix} 3 \\ -2 \\ 4 \end{bmatrix} \]
Square Matrix	A matrix with the same number of rows and columns (n × n).	\[ \begin{bmatrix} 1 & 2 &-11\\ 3 & 4 & 0 \\ -4 & 0 & 0.05 \end{bmatrix} \]
Zero Matrix	A matrix in which all entries are zero.	\[ \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \]
Diagonal Matrix	A square matrix with all non-diagonal entries equal to zero.	\[ \begin{bmatrix} 5 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 2 \end{bmatrix} \]
Identity Matrix	A square matrix with 1's on the main diagonal and 0's elsewhere.	\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Symmetric Matrix	A square matrix that is equal to its transpose, i.e. entries are mirrored across the diagonal.	\[ \begin{bmatrix} 2 & 3 & 1 \\ 3 & 4 & -2 \\ 1 & -2 & 5 \end{bmatrix} \]
Triangular Matrix	A square matrix is called upper triangular if all the entries below the main diagonal are zero. It is called lower triangular if all the entries above the main diagonal are zero.	Upper Triangle: \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \] Lower Triangle: \[ \begin{bmatrix} 7 & 0 & 0 \\ 8 & 9 & 0 \\ 1 & 4 & 2 \end{bmatrix} \]

 

Example

Identify the type of each of the following matrices 
\[
A = \begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}, \quad
B = \begin{bmatrix}
4 & 0 & 0 \\
0 & 7 & 0 \\
0 & 0 & 1
\end{bmatrix}, \quad
C = \begin{bmatrix}
5 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}.
\]

Solution

•  Matrix \(A\) is a square matrix of size \(2 \times 2\). It is also a zero matrix, since all entries are zero, and it can be considered a diagonal matrix because all off-diagonal entries are zero.  
•  Matrix \(B\) is a square matrix of size \(3 \times 3\). It is diagonal, since only the main diagonal entries are nonzero, and it is also symmetric.
• Matrix \(C\) is a square matrix of size \(3 \times 3\). It is diagonal, and at the same time both upper triangular and lower triangular, since all non-diagonal entries are zero.

Now that we know when two matrices are considered equal, we can move on to operations that combine or modify them. Just as we add or scale numbers, we can also add matrices and multiply a matrix by a scalar (a single number). However, with matrices there are specific constraints that we need to investigate. These operations will allow us to manipulate matrices in useful ways and prepare us for more advanced topics such as matrix multiplication and linear transformations.  

Matrix Addition

Definition: Matrix Addition

If \(A\) and \(B\) are matrices of the same size, their sum \(A + B\) is the matrix formed by adding corresponding entries.

If \(A = \left[ a_{ij} \right]\) and \(B = \left[ b_{ij} \right]\), this takes the form

\[A + B = \left[ a_{ij} + b_{ij} \right] \nonumber \]

Note that addition is not defined for matrices of different sizes.

Example \(\PageIndex{2}\)

If \(A = \left[ \begin{array}{rrr} 2 & 1 & 3 \\ -1 & 2 & 0 \end{array} \right]\) and \(B = \left[ \begin{array}{rrr} 1 & 1 & -1 \\ 2 & 0 & 6 \end{array} \right]\), compute \(A + B\).

Solution

\[A + B = \left[ \begin{array}{rrr} 2 + 1 & 1 + 1 & 3 - 1 \\ -1 + 2 & 2 + 0 & 0 + 6 \end{array} \right] = \left[ \begin{array}{rrr} 3 & 2 & 2 \\ 1 & 2 & 6 \end{array} \right] \nonumber \]

Example \(\PageIndex{3}\)

Find \(a\), \(b\), and \(c\) if \(\left[ \begin{array}{ccc} a & b & c \end{array} \right] + \left[ \begin{array}{ccc} c & a & b \end{array} \right] = \left[ \begin{array}{ccc} 3 & 2 & -1 \end{array} \right]\).

Solution

Add the matrices on the left side to obtain

\[\left[ \begin{array}{ccc} a + c & b + a & c + b \end{array} \right] = \left[ \begin{array}{rrr} 3 & 2 & -1 \end{array} \right] \nonumber \]

Because corresponding entries must be equal, this gives three equations: \(a + c = 3\), \(b + a = 2\), and \(c + b = -1\). Solving these yields \(a = 3\), \(b = -1\), \(c = 0\).

Proposition: Properties of Matrix Addition

Let \(A,B\) and \(C\) be matrices. Then, the following properties hold.

• Commutative Law of Addition \[A+B=B+A \label{mat1}\]
• Associative Law of Addition \[\left( A+B\right) +C=A+\left( B+C\right) \label{mat2}\]
• Existence of an Additive Identity \[\begin{array}{c} \mbox{There exists a zero matrix 0 such that}\\ A+0=A \label{mat3} \end{array}\]
• Existence of an Additive Inverse \[\begin{array}{c} \mbox{There exists a matrix $-A$ such that} \\ A+\left( -A\right) =0 \label{mat4} \end{array}\]

We call the zero matrix in \(\eqref{mat3}\) the additive identity. Similarly, we call the matrix \(-A\) in \(\eqref{mat4}\) the additive inverse. \(-A\) is defined to equal \(\left( -1\right) A = [-a_{ij}].\) In other words, every entry of \(A\) is multiplied by \(-1\), we will discuss in scalar multiplication. It si not hard to prove the above proberties as they are derived from the addition Properties  of real numbers. 

Scalar Multiplication of Matrices

Recall that we use the word scalar when referring to numbers. Therefore, scalar multiplication of a matrix is the multiplication of a matrix by a number. To illustrate this concept, consider the following example in which a matrix is multiplied by the scalar \(3\). \[3\left[ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 5 & 2 & 8 & 7 \\ 6 & -9 & 1 & 2 \end{array} \right] = \left[ \begin{array}{rrrr} 3 & 6 & 9 & 12 \\ 15 & 6 & 24 & 21 \\ 18 & -27 & 3 & 6 \end{array} \right]\nonumber \]

The new matrix is obtained by multiplying every entry of the original matrix by the given scalar.

The formal definition of scalar multiplication is as follows.

Definition: Scalar Multiplication of Matrices

If \(A=\left[ a_{ij}\right]\) and \(k\) is a scalar, then \(kA=\left[ ka_{ij}\right] .\)

Consider the following example.

Example: Effect of Multiplication by a Scalar

Find the result of multiplying the following matrix \(A\) by \(7\). \[A=\left[ \begin{array}{rr} 2 & 0 \\ 1 & -4 \end{array} \right]\nonumber \]

Solution

By the scalar multiplication definition, we multiply each element of \(A\) by \(7\). Therefore, \[7A = 7\left[ \begin{array}{rr} 2 & 0 \\ 1 & -4 \end{array} \right] = \left[ \begin{array}{rr} 7(2) & 7(0) \\ 7(1) & 7(-4) \end{array} \right] = \left[ \begin{array}{rr} 14 & 0 \\ 7 & -28 \end{array} \right]\nonumber \]

Similarly to addition of matrices, there are several properties of scalar multiplication which hold.

Proposition \(\PageIndex{2}\): Properties of Scalar Multiplication

Let \(A, B\) be matrices, and \(k, p\) be scalars. Then, the following properties hold.

• Distributive Law over Matrix Addition \[k \left( A+B\right) =k A+ kB\nonumber \]
• Distributive Law over Scalar Addition \[\left( k +p \right) A= k A+p A\nonumber \]
• Associative Law for Scalar Multiplication \[k \left( p A\right) = \left( k p \right) A\nonumber \]
• Rule for Multiplication by \(1\) \[1A=A\nonumber \]

Example \(\PageIndex{4}\)

Let \(A = \left[ \begin{array}{rrr} 3 & -1 & 0 \\ 1 & 2 & -4 \end{array} \right]\), \(B = \left[ \begin{array}{rrr} 1 & -1 & 1 \\ -2 & 0 & 6 \end{array} \right]\), \(C = \left[ \begin{array}{rrr} 1 & 0 & -2 \\ 3 & 1 & 1 \end{array} \right]\). Compute \(-A\), \(A - B\), and
\(A + B - C\).

Solution

\[\begin{aligned} -A &= \left[ \begin{array}{rrr} -3 & 1 & 0 \\ -1 & -2 & 4 \end{array} \right] \\ A - B &= \left[ \begin{array}{lcr} 3 - 1 & -1 - (-1) & 0 - 1 \\ 1 - (-2) & 2 - 0 & -4 - 6 \end{array} \right] = \left[ \begin{array}{rrr} 2 & 0 & -1 \\ 3 & 2 & -10 \end{array} \right] \\ A + B - C &= \left[ \begin{array}{rrl} 3 + 1 - 1 & -1 - 1 - 0 & 0 + 1 -(-2) \\ 1 - 2 - 3 & 2 + 0 - 1 & -4 + 6 -1 \end{array} \right] = \left[ \begin{array}{rrr} 3 & -2 & 3 \\ -4 & 1 & 1 \end{array} \right]\end{aligned} \nonumber \]

Example \(\PageIndex{5}\)

Solve \(\left[ \begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array} \right] + X = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right]\) where \(X\) is a matrix.

Solution

We solve a numerical equation \(a + x = b\) by subtracting the number \(a\) from both sides to obtain \(x = b - a\). This also works for matrices. To solve \(\left[ \begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array} \right] + X = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right]\) simply subtract the matrix \(\left[ \begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array} \right]\) from both sides to get

\[X = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right] - \left[ \begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array} \right] = \left[ \begin{array}{cr} 1 - 3 & 0 - 2 \\ -1 - (-1) & 2 - 1 \end{array} \right] = \left[ \begin{array}{rr} -2 & -2 \\ 0 & 1 \end{array} \right] \nonumber \]

The reader should verify that this matrix \(X\) does indeed satisfy the original equation.

 

Example \(\PageIndex{6}\)

If \(A = \left[ \begin{array}{rrr} 3 & -1 & 4 \\ 2 & 0 & 6 \end{array} \right]\) and \(B = \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 0 & 3 & 2 \end{array} \right]\) compute \(5A\), \(\frac{1}{2}B\), and \(3A - 2B\).

Solution

\[\begin{aligned} 5A &= \left[ \begin{array}{rrr} 15 & -5 & 20 \\ 10 & 0 & 30 \end{array} \right], \quad \frac{1}{2}B = \left[ \begin{array}{rrr} \frac{1}{2} & 1 & -\frac{1}{2} \\ 0 & \frac{3}{2} & 1 \end{array} \right] \\ 3A - 2B &= \left[ \begin{array}{rrr} 9 & -3 & 12 \\ 6 & 0 & 18 \end{array} \right] - \left[ \begin{array}{rrr} 2 & 4 & -2 \\ 0 & 6 & 4 \end{array} \right] = \left[ \begin{array}{rrr} 7 & -7 & 14 \\ 6 & -6 & 14 \end{array} \right]\end{aligned} \nonumber \]

If \(A\) is any matrix, note that \(kA\) is the same size as \(A\) for all scalars \(k\). We also have

\[0A = 0 \quad \mbox{ and } \quad k0 = 0 \nonumber \]

because the zero matrix has every entry zero. In other words, \(kA = 0\) if either \(k = 0\) or \(A = 0\). The converse of this statement is also true, as Example [exa:002159] shows.

Example \(\PageIndex{7}\)

If \(kA = 0\), show that either \(k = 0\) or \(A = 0\).

Solution

Write \(A = \left[ a_{ij} \right]\) so that \(kA = 0\) means \(ka_{ij} = 0\) for all \(i\) and \(j\). If \(k = 0\), there is nothing to do. If \(k \neq 0\), then \(ka_{ij} = 0\) implies that \(a_{ij} = 0\) for all \(i\) and \(j\); that is, \(A = 0\).

 

Example \(\PageIndex{8}\)

Simplify \(2(A + 3C) - 3(2C - B) - 3 \left[ 2(2A + B - 4C) - 4(A - 2C) \right]\) where \(A\), \(B\), and \(C\) are all matrices of the same size.

Solution

The reduction proceeds as though \(A\), \(B\), and \(C\) were variables.

\[\begin{aligned} 2(A &+ 3C) - 3(2C - B) - 3 \left[ 2(2A + B - 4C) - 4(A - 2C) \right] \\ &= 2A + 6C - 6C + 3B - 3 \left[ 4A + 2B - 8C - 4A + 8C \right] \\ &= 2A + 3B - 3 \left[ 2B \right] \\ &= 2A - 3B\end{aligned} \nonumber \]

So far, we have learned how to add matrices and multiply them by scalars, which allow us to combine and scale information within a matrix. Another important operation is the \textbf{transpose} of a matrix. This operation does not change the numerical values of the entries but instead rearranges their positions, giving us a new perspective on the same data. Understanding the transpose will be essential when we discuss symmetric matrices, inner products, and many applications in linear algebra.

Transpose of a Matrix

Many results about a matrix \(A\) involve the rows of \(A\), and the corresponding result for columns is derived in an analogous way, essentially by replacing the word row by the word column throughout. The following definition is made with such applications in mind.

Definition: Transpose of a Matrix

If \(A\) is an \(m \times n\) matrix, the transpose of \(A\), written \(A^{T}\), is the \(n \times m\) matrix whose rows are just the columns of \(A\) in the same order.

In other words, the first row of \(A^{T}\) is formed from the first column of \(A\) (with the entries listed in order). Similarly, the second row of \(A^{T}\) comes from the second column of \(A\), and so on for the remaining rows and columns.  

Formally, if \(A = [a_{ij}]\) is a matrix and we write \(A^{T} = [b_{ij}]\), then the entry \(b_{ij}\) in \(A^{T}\) is equal to the entry \(a_{ji}\) in \(A\). In other words,  
\[
b_{ij} = a_{ji}.
\]

Example \(\PageIndex{9}\)

Write down the transpose of each of the following matrices.

\[A = \left[ \begin{array}{r} 1 \\ 3 \\ 2 \end{array} \right] \quad B = \left[ \begin{array}{rrr} 5 & 2 & 6 \end{array} \right] \quad C = \left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array} \right] \quad D = \left[ \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & 2 \\ -1 & 2 & 1 \end{array} \right] \nonumber \]

Solution

\[A^{T} = \left[ \begin{array}{rrr} 1 & 3 & 2 \end{array} \right],\ B^{T} = \left[ \begin{array}{r} 5 \\ 2 \\ 6 \end{array} \right],\ C^{T} = \left[ \begin{array}{rrr} 1 & 3 & 5 \\ 2 & 4 & 6 \end{array} \right], \mbox{ and } D^{T} = D. \nonumber \]

\[\label{eq:transpose} \mbox{If } A = \left[ a_{ij} \right] \mbox{, then } A^{T} = \left[ a_{ji} \right]. \]

This is useful in verifying the following properties of transposition.

Theorem \(\PageIndex{1}\)

Let \(A\) and \(B\) denote matrices of the same size, and let \(k\) denote a scalar.

1. If \(A\) is an \(m \times n\) matrix, then \(A^{T}\) is an \(n \times m\) matrix.
2. \((A^{T})^{T} = A\).
3. \((kA)^{T} = kA^{T}\).
4. \((A + B)^{T} = A^{T} + B^{T}\).

We can use the definition of transpose and the properties of matrix addition and scalar multiplication to verify the above theorem. Here is a example.

 

Example \(\PageIndex{1}\)

Consider the matrix  \(A\) , and let \(k\) denote a scalar. Verify the following equation.

 \((kA)^{T} = kA^{T}\).

Solution

Proof. Property 1 is part of the definition of \(A^{T}\), and Property 2 follows from ([eq:transpose]). As to Property 3: If \(A = \left[ a_{ij} \right]\), then \(kA = \left[ ka_{ij} \right]\), so ([eq:transpose]) gives

\[(kA)^{T} = \left[ ka_{ji} \right] = k \left[ a_{ji} \right] = kA^{T} \nonumber \]

Finally, if \(B = \left[ b_{ij} \right]\), then \(A + B = \left[ c_{ij} \right]\) where \(c_{ij} = a_{ij} + b_{ij}\) Then ([eq:transpose]) gives Property 4:

\[(A + B)^{T} = \left[ c_{ij} \right]^{T} = \left[ c_{ji} \right] = \left[ a_{ji} + b_{ji} \right] = \left[ a_{ji} \right] + \left[ b_{ji} \right] = A^{T} + B^{T} \nonumber \]

There is another useful way to think of transposition. If \(A = \left[ a_{ij} \right]\) is an \(m \times n\) matrix, the elements \(a_{11}, a_{22}, a_{33}, \dots\) are called the main diagonal of \(A\). Hence the main diagonal extends down and to the right from the upper left corner of the matrix \(A\); shown in bold in the following examples:

\[ \begin{pmatrix}
\mathbf{a_{11}} & a_{12} \\
a_{21} & \mathbf{ a_{22} } \\
a_{31} & a_{32}
\end{pmatrix} \begin{pmatrix}
\mathbf{a_{11}} & a_{12} & a_{13} \\
a_{21} & \mathbf{ a_{22}} & a_{23}
\end{pmatrix} \begin{pmatrix}
\mathbf{a_{11}} & a_{12} & a_{13} \\
a_{21} & \mathbf{a_{22}} & a_{23} \\
a_{31} & a_{32} & \mathbf{ a_{33}}
\end{pmatrix} \begin{pmatrix}
\mathbf{ a_{11}} \\
a_{21}
\end{pmatrix}\]

Thus forming the transpose of a matrix \(A\) can be viewed as “flipping” \(A\) about its main diagonal, or as “rotating” \(A\) through \(180^{\circ}\) about the line containing the main diagonal. This makes Property 2 in Theorem \(\PageIndex{2}\) transparent.

Example \(\PageIndex{10}\)

Solve for \(A\) if \(\left(2A^{T} - 3 \left[ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right] \right)^{T} = \left[ \begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array} \right]\).

Solution

Using Theorem \(\PageIndex{2}\), the left side of the equation is

\[\left(2A^{T} - 3 \left[ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right]\right)^{T} = 2\left(A^{T}\right)^{T} - 3 \left[ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right]^{T} = 2A - 3 \left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right] \nonumber \]

Hence the equation becomes

\[2A - 3 \left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right] = \left[ \begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array} \right] \nonumber \]

Thus \(2A = \left[ \begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array} \right] + 3 \left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right] = \left[ \begin{array}{rr} 5 & 0 \\ 5 & 5 \end{array} \right]\), so finally \(A = \frac{1}{2} \left[ \begin{array}{rr} 5 & 0 \\ 5 & 5 \end{array} \right] = \frac{5}{2} \left[ \begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right]\).

In solving problems, the transpose often appears as a useful tool. For example, it plays an important role when checking whether a matrix is symmetric. The transpose is an operation, and we describe it by saying we “take the transpose of \(A\).”

Reacll that a matrix \(A\) is called symmetric if \(A = A^{T}\). A symmetric matrix \(A\) is necessarily square (if \(A\) is \(m \times n\), then \(A^{T}\) is \(n \times m\), so \(A = A^{T}\) forces \(n = m\)). The name comes from the fact that these matrices exhibit a symmetry about the main diagonal. That is, entries that are directly across the main diagonal from each other are equal.

For example, \(\left[ \begin{array}{ccc} a & b & c \\ b^\prime & d & e \\ c^\prime & e^\prime & f \end{array} \right]\) is symmetric when \(b = b^\prime\), \(c = c^\prime\), and \(e = e^\prime\).

Example \(\PageIndex{11}\)

If \(A\) and \(B\) are symmetric \(n \times n\) matrices, show that \(A + B\) is symmetric.

Solution

We have \(A^{T} = A\) and \(B^{T} = B\), so, by Theorem [thm:002240], we have \((A + B)^{T} = A^{T} + B^{T} = A + B\). Hence \(A + B\) is symmetric.

Example \(\PageIndex{12}\)

Suppose a square matrix \(A\) satisfies \(A = 2A^{T}\). Show that necessarily \(A = 0\).

Solution

If we iterate the given equation, Theorem [thm:002240] gives

\[A = 2A^{T} = 2 {\left[ 2A^{T} \right]}^T = 2 \left[ 2(A^{T})^{T} \right] = 4A \nonumber \]

Subtracting \(A\) from both sides gives \(3A = 0\), so \(A = \frac{1}{3}(0) = 0\).

---

1. If \(p\) and \(q\) are statements, we say that \(p\) implies \(q\) if \(q\) is true whenever \(p\) is true. Then “\(p\) if and only if \(q\)” means that both \(p\) implies \(q\) and \(q\) implies \(p\). See Appendix [chap:appbproofs] for more on this.↩

Exercises

Exercise

Find \(a\), \(b\), \(c\), and \(d\) if

1. \(\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} c - 3d & -d \\ 2a + d & a + b \end{array} \right]\)
2. \(\left[ \begin{array}{cc} a - b & b - c \\ c - d & d - a \end{array} \right] = 2 \left[ \begin{array}{rr} 1 & 1 \\ -3 & 1 \end{array} \right]\)
3. \(3 \left[ \begin{array}{c} a \\ b \end{array} \right] + 2 \left[ \begin{array}{rr} b \\ a \end{array} \right] = \left[ \begin{array}{rr} 1 \\ 2 \end{array} \right]\)
4. \(\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} b & c \\ d & a \end{array} \right]\)

Answer

2. \((a\ b\ c\ d) = (-2, -4, -6, 0) + t(1, 1, 1, 1)\),
3. \(t\) arbitrary \(a = b = c = d = t\), \(t\) arbitrary

Exercise

Compute the following:

1. \( \left[\begin{array}{lll}3 & 2 & 1 \\5 & 1 & 0\end{array}\right]-5\left[\begin{array}{rrr}3 & 0 & -2 \\1 & -1 & 2\end{array}\right]\)
2. \( 3\left[\begin{array}{r}3 \\-1\end{array}\right]-5\left[\begin{array}{l}6 \\2\end{array}\right]+7\left[\begin{array}{r}1 \\-1 \end{array}\right] \)
3. \( \left[\begin{array}{rr}-2 & 1 \\3 & 2\end{array}\right]-4\left[\begin{array}{ll}1 & -2 \\0 & -1\end{array}\right]+3\left[\begin{array}{rr}2 & -3 \\-1 & -2\end{array}\right]\)
4. \( \left[\begin{array}{lll}3 & -1 & 2\end{array}\right]-2\left[\begin{array}{lll}9 & 3 & 4\end{array}\right]+\left[\begin{array}{lll}3 & 11 & -6\end{array}\right]\)
5. \( \left[\begin{array}{rrrr}1 & -5 & 4 & 0 \\2 & 1 & 0 & 6\end{array}\right]^T \)
6. \( \left[\begin{array}{rrr}0 & -1 & 2 \\1 & 0 & -4 \\-2 & 4 & 0\end{array}\right]^T \)
7. \( \left[\begin{array}{rr}3 & -1 \\2 & 1\end{array}\right]-2\left[\begin{array}{rr}1 & -2 \\1 & 1\end{array}\right]^T \)
8. \( 3\left[\begin{array}{rr}2 & 1 \\-1 & 0\end{array}\right]^T-2\left[\begin{array}{rr}1 & -1 \\2 & 3\end{array}\right] \)

Answer

2. \(\left[ \begin{array}{r} -14 \\ -20 \end{array} \right]\) \((-12, 4, -12)\)
3. \(\left[ \begin{array}{rrr} 0 & 1 & -2 \\ -1 & 0 & 4 \\ 2 & -4 & 0 \end{array} \right]\)
4. \(\left[ \begin{array}{rr} 4 & -1 \\ -1 & -6 \end{array} \right]\)

Exercise

Let \(A = \left[ \begin{array}{rr} 2 & 1 \\ 0 & -1 \end{array} \right]\),
\(B = \left[ \begin{array}{rrr} 3 & -1 & 2 \\ 0 & 1 & 4 \end{array} \right]\), \(C = \left[ \begin{array}{rr} 3 & -1 \\ 2 & 0 \end{array} \right]\),
\(D = \left[ \begin{array}{rr} 1 & 3 \\ -1 & 0 \\ 1 & 4 \end{array} \right]\), and \(E = \left[ \begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]\).

Compute the following (where possible).

1. \(3A - 2B\)
2. \(5C\)
3. \(3E^{T}\)
4. \(B + D\)
5. \(4A^{T} - 3C\)
6. \((A + C)^{T}\)
7. \(2B - 3E\)
8. \(A - D\)
9. ((B - 2E)^{T}\)

Answer

2. \(\left[ \begin{array}{rr} 15 & -5 \\ 10 & 0 \end{array} \right]\)
3. Impossible
4. \(\left[ \begin{array}{rr} 5 & 2 \\ 0 & -1 \end{array} \right]\)
5. Impossible

Exercise

Find \(A\) if:

1. \(5A - \left[ \begin{array}{rr} 1 & 0 \\ 2 & 3 \end{array} \right] = 3A - \left[ \begin{array}{rr} 5 & 2 \\ 6 & 1 \end{array} \right]\)
2. \(3A - \left[ \begin{array}{r} 2 \\ 1 \end{array} \right] = 5A - 2 \left[ \begin{array}{rr} 3 \\ 0 \end{array} \right]\)

Answer

b. \(\left[ \begin{array}{r} 2 \\ -\frac{1}{2} \end{array} \right]\)

Exercise

Find \(A\) in terms of \(B\) if:

1. \(A + B = 3A + 2B\)
2. \(2A - B = 5(A + 2B)\)

Answer

b. \(A = -\frac{11}{3}B\)

Exercise

If \(X\), \(Y\), \(A\), and \(B\) are matrices of the same size, solve the following systems of equations to obtain \(X\) and \(Y\) in terms of \(A\) and \(B\).

1. \(5X + 3Y = A \\ 2X + Y = B\)
2. \(4X + 3Y = A \\ 5X + 4Y = B\)

Answer

b. \(X = 4A - 3B\), \(Y = 4B - 5A\)

Example

Find all matrices \(X\) and \(Y\) such that:

1. \(3X - 2Y = \left[ \begin{array}{rr} 3 & - 1 \end{array} \right]\)
2. \(2X - 5Y = \left[ \begin{array}{rr} 1 & 2 \end{array} \right]\)

Solution

\(Y = (s, t)\), \(X = \frac{1}{2}(1 + 5s, 2 + 5t)\); \(s\) and \(t\) arbitrary

Exercise

Simplify the following expressions where \(A\), \(B\), and \(C\) are matrices.

1. \(2 \left[ 9(A - B) + 7(2B - A) \right]\)
   \(- 2 \left[ 3(2B + A) - 2(A + 3B) - 5(A + B) \right]\)
2. \(5 \left[ 3(A - B + 2C) - 2(3C - B) - A \right]\)
   \(+ 2 \left[ 3(3A - B + C) + 2(B - 2A) - 2C \right]\)

Answer

b. \(20A - 7B + 2C\)

Exercise

If \(A\) is any \(2 \times 2\) matrix, show that:

1. \(A = a \left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right] + b \left[ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right] + c \left[ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right] + d \left[ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array} \right]\) for some numbers \(a\), \(b\), \(c\), and \(d\).
2. \(A = p \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right] + q \left[ \begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right] + r \left[ \begin{array}{rr} 1 & 0 \\ 1 & 0 \end{array} \right] + s \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right]\) for some numbers \(p\), \(q\), \(r\), and \(s\).

Answer

b. If \(A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\), then \((p, q, r, s) = \frac{1}{2}(2d, a + b - c - d, a - b + c - d, -a + b + c + d)\).

Exercise

Let \(A = \left[ \begin{array}{rrr} 1 & 1 & -1 \end{array} \right]\),
\(B = \left[ \begin{array}{rrr} 0 & 1 & 2 \end{array} \right]\), and \(C = \left[ \begin{array}{rrr} 3 & 0 & 1 \end{array} \right]\). If
\(rA + sB + tC = 0\) for some scalars \(r\), \(s\), and \(t\), show that necessarily \(r = s = t = 0\).

 

Exercise

1. If \(Q + A = A\) holds for every \(m \times n\) matrix \(A\), show that \(Q = 0_{mn}\).
2. If \(A\) is an \(m \times n\) matrix and \(A + A^\prime = 0_{mn}\), show that \(A^\prime = -A\

Answer

b. If \(A + A^\prime = 0\) then \(-A = -A + 0 = -A + (A + A^\prime) = (-A + A) + A^\prime = 0 + A^\prime = A^\prime\)

Exercise

If \(A\) denotes an \(m \times n\) matrix, show that \(A = -A\) if and only if \(A = 0\).

Exercise \

In each case determine all \(s\) and \(t\) such that the given matrix is symmetric:

1. \(\left[ \begin{array}{rr} 1 & s \\ -2 & t \end{array} \right]\)
2. \(\left[ \begin{array}{cc} s & t \\ st & 1 \end{array} \right]\)
3. \(\left[ \begin{array}{crc} s & 2s & st \\ t & -1 & s \\ t & s^{2} & s \end{array} \right]\)
4. \(\left[ \begin{array}{ccc} 2 & s & t \\ 2s & 0 & s + t \\ 3 & 3 & t \end{array} \right]\)

Answer

2. \(s = 1\) or \(t = 0\)
3. \(s = 0\), and \(t = 3\)

Exercise

In each case find the matrix \(A\).

1. \(\left(A + 3 \left[ \begin{array}{rrr} 1 & -1 & 0 \\ 1 & 2 & 4 \end{array} \right] \right)^{T} = \left[ \begin{array}{rr} 2 & 1 \\ 0 & 5 \\ 3 & 8 \end{array} \right]\)
2. \(\left(3A^{T} + 2 \left[ \begin{array}{rr} 1 & 0 \\ 0 & 2 \end{array} \right] \right)^{T} = \left[ \begin{array}{rr} 8 & 0 \\ 3 & 1 \end{array} \right]\)
3. \(\left(2A - 3 \left[ \begin{array}{rrr} 1 & 2 & 0 \end{array} \right] \right)^{T} = 3A^{T} + \left[ \begin{array}{rrr} 2 & 1 & -1 \end{array} \right]^{T}\)
4. \(\left(2A^{T} - 5 \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right] \right)^{T} = 4A - 9 \left[ \begin{array}{rr} 1 & 1 \\ -1 & 0 \end{array} \right]\)

Answer

2. \(\left[ \begin{array}{rr} 2 & 0 \\ 1 & -1 \end{array} \right]\)
3. \(\left[ \begin{array}{rr} 2 & 7 \\ -\frac{9}{2} & -5 \end{array} \right]\)

Exercise

Let \(A\) and \(B\) be symmetric (of the same size). Show that each of the following is symmetric.

1. \((A - B)\)
2. \(kA\) for any scalar \(k\)

Answer

1. \(A = A^{T}\),
2. \((kA)^{T} = kA^{T} = kA\).

Exercise

Show that \(A + A^{T}\) is symmetric for any square matrix \(A\).

Exercise

If \(A\) is a square matrix and \(A = kA^{T}\) where \(k \neq \pm 1\), show that \(A = 0\).

Exercise

In each case either show that the statement is true or give an example showing it is false.

1. If \(A + B = A + C\), then \(B\) and \(C\) have the same size.
2. If \(A + B = 0\), then \(B = 0\).
3. If the \((3, 1)\)-entry of \(A\) is \(5\), then the \((1, 3)\)-entry of \(A^{T}\) is \(-5\).
4. \(A\) and \(A^{T}\) have the same main diagonal for every matrix \(A\).
5. If \(B\) is symmetric and \(A^{T} = 3B\), then \(A = 3B\).
6. If \(A\) and \(B\) are symmetric, then \(kA + mB\) is symmetric for any scalars \(k\) and \(m\).

Answer

1. False. Take \(B = -A\) for any \(A \neq 0\).
2. True. Transposing fixes the main diagonal.
3. True. \((kA + mB)^{T} = (kA)^{T} + (mB)^{T} = kA^{T} + mB^{T} = kA + mB\)

Exercise

A square matrix \(W\) is called skew-symmetric if \(W^{T} = -W\). Let \(A\) be any square matrix.

1. Show that \(A - A^{T}\) is skew-symmetric.
2. Find a symmetric matrix \(S\) and a skew-symmetric matrix \(W\) such that \(A = S + W\).
3. Show that \(S\) and \(W\) in part (b) are uniquely determined by \(A\).

Answer

c. Suppose \(A = S + W\), where \(S = S^{T}\) and \(W = -W^{T}\). Then \(A^{T} = S^{T} + W^{T} = S - W\), so \(A + A^{T} = 2S\) and \(A - A^{T} = 2W\). Hence \(S = \frac{1}{2}(A + A^{T})\) and \(W = \frac{1}{2}(A - A^{T})\) are uniquely determined by \(A\).

Exercise

If \(W\) is skew-symmetric (Exercise \(\PageIndex{20}\)), show that the entries on the main diagonal are zero.

2.  

Exercise

Prove the following parts of Theorem [thm:002170].

1. \((k + p)A = kA + pA\)
2. \((kp)A = k(pA)\)

Answer

b. If \(A = \left[ a_{ij} \right]\) then \((kp)A = \left[ (kp)a_{ij} \right] = \left[ k(pa_{ij}) \right] = k \left[ pa_{ij} \right] = k(pA)\).

1.  

Exercise

Let \(A, A_{1}, A_{2}, \dots, A_{n}\) denote matrices of the same size. Use induction on \(n\) to verify the following extensions of properties 5 and 6 of Theorem \(\PageIndex{1}\).

1. (k(A_{1} + A_{2} + \dots + A_{n}) = kA_{1} + kA_{2} + \dots + kA_{n}\) for any number \(k\)
2. \((k_{1} + k_{2} + \dots + k_{n})A = k_{1}A + k_{2}A + \dots + k_{n}A\) for any numbers \(k_{1}, k_{2}, \dots, k_{n}\)

Example

Let \(A\) be a square matrix. If \(A = pB^{T}\) and \(B = qA^{T}\) for some matrix \(B\) and numbers \(p\) and \(q\), show that either \(A = 0 = B\) or \(pq = 1\).