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Mathematics LibreTexts

4: Unary Matrix Operations

( \newcommand{\kernel}{\mathrm{null}\,}\)

Learning Objectives

After reading this chapter, you should be able to:

  1. know what unary operations are,
  2. fnd the transpose of a square matrix and its relationship to symmetric matrices,
  3. find the trace of a matrix, and
  4. find the determinant of a matrix by the cofactor method.

What is the transpose of a matrix?

Let \left\lbrack A \right\rbrack be a m \times n matrix. Then \left\lbrack B \right\rbrack is the transpose of \left\lbrack A \right\rbrack if b_{ji} = a_{ij} for all i and j. That is, the i^{th} row and the j^{th} column element of \left\lbrack A \right\rbrack is the j^{th} row and i^{th} column element of \left\lbrack B \right\rbrack. Note, \left\lbrack B \right\rbrack would be a n \times m matrix. The transpose of \left\lbrack A \right\rbrack is denoted by \left\lbrack A \right\rbrack^{T}.

Example 1

Find the transpose of

\left\lbrack A \right\rbrack = \begin{bmatrix} \begin{matrix} 25 & 20 \\ \end{matrix} & \begin{matrix} 3 & 2 \\ \end{matrix} \\ \begin{matrix} 5 & 10 \\ \end{matrix} & \begin{matrix} 15 & 25 \\ \end{matrix} \\ \begin{matrix} 6 & 16 \\ \end{matrix} & \begin{matrix} 7 & 27 \\ \end{matrix} \\ \end{bmatrix} \nonumber

Solution

The transpose of \left\lbrack A \right\rbrack is

\left\lbrack A \right\rbrack^{T} = \left\lbrack \begin{matrix} \begin{matrix} 25 \\ 20 \\ \end{matrix} \\ \begin{matrix} 3 \\ 2 \\ \end{matrix} \\ \end{matrix}\begin{matrix} \begin{matrix} 5 \\ 10 \\ \end{matrix} \\ \begin{matrix} 15 \\ 25 \\ \end{matrix} \\ \end{matrix}\begin{matrix} \begin{matrix} 6 \\ 16 \\ \end{matrix} \\ \begin{matrix} 7 \\ 27 \\ \end{matrix} \\ \end{matrix} \right\rbrack \nonumber

Note, the transpose of a row vector is a column vector and the transpose of a column vector is a row vector.

Also, note that the transpose of a transpose of a matrix is the matrix itself. That is,

\left( \left\lbrack A \right\rbrack^{T} \right)^{T} = \left\lbrack A \right\rbrack \nonumber

Also,

\left( \left\lbrack A \right\rbrack + \left\lbrack B \right\rbrack \right)^{T} = \left\lbrack A \right\rbrack^{T} + \left\lbrack B \right\rbrack^{T};\ \left( {cA} \right)^{T} = cA^{T} \nonumber

What is a symmetric matrix?

A square matrix \left\lbrack A \right\rbrack with real elements were a_{ij} = a_{ji} for i = 1,\ 2,\ \ldots\ ,\ n and j = 1,\ 2,\ \ldots\ ,\ n is called a symmetric matrix. This is the same as saying that if \left\lbrack A \right\rbrack = \left\lbrack A \right\rbrack^{T}, then \left\lbrack A \right\rbrack^{T} is a symmetric matrix.

Example 2

Give an example of a symmetric matrix

Solution

\left\lbrack A \right\rbrack = \begin{bmatrix} 21.2 & 3.2 & 6 \\ 3.2 & 21.5 & 8 \\ 6 & 8 & 9.3 \\ \end{bmatrix} \nonumber

Is a symmetric matrix as a_{12} = a_{21} = 3.2,\ \ a_{13} = a_{31} = 6 and a_{13} = a_{31} = 8.

What is a skew-symmetric matrix?

A n \times n matrix is skew-symmetric if a_{ij} = - a_{ji}, for i = 1,\ \ldots\ ,\ n and j = 1,\ \ldots\ ,\ n. This the same as

\left\lbrack A \right\rbrack = - \left\lbrack A \right\rbrack^{T} \nonumber

Example 3

Give an example of a skew-symmetric matrix

Solution

\begin{bmatrix} 0 & 1 & 2 \\ - 1 & 0 & - 5 \\ - 2 & 5 & 0 \\ \end{bmatrix} \nonumber

Is a skew symmetric matrix as a_{12} = - a_{21} = 1;\ \ a_{13} = - a_{31} = 2;\ a_{23} = - a_{32} = - 5. Since a_{ii} = - a_{ii} only if a_{ii} = 0, all the diagonal elements of a skew-symmetric matrix have to be zero.

What is the trace of a matrix?

The trace of a n \times n matrix \left\lbrack A \right\rbrack is the sum of the diagonal entries of \left\lbrack A \right\rbrack. That is

{tr}\left\lbrack A \right\rbrack = \sum_{i = 1}^{n}a_{ii} \nonumber

Example 4

Find the trace of:

\left\lbrack A \right\rbrack = \begin{bmatrix} 15 & 6 & 7 \\ 2 & - 4 & 2 \\ 3 & 2 & 6 \\ \end{bmatrix} \nonumber

Solution

\begin{split} {tr}\left\lbrack A \right\rbrack &= \sum_{i = 1}^{n}a_{ii}\\ &= 15 + \left( - 4 \right) + 6\\ &= 17 \end{split} \nonumber

Example 5

The sales of tires are given by make (rows) and quarters (columns) for Blowout r’us store location A, as shown below

\left\lbrack A \right\rbrack = \left\lbrack \begin{matrix} 25 \\ 5 \\ 6 \\ \end{matrix}\begin{matrix} 20 \\ 10 \\ 16 \\ \end{matrix}\begin{matrix} 3 \\ 15 \\ 7 \\ \end{matrix}\begin{matrix} 2 \\ 25 \\ 27 \\ \end{matrix} \right\rbrack \nonumber

Where the rows represent the sales of Tirestone, Michigan, and Copper tires, and the columns represent the quarter numbers 1, 2, 3, 4.

Find the total yearly revenue of store A if the prices of tires vary by quarters as follows

\left\lbrack B \right\rbrack = \left\lbrack \begin{matrix} 33.25 \\ 40.19 \\ 25.03 \\ \end{matrix}\begin{matrix} 30.01 \\ 38.02 \\ 22.02 \\ \end{matrix}\begin{matrix} 35.02 \\ 41.03 \\ 27.03 \\ \end{matrix}\begin{matrix} 30.05 \\ 38.23 \\ 22.95 \\ \end{matrix} \right\rbrack \nonumber

Where the rows represent the cost of each tire made by Tirestone, Michigan, and Copper, and the columns represent the quarter numbers.

Solution

\left\lbrack C \right\rbrack = \left\lbrack B \right\rbrack^{T} \nonumber

Example 6

Blowout r’us store location A and the sales of tires are given by make (in rows) and quarters (in columns) as shown below

\begin{split} \left\lbrack A \right\rbrack &= \left\lbrack \begin{matrix} 25 \\ 5 \\ 6 \\ \end{matrix}\begin{matrix} 20 \\ 10 \\ 16 \\ \end{matrix}\begin{matrix} 3 \\ 15 \\ 7 \\ \end{matrix}\begin{matrix} 2 \\ 25 \\ 27 \\ \end{matrix} \right\rbrack\\ &= \left\lbrack \begin{matrix} 33.25 \\ 40.19 \\ 25.03 \\ \end{matrix}\begin{matrix} 30.01 \\ 38.02 \\ 22.02 \\ \end{matrix}\begin{matrix} 35.02 \\ 41.03 \\ 27.03 \\ \end{matrix}\begin{matrix} 30.05 \\ 38.23 \\ 22.95 \\ \end{matrix} \right\rbrack\\ &= \left\lbrack \begin{matrix} \begin{matrix} 33.25 \\ 30.01 \\ \end{matrix} \\ \begin{matrix} 35.02 \\ 30.05 \\ \end{matrix} \\ \end{matrix}\begin{matrix} \begin{matrix} 40.19 \\ 38.02 \\ \end{matrix} \\ \begin{matrix} 41.03 \\ 38.23 \\ \end{matrix} \\ \end{matrix}\begin{matrix} \begin{matrix} 25.03 \\ 22.02 \\ \end{matrix} \\ \begin{matrix} 27.03 \\ 22.95 \\ \end{matrix} \\ \end{matrix} \right\rbrack \end{split} \nonumber

Recognize now that if we find \left\lbrack A \right\rbrack\left\lbrack C \right\rbrack, we get

\begin{split} \left\lbrack D \right\rbrack &= \left\lbrack A \right\rbrack\left\lbrack C \right\rbrack\\ &= \left\lbrack \begin{matrix} 25 \\ 5 \\ 6 \\ \end{matrix}\begin{matrix} 20 \\ 10 \\ 16 \\ \end{matrix}\begin{matrix} 3 \\ 15 \\ 7 \\ \end{matrix}\begin{matrix} 2 \\ 25 \\ 27 \\ \end{matrix} \right\rbrack\left\lbrack \begin{matrix} \begin{matrix} 33.25 \\ 30.01 \\ \end{matrix} \\ \begin{matrix} 35.02 \\ 30.05 \\ \end{matrix} \\ \end{matrix}\begin{matrix} \begin{matrix} 40.19 \\ 38.02 \\ \end{matrix} \\ \begin{matrix} 41.03 \\ 38.23 \\ \end{matrix} \\ \end{matrix}\begin{matrix} \begin{matrix} 25.03 \\ 22.02 \\ \end{matrix} \\ \begin{matrix} 27.03 \\ 22.95 \\ \end{matrix} \\ \end{matrix} \right\rbrack\\ &= \begin{bmatrix} 1597 & 1965 & 1193 \\ 1743 & 2152 & 1325 \\ 1736 & 2169 & 1311 \\ \end{bmatrix} \end{split} \nonumber

The diagonal elements give the sales of each brand of tire for the whole year. That is

d_{11} = \$ 1597 (Tirestone sales)

d_{22} = \$ 2152 (Michigan sales)

d_{33} = \$ 1597 (Copper sales)

The total yearly sales of all three brans of tires are

\begin{split} \sum_{i = 1}^{3}d_{ii} &= 1597 + 2152 + 1311\\ &= \text{\$} 5060 \end{split} \nonumber

And this is the trace of matrix \left\lbrack D \right\rbrack.

Define the determinant of a matrix.

The determinant of a square matrix is a single unique real number corresponding to a matrix. For a matrix \left\lbrack A \right\rbrack, determinant is denoted by \left| A \right| or \det\left( A \right). So do not use \left\lbrack A \right\rbrack and \left| A \right| interchangeably.

For a 2 \times 2 matrix

\left\lbrack A \right\rbrack = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \nonumber

\det\left( A \right) = a_{11}a_{22} - a_{12}a_{21} \nonumber

How does one calculate the determinant of any square matrix?

Let \left\lbrack A \right\rbrack be a n \times n matrix. The minor of entry a_{ij} is denoted by M_{ij} and is defined as the determinant of the \left( n - 1 \right) \times \left( n - 1 \right) submatrix of \left\lbrack A \right\rbrack, where the submatrix is obtained by deleting the i^{th} row and j^{th} column of the matrix \left\lbrack A \right\rbrack. The determinant is then given by

\det\left( A \right) = \sum_{j = 1}^{n}{\left( - 1 \right)^{i + j}a_{ij}M_{ij}}{\ for\ any\ }i = 1,\ 2,\ \ldots\ ,\ n \nonumber

or

\det\left( A \right) = \sum_{i = 1}^{n}{\left( - 1 \right)^{i + j}a_{ij}M_{ij}}{\ for\ any\ }j = 1,\ 2,\ \ldots\ ,\ n \nonumber

Couple that with \det\left( A \right) = a_{11} for a 1 \times 1 matrix \left\lbrack A \right\rbrack, we can always reduce the determinant of a matrix to determinants of 1 \times 1 matrices. The number \left( - 1 \right)^{i + j}M_{ij} is called the cofactor of a_{ij} and is denoted by c_{ij}. The formula for the determinant can then be written as

\det\left( A \right) = \sum_{j = 1}^{n}{a_{ij}C_{ij}}{\ for\ any\ }i = 1,\ 2,\ \ldots\ ,\ n \nonumber

or

\det\left( A \right) = \sum_{i = 1}^{n}{a_{ij}C_{ij}}{\ for\ any\ }j = 1,\ 2,\ \ldots\ ,\ n \nonumber

Determinants are not generally calculated using this method as it becomes computationally intensive for large matrices. For a n \times n matrix, it requires arithmetic operations proportional to n!.

Example 6

Find the determinant of

\lbrack A\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix} \nonumber

Solution

Method 1:

\det\left( A \right) = \sum_{j = 1}^{3}{\left( - 1 \right)^{i + j}a_{ij}M_{ij}{\ \ for\ \ any\ \ }i = 1,\ \ 2,\ \ 3} \nonumber

Let us choose i = 1 in the formula

\begin{split} \det\left( A \right) &= \sum_{j = 1}^{3}{\left( - 1 \right)^{1 + j}a_{1j}M_{1j}}\\ &= \left( - 1 \right)^{1 + 1}a_{11}M_{11} + \left( - 1 \right)^{1 + 2}a_{12}M_{12} + \left( - 1 \right)^{1 + 3}a_{13}^{\ }\ M_{13}\\ &= a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} \end{split} \nonumber

\begin{split} M_{11} &= \left| \begin{matrix} 8 & 1 \\ 12 & 1 \\ \end{matrix} \right|\\ &= - 4 \end{split} \nonumber

\begin{split} M_{12} &= \left| \begin{matrix} 64 & 1 \\ 144 & 1 \\ \end{matrix} \right|\\ &= - 80 \end{split} \nonumber

\begin{split} M_{13} &= \left| \begin{matrix} 64 & 8 \\ 144 & 12 \\ \end{matrix} \right|\\ &= - 384 \end{split} \nonumber

\begin{split} det(A) &= a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}\\ &= 25\left( - 4 \right) - 5\left( - 80 \right) + 1\left( - 384 \right)\\ &= - 100 + 400 - 384\\ &= - 84 \end{split} \nonumber

Also for i = 1,

\det\left( A \right) = \sum_{j = 1}^{3}{a_{1j}C_{1j}} \nonumber

\begin{split} C_{11} &= \left( - 1 \right)^{1 + 1}M_{11}\\ &= M_{11}\\ &= - 4 \end{split} \nonumber

\begin{split} C_{12} &= \left( - 1 \right)^{1 + 2}M_{12}\\ &= - M_{12}\\ &= 80 \end{split} \nonumber

\begin{split} C_{13} &= \left( - 1 \right)^{1 + 3}M_{13}\\ &= M_{13}\\ &= - 384 \end{split} \nonumber

\begin{split} \det\left( A \right) &= a_{11}C_{11} + a_{21}C_{21} + a_{31}C_{31}\\ &= (25)\left( - 4 \right) + (5)\left( 80 \right) + (1)\left( - 384 \right)\\ &= - 100 + 400 - 384\\ &= - 84 \end{split} \nonumber

Method 2:

\det\left( A \right) = \sum_{i = 1}^{3}{\left( - 1 \right)^{i + j}a_{ij}M_{ij}} \ for\ any\ j = 1,\ 2,\ 3 \nonumber

Let us choose j = 2 in the formula

\begin{split} \det\left( A \right) &= \sum_{i = 1}^{3}{\left( - 1 \right)^{i + 2}a_{i2}M_{i2}}\\ &= \left( - 1 \right)^{1 + 2}a_{12}M_{12} + \left( - 1 \right)^{2 + 2}a_{22}M_{22} + \left( - 1 \right)^{3 + 2}a_{32}M_{32}\\ &= - a_{12}M_{12} + a_{22}M_{22} - a_{32}M_{32} \end{split} \nonumber

\begin{split} M_{12} &= \left| \begin{matrix} 64 & 1 \\ 144 & 1 \\ \end{matrix} \right|\\ &= - 80 \end{split} \nonumber

\begin{split} M_{22} &= \left| \begin{matrix} 25 & 1 \\ 144 & 1 \\ \end{matrix} \right|\\ &= - 119 \end{split} \nonumber

\begin{split} M_{32} &= \left| \begin{matrix} 25 & 1 \\ 64 & 1 \\ \end{matrix} \right|\\ &= - 39 \end{split} \nonumber

\begin{split} det(A) &= - a_{12}M_{12} + a_{22}M_{22} - a_{32}M_{32}\\ &= - 5( - 80) + 8( - 119) - 12( - 39)\\ &= 400 - 952 + 468\\ &= - 84 \end{split} \nonumber

In terms of cofactors for j = 2,

\det\left( A \right) = \sum_{i = 1}^{3}{a_{i2}C_{i2}} \nonumber

\begin{split} C_{12} &= \left( - 1 \right)^{1 + 2}M_{12}\\ &= - M_{12}\\ &= 80 \end{split} \nonumber

\begin{split} C_{22} &= \left( - 1 \right)^{2 + 2}M_{22}\\ &= M_{22}\\ &= - 119 \end{split} \nonumber

\begin{split} C_{32} &= \left( - 1 \right)^{3 + 2}M_{32}\\ &= - M_{32}\\ &= 39 \end{split} \nonumber

\begin{split} \det\left( A \right) &= a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}\\ &= (5)\left( 80 \right) + (8)\left( - 119 \right) + (12)\left( 39 \right)\\ &= 400 - 952 + 468\\ &= - 84 \end{split} \nonumber

Is there a relationship between det(AB), and det(A) and det(B)?

Yes, if \lbrack A\rbrack and \lbrack B\rbrack are square matrices of same size, then

det({AB}) = det(A)det(B) \nonumber

Are there some other theorems that are important in finding the determinant of a square matrix?

Theorem 1: If a row or a column in a n \times n matrix \lbrack A\rbrack is zero, then det(A) = 0.

Theorem 2: Let \lbrack A\rbrack be a n \times n matrix. If a row is proportional to another row, then det(A) = 0.

Theorem 3: Let \lbrack A\rbrack be a n \times n matrix. If a column is proportional to another column, then det(A) = 0.

Theorem 4: Let \lbrack A\rbrack be a n \times nmatrix. If a column or row is multiplied by k to result in matrix k, then det(B) = kdet(A).

Theorem 5: Let \lbrack A\rbrack be a n \times n upper or lower triangular matrix, then det(A) = \overset{n}{\underset{i = 1}{\Pi}}a_{ii}.

Example 7

What is the determinant of

\lbrack A\rbrack = \begin{bmatrix} 0 & 2 & 6 & 3 \\ 0 & 3 & 7 & 4 \\ 0 & 4 & 9 & 5 \\ 0 & 5 & 2 & 1 \\ \end{bmatrix} \nonumber

Solution

Since one of the columns (first column in the above example) of \lbrack A\rbrack is a zero, det(A) = 0.

Example 8

What is the determinant of

\lbrack A\rbrack = \begin{bmatrix} 2 & 1 & 6 & 4 \\ 3 & 2 & 7 & 6 \\ 5 & 4 & 2 & 10 \\ 9 & 5 & 3 & 18 \\ \end{bmatrix} \nonumber

Solution

det(A) is zero because the fourth column

\begin{bmatrix} 4 \\ 6 \\ 10 \\ 18 \\ \end{bmatrix} \nonumber

is 2 times the first column

\begin{bmatrix} 2 \\ 3 \\ 5 \\ 9 \\ \end{bmatrix} \nonumber

Example 9

If the determinant of

\lbrack A\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix} \nonumber

is - 84, then what is the determinant of

\lbrack B\rbrack = \begin{bmatrix} 25 & 10.5 & 1 \\ 64 & 16.8 & 1 \\ 144 & 25.2 & 1 \\ \end{bmatrix} \nonumber

Solution

Since the second column of \lbrack B\rbrack is 2.1 times the second column of \lbrack A\rbrack

det(B) = 2.1det(A) \nonumber

= (2.1)( - 84) \nonumber

= - 176.4 \nonumber

Example 10

Given the determinant of

\lbrack A\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix} \nonumber

is - 84, what is the determinant of

\lbrack B\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 0 & - 4.8 & - 1.56 \\ 144 & 12 & 1 \\ \end{bmatrix} \nonumber

Solution

Since \lbrack B\rbrack is simply obtained by subtracting the second row of \lbrack A\rbrack by 2.56 times the first row of \lbrack A\rbrack,

\begin{split} det(B) &= det(A)\\ &= - 84 \end{split} \nonumber

Example 11

What is the determinant of

\lbrack A\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 0 & - 4.8 & - 1.56 \\ 0 & 0 & 0.7 \\ \end{bmatrix} \nonumber

Solution

Since \lbrack A\rbrack is an upper triangular matrix

\begin{split} \det\left( A \right) &= \prod_{i = 1}^{3}a_{ii}\\ &= a_{11} \times a_{22} \times a_{33}\\ &= 25 \times ( - 4.8) \times 0.7\\ &= - 84 \end{split} \nonumber

Unary Matrix Operations Quiz

Quiz 1

If the determinant of a 4 \times 4 matrix is given as 20, then the determinant of 5 is

(A) 100

(B) 12500

(C) 25

(D) 62500

Quiz 2

If the matrix product \left\lbrack A \right\rbrack\ \left\lbrack B \right\rbrack\ \left\lbrack C \right\rbrack is defined, then \left( \left\lbrack A \right\rbrack\left\lbrack B \right\rbrack\left\lbrack C \right\rbrack \right)^{T} is

(A) \left\lbrack C \right\rbrack^{T}\ \left\lbrack B \right\rbrack^{T}\ \left\lbrack A \right\rbrack^{T}

(B) \left\lbrack A \right\rbrack^{T}\ \left\lbrack B \right\rbrack^{T}\ \left\lbrack C \right\rbrack^{T}

(C) \left\lbrack A \right\rbrack\ \left\lbrack B \right\rbrack\ \left\lbrack C \right\rbrack^{T}

(D) \left\lbrack A \right\rbrack^{T}\ \left\lbrack B \right\rbrack\ \left\lbrack C \right\rbrack

Quiz 3

The trace of a matrix

\begin{bmatrix} 5 & 6 & - 7 \\ 9 & - 11 & 13 \\ - 17 & 19 & 23 \\ \end{bmatrix} \nonumber

is

(A) 17

(B) 39

(C) 40

(D) 110

Quiz 4

A square n \times n matrix \left\lbrack A \right\rbrack is symmetric if

(A) a_{ij} = a_{ji},\ i = j for all i,j

(B) a_{ij} = a_{ji},\ i \neq j for all i,j

(C) a_{ij} = - a_{ji},\ i = j for all i,j

(D) a_{ij} = - a_{ji},\ i \neq j for all i,j

Quiz 5

The determinant of the matrix

\begin{bmatrix} 25 & 5 & 1 \\ 0 & 3 & 8 \\ 0 & 9 & a \\ \end{bmatrix}

is 50. The value of a is then

(A) 0.6667

(B) 24.67

(C) -23.33

(D) 5.556

Quiz 6

\left\lbrack A \right\rbrack is a 5 \times 5 matrix and a matrix \left\lbrack B \right\rbrack is obtained by the row operations of replacing Row\ 1 with Row\ 3, and then Row\ 3 is replaced by a linear combination of 2 \times Row\ 3 + 4 \times Row\ 2. If \det\left( A \right) = 17, then \det\left( B \right) is equal to

(A) 12

(B) -34

(C) -112

(D) 112

Unary Matrix Operations Exercise

Exercise 1

Let

\lbrack A\rbrack = \begin{bmatrix} 25 & 3 & 6 \\ 7 & 9 & 2 \\ \end{bmatrix}.

Find \lbrack A\rbrack^{T}

Answer

\begin{bmatrix} 25 & 7 \\ 3 & 9 \\ 6 & 2 \\ \end{bmatrix}

Exercise 2

If \lbrack A\rbrack and \lbrack B\rbrack are two n \times n symmetric matrices, show that \lbrack A\rbrack + \lbrack B\rbrack is also symmetric. Hint: Let \left\lbrack C \right\rbrack = \left\lbrack A \right\rbrack + \left\lbrack B \right\rbrack

Answer

c_{ij} = a_{ij} + b_{ij} for all i, j.

and

c_{ji} = a_{ji} + b_{ji} for all i, j.
c_{ji} = a_{ij} + b_{ij} as \left\lbrack A \right\rbrack and \left\lbrack B \right\rbrack are symmetric

Hence c_{ji} = c_{ij}.

Exercise 3

Give an example of a 4 \times 4 symmetric matrix.

Exercise 4

Give an example of a 4 \times 4 skew-symmetric matrix.

Exercise 5

What is the trace of

  1. \left\lbrack A \right\rbrack = \begin{bmatrix} 7 & 2 & 3 & 4 \\ - 5 & - 5 & - 5 & - 5 \\ 6 & 6 & 7 & 9 \\ - 5 & 2 & 3 & 10 \\ \end{bmatrix}
  2. For

\left\lbrack A \right\rbrack = \begin{bmatrix} 10 & - 7 & 0 \\ - 3 & 2.099 & 6 \\ 5 & - 1 & 5 \\ \end{bmatrix}

Find the determinant of \lbrack A\rbrack using the cofactor method.

Answer

a) 19
b) - 150.05

Exercise 6

det(3\lbrack A\rbrack) of a n \times n matrix is

  1. 3det(A)
  2. 3det(A)
  3. 3^{n}det(A)
  4. 9det(A)
Answer

C

Exercise 7

For a 5 \times 5 matrix \lbrack A\rbrack, the first row is interchanged with the fifth row, the determinant of the resulting matrix \lbrack B\rbrackis

  1. det(A)
  2. - det(A)
  3. 5det(A)
  4. 2det(A)
Answer

A

Exercise 8

\det\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ \end{bmatrix} is

  1. 0
  2. 1
  3. -1
  4. \infty
Answer

C

Exercise 9

Without using the cofactor method of finding determinants, find the determinant of

\left\lbrack A \right\rbrack = \begin{bmatrix} 0 & 0 & 0 \\ 2 & 3 & 5 \\ 6 & 9 & 2 \\ \end{bmatrix}

Answer

0: Can you answer why?

Exercise 10

Without using the cofactor method of finding determinants, find the determinant of

\left\lbrack A \right\rbrack = \begin{bmatrix} 0 & 0 & 2 & 3 \\ 0 & 2 & 3 & 5 \\ 6 & 7 & 2 & 3 \\ 6.6 & 7.7 & 2.2 & 3.3 \\ \end{bmatrix}

Answer

0: Can you answer why?

Exercise 11

Without using the cofactor method of finding determinants, find the determinant of

\left\lbrack A \right\rbrack = \begin{bmatrix} 5 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 2 & 5 & 6 & 0 \\ 1 & 2 & 3 & 9 \\ \end{bmatrix}

Answer

5 \times 3 \times 6 \times 9 = 810: Can you answer why?

Exercise 12

Given the matrix

\left\lbrack A \right\rbrack = \begin{bmatrix} 125 & 25 & 5 & 1 \\ 512 & 64 & 8 & 1 \\ 1157 & 89 & 13 & 1 \\ 8 & 4 & 2 & 1 \\ \end{bmatrix}

and

det(A) = - 32400

find the determinant of

  1. \left\lbrack A \right\rbrack = \begin{bmatrix} 125 & 25 & 5 & 1 \\ 512 & 64 & 8 & 1 \\ 1141 & 81 & 9 & - 1 \\ 8 & 4 & 2 & 1 \\ \end{bmatrix}
  2. \left\lbrack A \right\rbrack = \begin{bmatrix} 125 & 25 & 1 & 5 \\ 512 & 64 & 1 & 8 \\ 1157 & 89 & 1 & 13 \\ 8 & 4 & 1 & 2 \\ \end{bmatrix}
  3. \left\lbrack B \right\rbrack = \begin{bmatrix} 125 & 25 & 5 & 1 \\ 1157 & 89 & 13 & 1 \\ 512 & 64 & 8 & 1 \\ 8 & 4 & 2 & 1 \\ \end{bmatrix}
  4. \left\lbrack C \right\rbrack = \begin{bmatrix} 125 & 25 & 5 & 1 \\ 1157 & 89 & 13 & 1 \\ 8 & 4 & 2 & 1 \\ 512 & 64 & 8 & 1 \\ \end{bmatrix}
  5. \left\lbrack D \right\rbrack = \begin{bmatrix} 125 & 25 & 5 & 1 \\ 512 & 64 & 8 & 1 \\ 1157 & 89 & 13 & 1 \\ 16 & 8 & 4 & 2 \\ \end{bmatrix}
Answer

A) –32400 B) 32400 C) 32400 D) -32400 E) -64800

Exercise 13

What is the transpose of

\lbrack A\rbrack = \begin{bmatrix} 25 & 20 & 3 & 2 \\ 5 & 10 & 15 & 25 \\ 6 & 16 & 7 & 27 \\ \end{bmatrix}

Answer

\left\lbrack A \right\rbrack^{T} = \begin{bmatrix} 25 & 5 & 6 \\ 20 & 10 & 16 \\ 3 & 15 & 7 \\ 2 & 25 & 27 \\ \end{bmatrix}

Exercise 14

What values of the missing numbers will make this a skew-symmetric matrix?

\lbrack A\rbrack = \begin{bmatrix} 0 & 3 & ? \\ ? & 0 & ? \\ 21 & ? & 0 \\ \end{bmatrix}

Answer

\begin{bmatrix} 0 & 3 & - 21 \\ - 3 & 0 & 4 \\ 21 & - 4 & 0 \\ \end{bmatrix}

Exercise 15

What values of the missing number will make this a symmetric matrix?

\lbrack A\rbrack = \begin{bmatrix} 2 & 3 & ? \\ ? & 6 & 7 \\ 21 & ? & 5 \\ \end{bmatrix}

Answer

\begin{bmatrix} 2 & 3 & 21 \\ 3 & 6 & 7 \\ 21 & 7 & 5 \\ \end{bmatrix}

Exercise 16

Find the determinant of
\left\lbrack A \right\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 5 \\ \end{bmatrix}

Answer

The determinant of \left\lbrack A \right\rbrack is, 25\begin{bmatrix} 8 & 1 \\ 12 & 5 \\ \end{bmatrix} - 5\begin{bmatrix} 64 & 1 \\ 144 & 5 \\ \end{bmatrix} + 1\begin{bmatrix} 64 & 8 \\ 144 & 12 \\ \end{bmatrix} \nonumber \begin{split} &=25(28) - 5(176) + 1( - 384)\\ &= -564 \end{split} \nonumber

Exercise 17

What is the determinant of an upper triangular matrix \lbrack A\rbrack that is of order n \times n?

Answer

The determinant of an upper triangular matrix is the product of its diagonal elements,\prod_{i = 1}^{n}a_{ii}

Exercise 18

Given the determinant of

\left\lbrack A \right\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & a \\ \end{bmatrix}

is- 564, find a.

Answer

 

det(A) = - 120a + 36

120a + 36 = 564

a = 5

Exercise 19

Why is the determinant of the following matrix zero?
\left\lbrack A \right\rbrack = \begin{bmatrix} 0 & 0 & 0 \\ 2 & 3 & 5 \\ 6 & 9 & 2 \\ \end{bmatrix}

Answer

The first row of the matrix is zero, hence, the determinant of the matrix is zero.

Exercise 20

Why is the determinant of the following matrix zero?
\left\lbrack A \right\rbrack = \begin{bmatrix} 0 & 0 & 2 & 3 \\ 0 & 2 & 3 & 5 \\ 6 & 7 & 2 & 3 \\ 6.6 & 7.7 & 2.2 & 3.3 \\ \end{bmatrix}

Answer

Row 4 of the matrix is 1.1 times Row 3. Hence, its determinant is zero.

Exercise 21

Show that if \lbrack A\rbrack\ \lbrack B\rbrack = \lbrack I\rbrack, where \lbrack A\rbrack\ , \ \lbrack B\rbrack and \lbrack I\rbrack are matrices of n \times n size and \lbrack I\rbrack is an identity matrix, then det(A) \neq 0 and det(B) \neq 0.

Answer

We know that det(AB) = det(A)det(B).

[A][B] = [I] \nonumber det(AB) = det(I) \nonumber

det(I) = \prod_{i = 1}^{n}{a_{ii} = \prod_{i = 1}^{n}1} = 1 \nonumber det(A)det(B) = 1 \nonumber

Therefore,

det(A) \neq 0 and

det(B) \neq 0.


This page titled 4: Unary Matrix Operations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Autar Kaw via source content that was edited to the style and standards of the LibreTexts platform.

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