# Chapter 4 Unary Matrix Operations

## 4.1 Learning Objectives

After reading this chapter, you should be able to:

(1). know what unary operations are,

(2). find 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.

## 4.2What 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 $$\left\lbrack A \right\rbrack^{T}$$.

### 4.2.1 Example 1

Find the transpose of

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

Solution

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

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

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,

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

Also,

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

## 4.3 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.

### 4.3.1 Example 2

Give an example of a symmetric matrix

Solution

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

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

## 4.4 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

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

### 4.4.1 Example 3

Give an example of a skew-symmetric matrix

Solution

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

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.

## 4.5 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

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

### 4.5.1 Example 4

Find the trace of:

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

Solution

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

### 4.5.2 Example 5

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

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

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

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

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

Solution

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

### 4.5.3 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

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

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

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

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

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

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

## 4.6 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

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

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

## 4.7 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

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

or

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

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

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

or

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

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 $$n!$$.

### 4.7.1 Example 6

Find the determinant of

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

Solution

#### 4.7.1.1 Method 1:

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

Let us choose $$i = 1$$ in the formula

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

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

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

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

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

Also for $$i = 1$$,

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

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

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

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

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

#### 4.7.1.2 Method 2:

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

Let us choose $$j = 2$$ in the formula

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

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

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

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

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

In terms of cofactors for $$j = 2$$,

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

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

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

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

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

## 4.8 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

$det({AB}) = det(A)det(B)$

## 4.9 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 $$n \times n$$matrix. 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 $$det(A) = \overset{n}{\underset{i = 1}{\Pi}}a_{{ii}}$$.

### 4.9.1 Example 7

What is the determinant of

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

Solution

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

### 4.9.2 Example 8

What is the determinant of

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

Solution

$$det(A)$$ is zero because the fourth column

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

is 2 times the first column

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

### 4.9.3 Example 9

If the determinant of

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

is $$- 84$$, then what is the determinant of

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

Solution

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

$det(B) = 2.1det(A)$

$= (2.1)( - 84)$

$= - 176.4$

### 4.9.4 Example 10

Given the determinant of

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

is $$- 84$$, what is the determinant of

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

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 $$\lbrack A\rbrack$$,

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

### 4.9.5 Example 11

What is the determinant of

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

Solution

Since $$\lbrack A\rbrack$$ is an upper triangular matrix

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

## 4.10 Unary Matrix Operations 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) $$62500$$

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

(3). The trace of a matrix

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

is

(A) $$17$$

(B) $$39$$

(C) $$40$$

(D) $$110$$

(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 $$i,j$$

(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) $$5.556$$

(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) $$112$$

## 4.11 Unary Matrix Operations Exercise

(1). Let

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

Find $$\lbrack A\rbrack^{T}$$

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

(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 $$c_{{ji}} = c_{{ij}}.$$

(3). Give an example of a $$4 \times 4$$ symmetric matrix.

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

(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) $$- 150.05$$

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

1. $$3det(A)$$

2. 3$$det(A)$$

3. $$3^{n}det(A)$$

4. $$9det(A)$$

(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 $$\lbrack B\rbrack$$is

1. $$det(A)$$

2. $$- det(A)$$

3. $$5det(A)$$

4. $$2det(A)$$

(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. $$\infty$$

(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?

(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?

(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?

(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

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

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

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

(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, $25\begin{bmatrix} 8 & 1 \\ 12 & 5 \\ \end{bmatrix} - 5\begin{bmatrix} 64 & 1 \\ 144 & 5 \\ \end{bmatrix} + 1\begin{bmatrix} 64 & 8 \\ 144 & 12 \\ \end{bmatrix}$ $\begin{split} &=25(28) - 5(176) + 1( - 384)\\ &= -564 \end{split}$

(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,$$\prod_{i = 1}^{n}a_{{ii}}$$

(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: $$det(A) = - 120a + 36$$

$$120a + 36 = 564$$

$$a = 5$$

(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.

(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.

(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 $$det(AB) = det(A)det(B)$$.

$[A][B] = [I]$ $det(AB) = det(I)$

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

Therefore,

$$det(A) \neq 0$$ and

$$det(B) \neq 0$$.