Chapter 04.04: Unary Matrix Operations

Lesson: Unary Matrix Operations

Learning Objectives

After successful completion of this lesson, 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.

  

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

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

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,

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

  

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

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

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

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

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.

  

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

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

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

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

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

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

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 \(\left\lbrack D \right\rbrack\).

  

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

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

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

  

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 \(n!\).

Example 6

Find the determinant of

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

Solution

Method 1:

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

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 \(i = 1\),

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

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

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

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

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

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 \(j = 2\),

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

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

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

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

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

  

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

\[det({AB}) = det(A)det(B)\]

  

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 \(det(A) = \overset{n}{\underset{i = 1}{\Pi}}a_{{ii}}\).

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, \(det(A) = 0\).

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

\[\begin{bmatrix} 2 \\ 3 \\ 5 \\ 9 \\ \end{bmatrix}\]

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 \(\lbrack A\rbrack\)

\[\begin{split} \det(B) &= 2.1\det(A)\\ &= (2.1)( - 84)\\ &= - 176.4 \end{split}\]

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 \(\lbrack A\rbrack\),

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

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

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

Key Terms:

Transpose

Symmetric Matrix

Skew-Symmetric Matrix

Trace of Matrix

Determinant

Multiple Choice Test

(1). If the determinant of a \(4 \times 4\) matrix is given as \(20\), then the determinant of \(5\left\lbrack A \right\rbrack\) 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) \(112\)

For complete solution, go to

https://ma.mathforcollege.com/mcquizzes/04sle/quiz_04sle_unarymatrixoperations_solution.pdf

Problem Set

(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. \(3\det(A)\)

  2. \(3\det(A)\)

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

  4. \(9\det(A)\)

Answer: C

  

(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. \(5\det(A)\)

  4. \(2\det(A)\)

Answer: 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. \(\infty\)

Answer: C

  

(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

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