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

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

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

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

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

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

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) $12500$

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

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

(3). The trace of a matrix

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

(A) $17$

(B) $39$

(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 $i,j$

(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) $24.67$

(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) $-34$

(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

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

$3\det(A)$
$3\det(A)$
$3^{n}\det(A)$
$9\det(A)$

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

$\det(A)$
$- \det(A)$
$5\det(A)$
$2\det(A)$

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

$0$
$1$
$-1$
$\infty$

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

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

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

$det(B) \neq 0$.