4: Unary Matrix Operations
( \newcommand{\kernel}{\mathrm{null}\,}\)
After reading this chapter, you should be able to:
- know what unary operations are,
- fnd the transpose of a square matrix and its relationship to symmetric matrices,
- find the trace of a matrix, and
- 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}.
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.
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
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
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
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
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!.
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}.
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.
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
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
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
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
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
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
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
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
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
\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
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}
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 symmetricHence c_{ji} = c_{ij}.
Give an example of a 4 \times 4 symmetric matrix.
Give an example of a 4 \times 4 skew-symmetric matrix.
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
det(3\lbrack A\rbrack) of a n \times n matrix is
- 3det(A)
- 3det(A)
- 3^{n}det(A)
- 9det(A)
- Answer
-
C
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
- det(A)
- - det(A)
- 5det(A)
- 2det(A)
- Answer
-
A
\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
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?
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?
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?
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
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}
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}
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}
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
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}
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
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
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The first row of the matrix is zero, hence, the determinant of the matrix is zero.
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
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Row 4 of the matrix is 1.1 times Row 3. Hence, its determinant is zero.
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
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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.