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 [A] be a m×n matrix. Then [B] is the transpose of [A] if bji=aij for all i and j. That is, the ith row and the jth column element of [A] is the jth row and ith column element of [B]. Note, [B] would be a n×m matrix. The transpose of [A] is denoted by [A]T.
Find the transpose of
[A]=[2520325101525616727]
Solution
The transpose of [A] is
[A]T=[2520325101525616727]
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,
([A]T)T=[A]
Also,
([A]+[B])T=[A]T+[B]T; (cA)T=cAT
What is a symmetric matrix?
A square matrix [A] with real elements were aij=aji for i=1, 2, … , n and j=1, 2, … , n is called a symmetric matrix. This is the same as saying that if [A]=[A]T, then [A]T is a symmetric matrix.
Give an example of a symmetric matrix
Solution
[A]=[21.23.263.221.58689.3]
Is a symmetric matrix as a12=a21=3.2, a13=a31=6 and a13=a31=8.
What is a skew-symmetric matrix?
A n×n matrix is skew-symmetric if aij=−aji, for i=1, … , n and j=1, … , n. This the same as
[A]=−[A]T
Give an example of a skew-symmetric matrix
Solution
[012−10−5−250]
Is a skew symmetric matrix as a12=−a21=1; a13=−a31=2; a23=−a32=−5. Since aii=−aii only if aii=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×n matrix [A] is the sum of the diagonal entries of [A]. That is
tr[A]=n∑i=1aii
Find the trace of:
[A]=[15672−42326]
Solution
tr[A]=n∑i=1aii=15+(−4)+6=17
The sales of tires are given by make (rows) and quarters (columns) for Blowout r’us store location A, as shown below
[A]=[2556201016315722527]
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
[B]=[33.2540.1925.0330.0138.0222.0235.0241.0327.0330.0538.2322.95]
Where the rows represent the cost of each tire made by Tirestone, Michigan, and Copper, and the columns represent the quarter numbers.
Solution
[C]=[B]T
Blowout r’us store location A and the sales of tires are given by make (in rows) and quarters (in columns) as shown below
[A]=[2556201016315722527]=[33.2540.1925.0330.0138.0222.0235.0241.0327.0330.0538.2322.95]=[33.2530.0135.0230.0540.1938.0241.0338.2325.0322.0227.0322.95]
Recognize now that if we find [A][C], we get
[D]=[A][C]=[2556201016315722527][33.2530.0135.0230.0540.1938.0241.0338.2325.0322.0227.0322.95]=[159719651193174321521325173621691311]
The diagonal elements give the sales of each brand of tire for the whole year. That is
d11=$1597 (Tirestone sales)
d22=$2152 (Michigan sales)
d33=$1597 (Copper sales)
The total yearly sales of all three brans of tires are
3∑i=1dii=1597+2152+1311=$5060
And this is the trace of matrix [D].
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 [A], determinant is denoted by |A| or det. 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
-
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
-
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
-
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.