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Find the determinant of A= 0203 −140−1 3110 102−2
Related Topics
Wize University Linear Algebra Textbook > Determinants
Basics of Determinants
4 Activities
Find the determinant of
A
=
[
0
2
0
3
−
1
4
0
−
1
3
1
1
0
1
0
2
−
2
]
A=\begin{bmatrix} 0&2&0&3\\ −1&4&0&−1\\ 3&1&1&0\\ 1&0&2&−2 \end{bmatrix}
A
=
0
−
1
3
1
2
4
1
0
0
0
1
2
3
−
1
0
−
2
Answer
I don't know
Check Submission
More Basics of Determinants Questions:
133 - FML 3 - 18.1W e.g. 17.3
If
A
‾
\bcb{\boldsymbol{ \ul{A} }}
A
is a
3
×
3
\bcb{\boldsymbol{ 3 \times 3}}
3
×
3
matrix with
det
(
A
‾
)
=
−
4
\bcb{\boldsymbol{ \detp{\ul{A}} = -4}}
det
(
A
)
=
−
4
, find:
det
(
−
A
T
)
\boldsymbol{ \detp{-\mtran{A}} }
det
(
−
A
T
)
133 - FML 3 - 18.1W e.g. 11
Find all possible values of
det
(
A
‾
)
\bcb{\boldsymbol{ \detp{\ul{A}} }}
det
(
A
)
given that
a
d
j
(
A
)
=
[
1
2
4
−
1
2
2
0
1
8
]
\bcb{\boldsymbol{ adj(A) = \begin{bmatrix} 1 & 2 & 4 \\ -1 & 2 & 2 \\ 0 & 1 & 8 \end{bmatrix} }}
adj
(
A
)
=
1
−
1
0
2
2
1
4
2
8
133 - FML 3 - 18.1W e.g. 71
B
‾
=
[
1
2
3
4
−
1
1
2
x
2
]
\bcb{\boldsymbol{ \ul{B} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & -1 & 1 \\ 2 & x & 2 \end{bmatrix} }}
B
=
1
4
2
2
−
1
x
3
1
2
such that
det
(
B
‾
)
=
−
8
\bcb{\boldsymbol{ \detp {\ul{B}} = -8 }}
det
(
B
)
=
−
8
, then
x
\bcb{\boldsymbol{ x }}
x
is:
(a)
1
11
(b)
−
1
(c)
0
(d)
None
of
the
above
\text{(a)}\;\boldsymbol{ \frac{1}{11} } \qquad\qquad\qquad \text{(b)}\; \boldsymbol{ -1}\qquad\qquad\qquad \text{(c)}\; \boldsymbol{ 0}\qquad\qquad\qquad \text{(d)}\; \text{\bf{None of the above}}
(a)
11
1
(b)
−
1
(c)
0
(d)
None of the above
Basics of Determinants
Answer the following true/false questions.
Compute the determinant of
[
3
1
−
1
1
1
0
2
3
−
2
0
3
4
1
0
2
2
]
\begin{bmatrix} 3&1&−1&1\\ 1&0&2&3\\ −2&0&3&4\\ 1&0&2&2 \end{bmatrix}
3
1
−
2
1
1
0
0
0
−
1
2
3
2
1
3
4
2
Basics of Determinants
Let
C
C
C
be an invertible matrix with
det
C
=
3
\det C = 3
det
C
=
3
, and let
B
=
[
−
2
−
1
0
0
1
0
0
0
−
5
]
B= \left[ \begin{array}{rrr} -2 & -1 & 0\\ 0 & 1 & 0\\ 0 & 0 & -5\\ \end{array} \right]
B
=
−
2
0
0
−
1
1
0
0
0
−
5
.
If
A
=
C
−
1
B
C
A = C^{-1}BC
A
=
C
−
1
B
C
, calculate
det
A
\det A
det
A
.
Determinant Properties Practice Question
If 𝑨 = 𝑪
-1
𝟏𝑩𝑪, where C is an invertible matrix, and
d
e
t
(
𝑩
)
=
12
;
d
e
t
(
𝑪
)
=
3.
Calculate
d
e
t
(
𝑨
)
.
det(𝑩) = 12; det(𝑪) = 3. \text{Calculate}\ \ det(𝑨).
d
e
t
(
B
)
=
12
;
d
e
t
(
C
)
=
3.
Calculate
d
e
t
(
A
)
.
133 - FML 1 - 18.1W e.g. 7
Recall that diagonal matrices are substantially easier to work with than non-diagonal matrices:
e.g. If
A
=
diag
(
a
,
b
,
c
)
=
[
a
0
0
0
b
0
0
0
c
]
\text{A}=\text{diag}\left(a,b,c\right) = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}
A
=
diag
(
a
,
b
,
c
)
=
a
0
0
0
b
0
0
0
c
Practice: Determinant Property
Practice Question: Determinant Property
Let 𝐶 be an invertible matrix, det (𝐵) = 10, and det (𝐶) = 3.
If 𝐴 = 𝐶
−1
𝐵𝐶, calculate det (𝐴).
Determinant Properties Practice Question
If 𝑨 = 𝑪
-1
𝟏𝑩𝑪, where C is an invertible matrix, and
d
e
t
(
𝑩
)
=
12
;
d
e
t
(
𝑪
)
=
3.
Calculate
d
e
t
(
𝑨
)
.
det(𝑩) = 12; det(𝑪) = 3. \text{Calculate}\ \ det(𝑨).
d
e
t
(
B
)
=
12
;
d
e
t
(
C
)
=
3.
Calculate
d
e
t
(
A
)
.
Basics of Determinants
Find
det
[
3
2
−
4
1
0
1
−
1
2
−
1
−
7
5
0
−
6
−
4
8
−
2
]
\text{det}\begin{bmatrix} 3&2&-4&1\\ 0&1&-1&2\\ -1&-7&5&0\\ -6&-4&8&-2 \end{bmatrix}
det
3
0
−
1
−
6
2
1
−
7
−
4
−
4
−
1
5
8
1
2
0
−
2
.
Determinants: Basics
If
A
=
[
1
2
3
4
0
2
1
−
5
0
0
−
3
0
0
0
0
−
1
]
A= \begin{bmatrix} 1&2&3&4\\ 0&2&1&-5\\ 0&0&-3&0\\ 0&0&0&-1 \end{bmatrix}
A
=
1
0
0
0
2
2
0
0
3
1
−
3
0
4
−
5
0
−
1
, calculate
d
e
t
(
A
d
j
A
)
det(Adj\ A)
d
e
t
(
A
d
j
A
)
.
Determinants: Basics
A
A
A
is a
3
×
3
3\times 3
3
×
3
matrix where
d
e
t
A
=
2
det\ A=2
d
e
t
A
=
2
.
Find
d
e
t
(
5
A
2
A
−
1
A
T
)
det(5A2A^{-1}A^T)
d
e
t
(
5
A
2
A
−
1
A
T
)
.
Determinants: Basics
A
=
[
4
5
−
1
2
]
,
B
=
[
−
1
0
0
0
2
1
0
0
30
20
10
0
1
3
−
4
−
10
]
,
C
=
[
−
1
5
5
−
25
]
,
I
10
=
10
×
10
identity matrix
A= \begin{bmatrix} 4&5\\ -1&2 \end{bmatrix}, \ B= \begin{bmatrix} -1&0&0&0\\ 2&1&0&0\\ 30&20&10&0\\ 1&3&-4&-10 \end{bmatrix}, \ C= \begin{bmatrix} -1&5\\ 5&-25 \end{bmatrix}, \ I_{10}=10\times 10\ \text{ identity matrix}
A
=
[
4
−
1
5
2
]
,
B
=
−
1
2
30
1
0
1
20
3
0
0
10
−
4
0
0
0
−
10
,
C
=
[
−
1
5
5
−
25
]
,
I
10
=
10
×
10
identity matrix
Compute
d
e
t
(
A
C
)
−
d
e
t
(
10
B
T
)
+
d
e
t
(
2
I
10
)
det(AC)-det(10B^T)+det(2I_{10})
d
e
t
(
A
C
)
−
d
e
t
(
10
B
T
)
+
d
e
t
(
2
I
10
)
.
Determinants: Basics
Find the determinant of the matrix
[
2
−
4
1
−
3
]
\begin{bmatrix}2&-4\\ 1&-3\end{bmatrix}
[
2
1
−
4
−
3
]
.