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133 - FML 3 - 18.1W e.g. 15
Related Topics
Wize University Linear Algebra Textbook > Determinants
Determinants and Inverse
3 Activities
A
‾
2
=
4
A
‾
and
A
‾
\bcb{\boldsymbol{ \ul{A}^2 = 4\ul{A} \text{ and } \ul{A} }}
A
2
=
4
A
and
A
is an invertible matrix, then a possible value for
det
(
A
‾
)
\bcb{\boldsymbol{\detp{\ul{A}}}}
det
(
A
)
is:
(a) 1
(b) 16
(c) 0
(d) -4
I don't know
Check Submission
More Determinants and Inverse Questions:
Practice Question: Determinant Properties
True or False?
If 𝐴 is a 4 × 4 matrix and det(𝐴) = −3, then the homogeneous system of linear equations
𝐴
𝑥
⃗
=
0
⃗
𝐴𝑥⃗=\vec{0}
A
x
⃗
=
0
has at least one non-trivial solution.
For what value of d, the following linear system has at least one solution:
{
𝑥
+
𝑦
+
𝑧
=
0
−
𝑥
+
𝑧
=
1
𝑥
+
3
𝑧
=
𝑑
\begin{cases} 𝑥 + 𝑦 + 𝑧 = 0\\ −𝑥 + 𝑧 = 1\\ 𝑥 + 3𝑧 = 𝑑 \end{cases}
⎩
⎨
⎧
x
+
y
+
z
=
0
−
x
+
z
=
1
x
+
3
z
=
d
133 - FML 3 - 18.1W e.g. 17.2
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
(
(
2
A
)
−
1
)
\boldsymbol{ \detp{(2A)^{-1}}}
det
((
2
A
)
−
1
)
133 - FML 3 - 18.1W e.g. 69
Given
A
‾
and
B
‾
3
×
3
\bcb{\boldsymbol{ \ul{A} \text{ and } \ul{B} ~ 3 \times 3}}
A
and
B
3
×
3
matrices with
det
(
A
‾
)
=
−
4
and det
(
B
T
)
=
3
\bcb{\boldsymbol{\detp{\ul{A}} = -4 \text{ and } \detp{ \mtran{B} } = 3}}
det
(
A
)
=
−
4
and
det
(
B
T
)
=
3
find
det
(
2
A
‾
B
T
A
‾
−
2
B
‾
A
T
B
−
1
)
\bcb{\boldsymbol{ \detp { 2\ul{A} \, \mtran{B} \, \ul{A}^{-2} \ul{B} \, \mtran{A} \, \minv{B} } }}
det
(
2
A
B
T
A
−
2
B
A
T
B
−
1
)
.
133 - FML 3 - 18.1W e.g. 63
A
‾
\bcb{\boldsymbol{ \ul{A} }}
A
is a
3
×
3
\bcb{\boldsymbol{ 3 \times 3}}
3
×
3
matrix such that
det
(
−
2
(
A
T
)
−
1
)
=
8
then det
(
A
‾
)
\bcb{\boldsymbol{ \detp{-2(\mtran{A})^{-1} }= 8 \text{ then } \detp{\ul{A}}}}
det
(
−
2
(
A
T
)
−
1
)
=
8
then
det
(
A
)
is:
Shortcut for 3x3 Determinants
Let
A
=
[
−
6
1
a
a
−
2
0
0
1
−
2
]
A=\left[\begin{array}{rrr} -6&1& a\\ a&-2&0\\ 0&1& -2 \end{array}\right]
A
=
−
6
a
0
1
−
2
1
a
0
−
2
. Find the value(s) of
a
a
a
such that
A
A
A
is
not invertible
.
Practice: Determinant of $3\times3$ Matrix
Let
A
=
[
6
1
a
a
−
2
0
0
1
2
]
A=\left[\begin{array}{rrr} 6&1&a\\ a&-2&0\\ 0&1&2 \end{array}\right]
A
=
6
a
0
1
−
2
1
a
0
2
using cofactor expansion along any row or column
Find the value(s) of
a
a
a
that will make
A
A
A
invertible
Let
B
=
[
𝑎
𝑏
𝑐
𝑑
𝑒
𝑓
𝑔
h
𝑖
]
B=\begin{bmatrix} 𝑎&𝑏&𝑐\\ 𝑑&𝑒&𝑓\\ 𝑔&ℎ&𝑖 \end{bmatrix}
B
=
a
d
g
b
e
h
c
f
i
be a
3
×
3
3\times3
3
×
3
matrix. Assume that
det
B
=
20
\det B = 20
det
B
=
20
. Determine if the matrix A that appears below is invertible by computing its determinant.
A
=
[
5
𝑎
5
𝑏
0
5
𝑐
0
0
1
0
−
𝑔
−
h
0
−
i
𝑑
−
10
𝑎
𝑒
−
10
𝑏
0
𝑓
−
10
𝑐
]
A=\begin{bmatrix} 5𝑎&5𝑏&0&5𝑐\\ 0&0&1&0\\ −𝑔&-h&0&-i\\ 𝑑 − 10𝑎&𝑒 − 10𝑏&0&𝑓 − 10𝑐 \end{bmatrix}
A
=
5
a
0
−
g
d
−
10
a
5
b
0
−
h
e
−
10
b
0
1
0
0
5
c
0
−
i
f
−
10
c
L
e
t
B
=
[
𝑎
𝑏
𝑐
𝑑
𝑒
𝑓
𝑔
h
𝑖
]
b
e
a
3
X
3
m
a
t
r
i
x
.
A
s
s
u
m
e
t
h
a
t
d
e
t
(
B
)
=
20.
Let\ B=\begin{bmatrix} 𝑎&𝑏&𝑐\\ 𝑑&𝑒&𝑓\\ 𝑔&ℎ&𝑖 \end{bmatrix}be\ a\ 3X3\ matrix.\ Assume\ that\ det(B) = 20.
L
e
t
B
=
a
d
g
b
e
h
c
f
i
b
e
a
3
X
3
ma
t
r
i
x
.
A
ss
u
m
e
t
ha
t
d
e
t
(
B
)
=
20.
Determine if the matrix A that appears below is invertible by computing its determinant.
Practice: Determinant Properties
Practice: Determinant Properties
True or False?
If 𝐴 is a 4 × 4 matrix and det(𝐴) = −3, then
d
e
t
(
A
T
)
=
1
d
e
t
(
A
−
1
)
det\left(A^T\right)=\frac{1}{det\left(A^{-1}\right)}
d
e
t
(
A
T
)
=
d
e
t
(
A
−
1
)
1
Determinants and Inverses
Let
A
=
[
a
b
c
d
e
f
1
2
−
3
]
A=\begin{bmatrix} a&b&c\\ d&e&f\\ 1&2&-3 \end{bmatrix}
A
=
a
d
1
b
e
2
c
f
−
3
, where it is known that
det
[
a
b
d
e
]
=
5
\text{det}\begin{bmatrix} a&b\\d&e \end{bmatrix}=5
det
[
a
d
b
e
]
=
5
,
det
[
a
c
d
f
]
=
−
2
\text{det}\begin{bmatrix} a&c\\d&f \end{bmatrix}=-2
det
[
a
d
c
f
]
=
−
2
and
det
[
b
c
e
f
]
=
3
\text{det}\begin{bmatrix} b&c\\e&f \end{bmatrix}=3
det
[
b
e
c
f
]
=
3
.
(a) Find
det
A
\text{det}A
det
A
.
(b) Find the (2,3)-entry of
Adj
A
\text{Adj}A
Adj
A
.
Cramer's Rule for Solving Linear Systems: Determinants and Inverses
Let
A
A
A
be a 3 x 3 matrix with
det
A
=
−
18
\text{det}A=-18
det
A
=
−
18
and
Adj
A
=
[
−
9
3
4
0
−
6
−
2
0
0
6
]
\text{Adj}\ A= \begin{bmatrix} -9&3&4\\ 0&-6&-2\\ 0&0&6 \end{bmatrix}
Adj
A
=
−
9
0
0
3
−
6
0
4
−
2
6
(a) Find
A
−
1
A^{-1}
A
−
1
.
(b) Find the solution to
A
x
=
b
A\bm{x}=\bm{b}
A
x
=
b
, where
x
=
[
x
y
z
]
\bm{x}=\begin{bmatrix} x\\y\\z \end{bmatrix}
x
=
x
y
z
and
b
=
[
−
1
2
0
]
\bm{b}=\begin{bmatrix} -1\\2\\0 \end{bmatrix}
b
=
−
1
2
0
Determinants and Inverses
Find the value(s) of
k
k
k
such that the matrix
[
1
0
−
2
1
2
3
k
0
4
]
\begin{bmatrix} 1&0&-2\\ 1&2&3\\ k&0&4 \end{bmatrix}
1
1
k
0
2
0
−
2
3
4
has no inverse.
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
)
.