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Suppose that A=1 -1031-4 and Ris the reduced row echelon form of A. Find the in…
Related Topics
Wize University Linear Algebra Textbook > Matrices
Matrix Inverse Algorithm
4 Activities
Suppose that
A
=
[
1
−
1
0
3
1
−
4
]
A=\begin{bmatrix}1 &-1&0\\3&1&-4\end{bmatrix}
A
=
[
1
3
−
1
1
0
−
4
]
and
R
R
R
is the reduced row echelon form of
A
A
A
. Find the invertible matrix
U
U
U
such that
R
=
U
A
R=UA
R
=
U
A
.
[
−
1
/
4
−
1
/
4
3
/
4
1
/
4
]
\begin{bmatrix}-1/4 &-1/4\\3/4&1/4\end{bmatrix}
[
−
1/4
3/4
−
1/4
1/4
]
[
1
/
4
1
/
4
−
3
/
4
1
/
4
]
\begin{bmatrix}1/4 &1/4\\-3/4&1/4\end{bmatrix}
[
1/4
−
3/4
1/4
1/4
]
[
−
1
/
2
−
1
/
2
3
/
2
1
/
2
]
\begin{bmatrix}-1/2 &-1/2\\3/2&1/2\end{bmatrix}
[
−
1/2
3/2
−
1/2
1/2
]
[
1
/
2
3
/
2
1
/
2
−
1
/
2
]
\begin{bmatrix}1/2 &3/2\\1/2&-1/2\end{bmatrix}
[
1/2
1/2
3/2
−
1/2
]
None of the above
I don't know
Check Submission
More Matrix Inverse Algorithm Questions:
Practice: Method of Inverse
Practice Question: Method of Inverse
Given the matrix
A
=
[
2
1
3
0
1
0
1
−
1
2
]
A=\left[\begin{array}{rrr} 2&1&3\\ 0&1&0\\ 1&-1&2 \end{array}\right]
A
=
2
0
1
1
1
−
1
3
0
2
,
Practice: Method of Inverse
Practice Question: Method of Inverse
Given the matrix
A
=
[
2
1
3
0
1
0
1
−
1
2
]
A=\left[\begin{array}{rrr} 2&1&3\\ 0&1&0\\ 1&-1&2 \end{array}\right]
A
=
2
0
1
1
1
−
1
3
0
2
,
Practice: Matrix Inverse
Practice Question: Matrix Inverse
For what value(s) of
k
will the matrix
[
1
2
1
0
1
5
0
k
1
]
\begin{bmatrix} 1&2&1\\ 0&1&5\\ 0&k&1 \end{bmatrix}
1
0
0
2
1
k
1
5
1
be invertible?
Practice: Matrix Inverse
Practice Question: Matrix Inverse
For what value(s) of
k
will the matrix
[
1
2
1
0
1
5
0
k
1
]
\begin{bmatrix} 1&2&1\\ 0&1&5\\ 0&k&1 \end{bmatrix}
1
0
0
2
1
k
1
5
1
be invertible?
Practice: Matrix Inverse
Practice Question: Matrix Inverse
For what value(s) of
k
will the matrix
[
1
2
1
0
1
5
0
k
1
]
\begin{bmatrix} 1&2&1\\ 0&1&5\\ 0&k&1 \end{bmatrix}
1
0
0
2
1
k
1
5
1
be invertible?
Practice: Matrix Inverse
Practice Question: Matrix Inverse
For what value(s) of
k
will the matrix
[
1
2
1
0
1
5
0
k
1
]
\begin{bmatrix} 1&2&1\\ 0&1&5\\ 0&k&1 \end{bmatrix}
1
0
0
2
1
k
1
5
1
be invertible?
Practice: Matrix Inverse
Practice Question: Matrix Inverse
For what value(s) of
k
will the matrix
[
1
2
1
0
1
5
0
k
1
]
\begin{bmatrix} 1&2&1\\ 0&1&5\\ 0&k&1 \end{bmatrix}
1
0
0
2
1
k
1
5
1
be invertible?
Practice: Matrix Inverse
Practice Question: Matrix Inverse
For what value(s) of
k
will the matrix
[
1
2
1
0
1
5
0
k
1
]
\begin{bmatrix} 1&2&1\\ 0&1&5\\ 0&k&1 \end{bmatrix}
1
0
0
2
1
k
1
5
1
be invertible?
133 - FML 3 - 18.1W e.g. 10
Find the value(s) of c such that A is NOT invertible given
A
=
[
c
−
c
1
−
1
−
1
−
1
1
−
c
c
]
\bcb{\boldsymbol{ A = \begin{bmatrix} c & -c & 1 \\ -1 & -1 & -1 \\ 1 & -c & c \end{bmatrix} }}
A
=
c
−
1
1
−
c
−
1
−
c
1
−
1
c
Practice: Matrix Inverse
Find all values of k so that the matrix
[
−
2
0
0
3
1
0
2
−
1
k
]
\begin{bmatrix} -2&0&0\\ 3&1&0\\ 2&-1&k \end{bmatrix}
−
2
3
2
0
1
−
1
0
0
k
has an inverse.
Find the inverse of
A
=
[
1
0
1
0
1
1
1
1
0
]
A=\begin{bmatrix} 1&0&1\\ 0&1&1\\1&1&0 \end{bmatrix}
A
=
1
0
1
0
1
1
1
1
0
.
For each of the following matrices, find the inverse or state why the inverse does not exist.
A
=
[
1
−
1
−
2
1
0
0
0
1
2
]
A=\begin{bmatrix} 1&-1&-2\\ 1&0&0\\ 0&1&2 \end{bmatrix}
A
=
1
1
0
−
1
0
1
−
2
0
2
B
=
[
3
1
9
15
0
6
]
B=\begin{bmatrix} 3&1&9\\ 15&0&6 \end{bmatrix}
B
=
[
3
15
1
0
9
6
]
Practice: Matrix Inverse
Find all values of k for which the matrix
[
2
−
6
−
1
k
2
−
22
]
\begin{bmatrix} 2&-6\\ -1&k^2-22 \end{bmatrix}
[
2
−
1
−
6
k
2
−
22
]
has no inverse.
Practice: Matrix Inverse
Find the inverse of the following matrices:
a.)
A
=
[
4
5
3
4
]
A=\begin{bmatrix} 4&5\\ 3&4 \end{bmatrix}
A
=
[
4
3
5
4
]
b.)
B
=
[
−
1
−
2
0
2
4
1
1
3
−
2
]
B=\begin{bmatrix} -1&-2&0\\ 2&4&1\\ 1&3&-2 \end{bmatrix}
B
=
−
1
2
1
−
2
4
3
0
1
−
2
Matrix Inverse
Practice: Matrix Inverse
If
[
1
a
x
2
b
y
0
c
z
]
\left[\begin{array}{c} 1&a&x\\ 2&b&y\\ 0&c&z \end{array}\right]
1
2
0
a
b
c
x
y
z
is the
inverse
of the matrix
[
1
p
0
−
3
q
4
2
r
1
]
\left[\begin{array}{c} 1&p&0\\ -3&q&4\\ 2&r&1 \end{array}\right]
1
−
3
2
p
q
r
0
4
1
, find the value of
r
r
r
.
Let
A
=
[
2
1
4
1
0
3
4
2
1
]
A=\begin{bmatrix}2&1&4\\1&0&3\\4&2&1\end{bmatrix}
A
=
2
1
4
1
0
2
4
3
1
. Find
A
−
1
A^{-1}
A
−
1
if possible.
Find the inverse of
[
0
2
1
2
5
5
1
2
1
]
\begin{bmatrix} 0 &2& 1\\ 2 &5 &5\\ 1 &2 &1 \end{bmatrix}
0
2
1
2
5
2
1
5
1
The following matrix A depends on a parameter p. Find all the values of the parameter p for which the
corresponding matrix is not invertible.
A
=
[
1
2
3
4
1000
2000
3000
4001
0
𝑝
𝑝
+
1
𝑝
+
2
0
−
𝑝
−
2
𝑝
+
1
−
3
𝑝
+
2
]
A=\begin{bmatrix} 1&2&3&4\\ 1000&2000&3000&4001\\ 0&𝑝&𝑝 + 1&𝑝 + 2\\ 0&−𝑝&−2𝑝 + 1&−3𝑝 + 2 \end{bmatrix}
A
=
1
1000
0
0
2
2000
p
−
p
3
3000
p
+
1
−
2
p
+
1
4
4001
p
+
2
−
3
p
+
2
For what value or values of a (if any) is the matrix below invertible?
[
1
2
1
0
3
5
0
𝑎
1
]
\begin{bmatrix} 1& 2& 1\\ 0 &3& 5\\ 0 &𝑎 &1 \end{bmatrix}
1
0
0
2
3
a
1
5
1
The following matrix A depends on a parameter p. Find all the values of the parameter
p for which the corresponding matrix is not invertible.
A
=
[
1
2
3
4
1000
2000
3000
4001
0
𝑝
𝑝
+
1
𝑝
+
2
0
−
𝑝
−
2
𝑝
+
1
−
3
𝑝
+
2
]
A=\begin{bmatrix} 1&2&3&4\\ 1000&2000&3000&4001\\ 0&𝑝&𝑝 + 1&𝑝 + 2\\ 0&−𝑝&−2𝑝 + 1&−3𝑝 + 2 \end{bmatrix}
A
=
1
1000
0
0
2
2000
p
−
p
3
3000
p
+
1
−
2
p
+
1
4
4001
p
+
2
−
3
p
+
2
Practice Question 3: Matrix Inverse
Practice Question: Matrix Inverse
For what value(s) of
k
will the matrix
[
1
2
1
0
1
5
0
k
1
]
\begin{bmatrix} 1&2&1\\ 0&1&5\\ 0&k&1 \end{bmatrix}
1
0
0
2
1
k
1
5
1
be invertible?
Find all the values of k for which the matrix A has an inverse
A
=
[
2
1
0
1
0
2
1
0
2
k
]
A=\begin{bmatrix} 2&1&0\\ 1&0&2\\ 1&0&2k \end{bmatrix}
A
=
2
1
1
1
0
0
0
2
2
k
Practice: Method of Inverse
Practice Question: Method of Inverse
Given the matrix
A
=
[
2
1
3
0
1
0
1
−
1
2
]
A=\left[\begin{array}{rrr} 2&1&3\\ 0&1&0\\ 1&-1&2 \end{array}\right]
A
=
2
0
1
1
1
−
1
3
0
2
,
Find all the values of 𝑘 such that the matrix 𝐴
=
[
3
3
−
6
10
−
k
2
]
has an inverse.
\text{Find all the values of 𝑘 such that the matrix 𝐴}=\begin{bmatrix} 3&3\\ -6&10-k^2 \end{bmatrix}\text{has an inverse.}
Find all the values of
k
such that the matrix
A
=
[
3
−
6
3
10
−
k
2
]
has an inverse.
Matrix Inverse
If
[
1
a
x
2
b
y
0
c
z
]
\left[\begin{array}{c} 1&a&x\\ 2&b&y\\ 0&c&z \end{array}\right]
1
2
0
a
b
c
x
y
z
is the
inverse
of the matrix
[
1
p
0
−
3
q
4
2
r
1
]
\left[\begin{array}{c} 1&p&0\\ -3&q&4\\ 2&r&1 \end{array}\right]
1
−
3
2
p
q
r
0
4
1
, find the value of
r
r
r
.
Practice: Adjoint Matrix
Let
B
=
[
1
−
1
0
2
1
1
1
0
0
0
−
1
1
−
2
1
−
2
−
1
]
B=\left[\begin{array}{rrrr} 1&-1&0&2\\ 1&1&1&0\\ 0&0&-1&1\\ -2&1&-2&-1 \end{array}\right]
B
=
1
1
0
−
2
−
1
1
0
1
0
1
−
1
−
2
2
0
1
−
1
, and note that
det
(
B
)
=
−
1
\text{det}(B)=-1
det
(
B
)
=
−
1
Use the determinant and the adjoint of
B
B
B
to find the entry in position
(
4
,
3
)
(4,3)
(
4
,
3
)
of the matrix
B
−
1
B^{-1}
B
−
1
Find the inverse of the matrix
[
1
0
2
2
1
−
1
3
−
2
15
]
\begin{bmatrix} 1&0&2\\ 2&1&-1\\ 3&-2&15 \end{bmatrix}
1
2
3
0
1
−
2
2
−
1
15
.
Matrix Inversion Algorithm
Example: Matrix Inverse Algorithm
Find the inverse of
A
=
[
1
3
2
5
]
A=\left[\begin{array}{rr}1 & 3\\2 & 5\end{array}\right]
A
=
[
1
2
3
5
]
using the formula for a
2
×
2
2\times2
2
×
2
matrix, and using the Gauss-Jordan elimination algorithm.
Find the inverse of the matrix
A
=
[
1
0
2
2
1
−
1
3
−
2
15
]
A=\left[\begin{array}{rrr} 1&0&2\\ 2&1&-1\\ 3&-2&15 \end{array}\right]
A
=
1
2
3
0
1
−
2
2
−
1
15
Matrix operations
Suppose that
A
=
[
1
−
1
0
3
−
2
0
0
4
1
]
A=\begin{bmatrix}1 & -1 & 0\\3 & -2 & 0\\0 & 4 & 1\end{bmatrix}
A
=
1
3
0
−
1
−
2
4
0
0
1
and
B
B
B
is a
3
×
2
3 \times 2
3
×
2
matrix. If the product
A
B
=
[
3
4
0
1
1
−
2
]
AB = \begin{bmatrix}3 & 4\\0 & 1\\1 & -2\end{bmatrix}
A
B
=
3
0
1
4
1
−
2
, find the (1,2)-entry of
B
B
B
(i.e.
B
12
B_{12}
B
12
).
Invertible matrix
For what value(s) of
a
a
a
will the matrix
[
2
−
4
3
a
]
\begin{bmatrix}2 & -4\\3 & a\end{bmatrix}
[
2
3
−
4
a
]
be invertible?
Matrix Inverse
Practice: Matrix Inverse
If
[
1
a
x
2
b
y
0
c
z
]
\left[\begin{array}{c} 1&a&x\\ 2&b&y\\ 0&c&z \end{array}\right]
1
2
0
a
b
c
x
y
z
is the
inverse
of the matrix
[
1
p
0
−
3
q
4
2
r
1
]
\left[\begin{array}{c} 1&p&0\\ -3&q&4\\ 2&r&1 \end{array}\right]
1
−
3
2
p
q
r
0
4
1
, find the value of
r
r
r
.
Practice Question: Matrix Inverse
Practice Question: Matrix Inverse
If
[
1
a
x
2
b
y
0
c
z
]
\left[\begin{array}{} 1&a&x\\ 2&b&y\\ 0&c&z \end{array}\right]
1
2
0
a
b
c
x
y
z
is the inverse of the matrix
[
1
p
0
−
3
q
4
2
r
1
]
\left[\begin{array}{} 1&p&0\\ -3&q&4\\ 2&r&1 \end{array}\right]
1
−
3
2
p
q
r
0
4
1
, find the value of
r
r
r
.
For what value(s) of
a
a
a
will the matrix
[
2
−
4
3
a
]
\begin{bmatrix}2 & -4\\3 & a\end{bmatrix}
[
2
3
−
4
a
]
be invertible?
Practice: Matrix Inverse
Find all values of k for which the matrix
[
2
−
6
−
1
k
2
−
22
]
\begin{bmatrix} 2&-6\\ -1&k^2-22 \end{bmatrix}
[
2
−
1
−
6
k
2
−
22
]
has no inverse.
Practice: Matrix Inverse
Practice Question: Matrix Inverse
Find all the values of
k
k
k
such that the matrix
[
2
1
0
1
0
2
1
0
2
k
]
\begin{bmatrix} 2&1&0\\ 1&0&2\\ 1&0&2k \end{bmatrix}
2
1
1
1
0
0
0
2
2
k
has an inverse.
Practice: Matrix Inverse
Practice Question: Matrix Inverse
For what value(s) of
k
will the matrix
[
1
2
1
0
1
5
0
k
1
]
\begin{bmatrix} 1&2&1\\ 0&1&5\\ 0&k&1 \end{bmatrix}
1
0
0
2
1
k
1
5
1
be invertible?
Practice: Matrix Inverse
Practice Question: Matrix Inverse
For what value(s) of
k
k
k
will the matrix
[
1
2
1
0
1
5
0
k
1
]
\left[\begin{array}{rrr} 1&2&1\\ 0&1&5\\ 0&k&1 \end{array}\right]
1
0
0
2
1
k
1
5
1
be invertible?
Practice Question 1: Matrix Inverse
Practice Question: Matrix Inverse
Find all the values of
k
k
k
such that the matrix
[
2
1
0
1
0
2
1
0
2
k
]
\begin{bmatrix} 2&1&0\\ 1&0&2\\ 1&0&2k \end{bmatrix}
2
1
1
1
0
0
0
2
2
k
has an inverse.
Practice: Matrix Inverse
Find all values of k so that the matrix
[
−
2
0
0
3
1
0
2
−
1
k
]
\begin{bmatrix} -2&0&0\\ 3&1&0\\ 2&-1&k \end{bmatrix}
−
2
3
2
0
1
−
1
0
0
k
has an inverse.
Practice: Matrix Inverse (~2016 Test2 #15)
Practice: Matrix Inverse
Find all the values of
k
k
k
such that the matrix
[
2
1
0
1
0
2
1
0
2
k
]
\begin{bmatrix} 2&1&0\\ 1&0&2\\ 1&0&2k \end{bmatrix}
2
1
1
1
0
0
0
2
2
k
has an inverse.