High School
SAT
SAT Elite 1500
SAT Tutoring
ACT
ACT Elite 33
ACT Tutoring
University
MCAT
MCAT Elite 515
Med-School Admissions
Pre-Med Tutoring
Pre-Med Plus
LSAT
LSAT Elite 170
LSAT Self-Paced
LSAT Tutoring
DAT
DAT Elite
DAT Tutoring
Log in
Get Started for Free
Practice: Method of Inverse
Related Topics
Wize University Linear Algebra Textbook > Matrices
Solving Matrix Equations
6 Activities
Wize University Linear Algebra Textbook > Matrices
Matrix Inverse Algorithm
4 Activities
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
,
Part 1
Part 2
a) find the inverse
A
−
1
A^{-1}
A
−
1
.
A
−
1
=
[
2
−
7
5
0
1
0
−
1
3
−
2
]
A^{-1}= \left[\begin{array}{rrr} 2&-7&5\\ 0&1&0\\ -1&3&-2 \end{array}\right]
A
−
1
=
2
0
−
1
−
7
1
3
5
0
−
2
A
−
1
=
[
−
2
−
7
5
0
1
0
−
1
3
2
]
A^{-1}= \left[\begin{array}{rrr} -2&-7&5\\ 0&1&0\\ -1&3&2 \end{array}\right]
A
−
1
=
−
2
0
−
1
−
7
1
3
5
0
2
A
−
1
=
[
2
−
5
−
3
0
1
0
−
1
3
2
]
A^{-1}= \left[\begin{array}{rrr} 2&-5&-3\\ 0&1&0\\ -1&3&2 \end{array}\right]
A
−
1
=
2
0
−
1
−
5
1
3
−
3
0
2
I don't know
Previous Part
Check Part 1
Next Part
Check Submission
More Solving Matrix Equations 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
,
133 - FML 3 - 18.1W e.g. 16
Let
M
‾
=
[
k
3
3
0
(
k
−
1
)
3
0
0
(
k
−
2
)
]
\bcb{\boldsymbol{ \ul{M} = \begin{bmatrix} k & 3 & 3 \\ 0 & (k-1) & 3 \\ 0 & 0 & (k-2) \end{bmatrix}}}
M
=
k
0
0
3
(
k
−
1
)
0
3
3
(
k
−
2
)
, for what value(s) of
k
\bcb{\boldsymbol{ k}}
k
does the system
M
‾
x
⃗
=
b
⃗
\bcb{\boldsymbol{ \ul{M} \vec{x} = \vec{b} }}
M
x
=
b
have a unique solution?
(a)
k
=
0
,
1
,
2
(b)
k
≠
0
,
1
,
2
(c)
k
≠
0
(d)
None
of
the
above
\text{(a)}\;\bcb{\boldsymbol{ k = 0, \, \, 1,\, 2}}\qquad\qquad\qquad \text{(b)}\;\bcb{\boldsymbol{ k \ne 0,\, 1\, ,2}}\qquad\qquad\qquad \text{(c)}\;\bcb{\boldsymbol{ k \ne 0}}\qquad\qquad\qquad \text{(d)}\; \text{\bf{None of the above}}
(a)
k
=
0
,
1
,
2
(b)
k
=
0
,
1
,
2
(c)
k
=
0
(d)
None of the above
~
Practice: Solving Matrix Equations
Solve for
A
A
A
in the following equation:
A
[
−
1
2
2
−
5
]
=
I
2
−
[
2
−
3
0
2
]
A \left[ \begin{array}{rr} -1&2\\ 2&-5 \end{array} \right] = I_2- \left[ \begin{array}{rr} 2&-3\\ 0&2 \end{array} \right]
A
[
−
1
2
2
−
5
]
=
I
2
−
[
2
0
−
3
2
]
Find the inverse of
[
1
0
2
2
1
−
1
3
−
2
15
]
and use it to solve the system
\text{Find the inverse of}\begin{bmatrix} 1&0&2\\ 2&1&-1\\ 3&-2&15 \end{bmatrix}\text{and use it to solve the system}
Find the inverse of
1
2
3
0
1
−
2
2
−
1
15
and use it to solve the system
𝑥
+
2
𝑧
=
3
𝑥+2𝑧=3
x
+
2
z
=
3
2
𝑥
+
𝑦
−
𝑧
=
4
2𝑥+𝑦−𝑧=4
2
x
+
y
−
z
=
4
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
Given the matrix 𝐴
=
[
2
1
3
0
1
0
1
−
1
2
]
.
\text{Given the matrix 𝐴}=\begin{bmatrix} 2&1&3\\ 0&1&0\\ 1&-1&2\\ \end{bmatrix}.
Given the matrix
A
=
2
0
1
1
1
−
1
3
0
2
.
a) find the inverse of 𝐴.
b) solve the system of linear equations
2
𝑥
+
𝑦
+
3
𝑧
=
3
2𝑥+𝑦+3𝑧=3
2
x
+
y
+
3
z
=
3
Find the general solution the following system of equations using inverse of the coefficient matrix
2
x
+
y
+
4
z
=
−
3
x
+
3
z
=
−
1
4
x
+
2
y
+
z
=
2
\begin{array}{l}2x+y+4z=-3\\x+3z=-1\\4x+2y+z=2\end{array}
2
x
+
y
+
4
z
=
−
3
x
+
3
z
=
−
1
4
x
+
2
y
+
z
=
2
Use the method of inverse to solve the system of linear equations
x
+
2
z
=
3
2
x
+
y
−
z
=
4
3
x
−
2
y
+
15
z
=
0
\begin{array}{c} x+2z&=&3\\ 2x+y-z&=&4\\ 3x-2y+15z&=&0 \end{array}
x
+
2
z
2
x
+
y
−
z
3
x
−
2
y
+
15
z
=
=
=
3
4
0
Use the fact that the matrix
A
=
[
4
1
−
4
2
0
−
1
−
3
−
1
4
]
A=\begin{bmatrix} 4&1&-4\\ 2&0&-1\\ -3&-1&4 \end{bmatrix}
A
=
4
2
−
3
1
0
−
1
−
4
−
1
4
has inverse matrix
Practice: Matrix Operations
Find the values of
a
a
a
and
b
b
b
such that
[
a
2
−
1
1
]
[
2
3
]
=
[
10
b
]
T
\left[\begin{array}{c} a&2\\ -1&1 \end{array}\right] \left[\begin{array}{c} 2\\ 3 \end{array}\right] = \left[\begin{array}{c} 10&b \end{array}\right]^T
[
a
−
1
2
1
]
[
2
3
]
=
[
10
b
]
T
Practice: Matrix Operations
Solve for
a
,
b
,
c
a,\ b,\ c
a
,
b
,
c
in the following matrix equation:
[
1
a
−
1
2
−
2
1
3
0
−
1
]
[
−
1
2
b
]
−
[
1
9
−
1
]
T
=
[
−
8
c
−
2
]
\begin{bmatrix} 1 &a&-1\\ 2&-2&1\\ 3&0&-1 \end{bmatrix} \begin{bmatrix} -1\\2\\b \end{bmatrix}- \begin{bmatrix} 1&9&-1 \end{bmatrix}^T =\begin{bmatrix} -8\\c\\-2 \end{bmatrix}
1
2
3
a
−
2
0
−
1
1
−
1
−
1
2
b
−
[
1
9
−
1
]
T
=
−
8
c
−
2
Practice: Matrix Inverse
Practice: Solving Matrix Equations
Solve for
A
A
A
in the following equation:
A
[
−
1
2
2
−
5
]
=
I
2
−
[
2
−
3
0
2
]
A \left[ \begin{array}{rr} -1&2\\ 2&-5 \end{array} \right] = I_2- \left[ \begin{array}{rr} 2&-3\\ 0&2 \end{array} \right]
A
[
−
1
2
2
−
5
]
=
I
2
−
[
2
0
−
3
2
]
Practice Question: Using Matrix inverse
Practice Question: Using Matrix Inverse
Let
A
=
[
1
2
0
−
1
]
A=\left[\begin{array}{} 1&2\\0&-1 \end{array}\right]
A
=
[
1
0
2
−
1
]
and suppose that
3
A
B
−
[
1
−
1
2
0
]
=
I
2
3AB- \left[\begin{array}{} 1&-1\\ 2&0 \end{array}\right] =I_2
3
A
B
−
[
1
2
−
1
0
]
=
I
2
.
Find the matrix
B
B
B
.
Solving Matrix Equations
Let
[
3
−
1
0
4
]
\begin{bmatrix} 3&-1\\ 0&4 \end{bmatrix}
[
3
0
−
1
4
]
be the inverse of the matrix
[
a
b
c
d
]
\begin{bmatrix} a&b\\c&d \end{bmatrix}
[
a
c
b
d
]
.
Find the solution to the following system of linear equations:
a
x
+
b
y
=
−
2
c
x
+
d
y
=
2
\begin{array}{rrr} ax+by&=&-2\\ cx+dy&=&2 \end{array}
a
x
+
b
y
c
x
+
d
y
=
=
−
2
2
Solving Matrix Equations
Solve the following system of linear equations using the inverse of the coefficient matrix.
3
x
+
4
y
=
1
−
2
x
−
y
=
0
\begin{array}{rrr} 3x+4y&=&1\\ -2x-y&=&0 \end{array}
3
x
+
4
y
−
2
x
−
y
=
=
1
0
Practice: Matrix Inverse
Solve for
A
A
A
in the following equation:
A
[
−
1
2
2
−
5
]
=
I
2
−
[
2
−
3
0
2
]
A\begin{bmatrix} -1&2\\ 2&-5 \end{bmatrix} = I_2-\begin{bmatrix} 2&-3\\ 0&2 \end{bmatrix}
A
[
−
1
2
2
−
5
]
=
I
2
−
[
2
0
−
3
2
]
Inverse of a Matrix
e.g. In each case find the matrix A:
(
a
)
[
2
[
1
1
−
2
3
]
−
5
A
‾
−
1
]
T
=
(
4
A
‾
T
)
−
1
;
(
b
)
[
2
A
‾
T
−
3
I
‾
]
−
1
=
[
3
1
1
1
]
\!\begin{array}{rr} (a) \left[2\begin{bmatrix} 1&1\\-2&3 \end{bmatrix}-5\underline{A}^{-1}\right]^T=\left( 4\underline{A}^T \right)^{-1}\!;& \hspace{1cm} (b) \left[ 2\underline{A}^T -3\underline{I} \right]^{-1} =\begin{bmatrix} 3&1\\1&1 \end{bmatrix}\end{array}
(
a
)
[
2
[
1
−
2
1
3
]
−
5
A
−
1
]
T
=
(
4
A
T
)
−
1
;
(
b
)
[
2
A
T
−
3
I
]
−
1
=
[
3
1
1
1
]
Practice: Solving Matrix Equations
Find the matrix
C
C
C
such that:
(
I
−
2
C
[
2
4
−
1
−
3
]
)
−
1
=
[
3
1
1
1
]
\left(I-2C\left[\begin{array}{rr}2&4\\-1&-3\end{array}\right]\right)^{-1} =\left[\begin{array}{rr}3&1\\1&1\end{array}\right]
(
I
−
2
C
[
2
−
1
4
−
3
]
)
−
1
=
[
3
1
1
1
]
Use the fact that the matrix
A
=
[
4
1
−
4
2
0
−
1
−
3
−
1
4
]
A=\begin{bmatrix} 4&1&-4\\ 2&0&-1\\ -3&-1&4 \end{bmatrix}
A
=
4
2
−
3
1
0
−
1
−
4
−
1
4
has inverse matrix
Cramer's Rule
Find
y
y
y
given the system of linear equations:
[
0
−
1
1
4
2
0
2
1
−
2
]
[
x
y
z
]
=
[
2
1
4
]
\left[ \begin{array}{rrr} 0&-1&1\\ 4&2&0\\ 2&1&-2 \end{array} \right] \begin{bmatrix} x\\y\\z\\ \end{bmatrix} = \begin{bmatrix} 2\\1\\4\\ \end{bmatrix}
0
4
2
−
1
2
1
1
0
−
2
x
y
z
=
2
1
4
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
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
.
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
.
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.