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Solving Matrix Equations
Related Topics
Wize University Linear Algebra Textbook > Matrices
Solving Matrix Equations
6 Activities
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
⟨
−
1
5
,
2
5
⟩
\left\lang -\dfrac{1}{5},\ \dfrac{2}{5}\right\rang
⟨
−
5
1
,
5
2
⟩
⟨
1
5
,
−
2
5
⟩
\left\lang \dfrac{1}{5},\ -\dfrac{2}{5}\right\rang
⟨
5
1
,
−
5
2
⟩
⟨
−
1
5
,
−
1
5
⟩
\left\lang-\dfrac{1}{5},\ -\dfrac{1}{5}\right\rang
⟨
−
5
1
,
−
5
1
⟩
⟨
2
5
,
2
5
⟩
\left\lang\dfrac{2}{5},\ \dfrac{2}{5}\right\rang
⟨
5
2
,
5
2
⟩
None of the above
I don't know
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: 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 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
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