Cramer's Rule

Solving Matrix Equations

6 Activities

Wize University Linear Algebra Textbook > Determinants

Cramer's Rule for Solving Linear Systems

3 Activities

Find yyy given the system of linear equations:

[0−1142021−2][xyz]=[214]\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}​042​−121​10−2​​​xyz​​=​214​​

y=

I don't know

Practice: Method of Inverse

Practice Question: Method of Inverse
Given the matrix A=[2130101−12]A=\left[\begin{array}{rrr}
2&1&3\\
0&1&0\\
1&-1&2
\end{array}\right]A=​201​11−1​302​​,

Practice: Method of Inverse

Practice Question: Method of Inverse
Given the matrix A=[2130101−12]A=\left[\begin{array}{rrr}
2&1&3\\
0&1&0\\
1&-1&2
\end{array}\right]A=​201​11−1​302​​,

133 - FML 3 - 18.1W e.g. 16

 Let   ⁣M‾=[k330(k−1)300(k−2)]\bcb{\boldsymbol{ \ul{M} = \begin{bmatrix} k & 3 & 3 \\ 0 & (k-1) & 3 \\ 0 & 0 & (k-2) \end{bmatrix}}}M​=​k00​3(k−1)0​33(k−2)​​, for what value(s) of k\bcb{\boldsymbol{ k}}k does the system   ⁣M‾x⃗=b⃗\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 AAA in the following equation:
A[−122−5]=I2−[2−302]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[−12​2−5​]=I2​−[20​−32​]

Find the inverse of[10221−13−215]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​123​01−2​2−115​​and use it to solve the system
𝑥+2𝑧=3𝑥+2𝑧=3x+2z=3
2𝑥+𝑦−𝑧=42𝑥+𝑦−𝑧=42x+y−z=4

Practice: Matrix Inverse

Find the inverse of the following matrices:
a.) A=[4534]A=\begin{bmatrix}
4&5\\
3&4
\end{bmatrix}A=[43​54​]
b.) B=[−1−2024113−2]B=\begin{bmatrix}
-1&-2&0\\
2&4&1\\
1&3&-2
\end{bmatrix}B=​−121​−243​01−2​​

Given the matrix 𝐴=[2130101−12].\text{Given the matrix 𝐴}=\begin{bmatrix}
2&1&3\\
0&1&0\\
1&-1&2\\
\end{bmatrix}.Given the matrix A=​201​11−1​302​​.
a) find the inverse of 𝐴. 
b) solve the system of linear equations          2𝑥+𝑦+3𝑧=32𝑥+𝑦+3𝑧=32x+y+3z=3                                                                  

Find the general solution the following system of equations using inverse of the coefficient matrix
2x+y+4z=−3x+3z=−14x+2y+z=2\begin{array}{l}2x+y+4z=-3\\x+3z=-1\\4x+2y+z=2\end{array}2x+y+4z=−3x+3z=−14x+2y+z=2​

Use the method of inverse to solve the system of linear equations x+2z=32x+y−z=43x−2y+15z=0\begin{array}{c}
x+2z&=&3\\
2x+y-z&=&4\\
3x-2y+15z&=&0
\end{array}x+2z2x+y−z3x−2y+15z​===​340​

Use the fact that the matrix
                                                 A=[41−420−1−3−14]A=\begin{bmatrix}
4&1&-4\\
2&0&-1\\
-3&-1&4
\end{bmatrix}A=​42−3​10−1​−4−14​​  
has inverse matrix

Practice: Method of Inverse

Practice Question: Method of Inverse
Given the matrix A=[2130101−12]A=\left[\begin{array}{rrr}
2&1&3\\
0&1&0\\
1&-1&2
\end{array}\right]A=​201​11−1​302​​,

Practice: Matrix Operations

Find the values of aaa and bbb such that [a2−11][23]=[10b]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​21​][23​]=[10​b​]T

Practice: Matrix Operations

Solve for a, b, ca,\ b,\ ca, b, c in the following matrix equation:
[1a−12−2130−1][−12b]−[19−1]T=[−8c−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}​123​a−20​−11−1​​​−12b​​−[1​9​−1​]T=​−8c−2​​

Practice: Matrix Inverse

Practice: Solving Matrix Equations
Solve for AAA in the following equation:
A[−122−5]=I2−[2−302]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[−12​2−5​]=I2​−[20​−32​]

Practice Question: Using Matrix inverse

Practice Question: Using Matrix Inverse
Let A=[120−1]A=\left[\begin{array}{}
1&2\\0&-1
\end{array}\right]A=[10​2−1​] and suppose that 3AB−[1−120]=I23AB-

\left[\begin{array}{}
1&-1\\
2&0
\end{array}\right]

=I_23AB−[12​−10​]=I2​. 
Find the matrix BBB.

Solving Matrix Equations

Let [3−104]\begin{bmatrix}
3&-1\\
0&4
\end{bmatrix}[30​−14​] be the inverse of the matrix [abcd]\begin{bmatrix}
a&b\\c&d
\end{bmatrix}[ac​bd​].
Find the solution to the following system of linear equations:
ax+by=−2cx+dy=2\begin{array}{rrr}
ax+by&=&-2\\
cx+dy&=&2
\end{array}ax+bycx+dy​==​−22​

Solving Matrix Equations

Solve the following system of linear equations using the inverse of the coefficient matrix.
3x+4y=1−2x−y=0\begin{array}{rrr}
3x+4y&=&1\\
-2x-y&=&0
\end{array}3x+4y−2x−y​==​10​

Practice: Matrix Inverse

Solve for AAA in the following equation:
A[−122−5]=I2−[2−302]A\begin{bmatrix}
-1&2\\
2&-5
\end{bmatrix}
=
I_2-\begin{bmatrix}
2&-3\\
0&2
\end{bmatrix}A[−12​2−5​]=I2​−[20​−32​]

Inverse of a Matrix

e.g. In each case find the matrix A:
 ⁣(a)[2[11−23]−5A‾−1]T=(4A‾T)−1 ⁣;(b)[2A‾T−3I‾]−1=[3111]\!\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​13​]−5A​−1]T=(4A​T)−1;​(b)[2A​T−3I​]−1=[31​11​]​

Practice: Solving Matrix Equations

Find the matrix CCC such that:
(I−2C[24−1−3])−1=[3111]\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−2C[2−1​4−3​])−1=[31​11​]

Use the fact that the matrix
A=[41−420−1−3−14]A=\begin{bmatrix}
4&1&-4\\
2&0&-1\\
-3&-1&4
\end{bmatrix}A=​42−3​10−1​−4−14​​
has inverse matrix

More Cramer's Rule for Solving Linear Systems Questions:

Solving System of Linear Equations using Cramer's Rule

Solve for zzz if [4−11023201][xyz]=[123]\begin{bmatrix}4&-1&1\\0&2&3\\2&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}​402​−120​131​​​xyz​​=​123​​

Cramer Rule to solve Linear System of equations

Cramer’s rule
If AAA is an n×nn\times nn×n invertible matrix and we wish to solve AX=BAX=BAX=B for X=[x1, …, xn]TX=\left[x_1,\ \ldots,\ x_n\right]^TX=[x1​, …, xn​]T, then Cramer's Rule gives
xi=det(Ai)det(A)x_i=\frac{det\left(A_i\right)}{det\left(A\right)}xi​=det(A)det(Ai​)​

Cramer's Rule: Linear Systems

Let  A=[−1−2430−215−1]A=\begin{bmatrix}
-1&-2&4\\
3&0&-2\\
1&5&-1
\end{bmatrix}A=​−131​−205​4−2−1​​.
(a) Find detA\text{det}AdetA.
(b) User Cramer's Rule to find the value of xxx in the unique solution to Ax=bA\bm{x}=\bm{b}Ax=b, where x=[xyz]\bm{x}=\begin{bmatrix}
x\\y\\z
\end{bmatrix}x=​xyz​​ and b=[00−2]\bm{b}=\begin{bmatrix}
0\\0\\-2
\end{bmatrix}b=​00−2​​

Cramer's Rule for Solving Linear Systems: Determinants and Inverses

Let AAA be a 3 x 3 matrix with detA=−18\text{det}A=-18detA=−18 and  Adj A=[−9340−6−2006]\text{Adj}\ A=
\begin{bmatrix}
-9&3&4\\
0&-6&-2\\
0&0&6
\end{bmatrix}Adj A=​−900​3−60​4−26​​
(a) Find A−1A^{-1}A−1.
(b) Find the solution to Ax=bA\bm{x}=\bm{b}Ax=b, where x=[xyz]\bm{x}=\begin{bmatrix}
x\\y\\z
\end{bmatrix}x=​xyz​​ and b=[−120]\bm{b}=\begin{bmatrix}
-1\\2\\0
\end{bmatrix}b=​−120​​

Cramer's Rule: Linear Systems

Given that det[a3−12−bc1−1d]=10,  det[p3−1q−bc5−1d]=3,  det[ap−12qc15d]=7det
\begin{bmatrix}
a&3&-1\\
2&-b&c\\
1&-1&d
\end{bmatrix}=10,\ 
\ 
det
\begin{bmatrix}
p&3&-1\\
q&-b&c\\
5&-1&d
\end{bmatrix}=3,\ 
\ 
det
\begin{bmatrix}
a&p&-1\\
2&q&c\\
1&5&d
\end{bmatrix}=7

det​a21​3−b−1​−1cd​​=10,  det​pq5​3−b−1​−1cd​​=3,  det​a21​pq5​−1cd​​=7, find the value of yyy in the unique solution to the following system of linear equations"
ax+3y−z=p2x−by+cz=qx−y+dz=5\begin{array}{c}
ax+3y-z=p\\
2x-by+cz=q\\
x-y+dz=5
\end{array}ax+3y−z=p2x−by+cz=qx−y+dz=5​

Cramer's Rule

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133 - FML 3 - 18.1W e.g. 16

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