Download Use Inverse Matrices to Solve Linear Systems

Use Inverse Matrices to
Solve Linear Systems
Chapter 3.8
Square Matrix
• Although a matrix may have any number of rows and columns, square
matrices have properties that we can use to solve systems of equations
• A square matrix is one of the form 𝑛 × 𝑛, where n is a positive whole
number
• That is, the number of rows equals the number of columns
• Like the real numbers, square matrices have a special matrix called the
identity matrix
• In the same way that, for the real numbers, 1 ⋅ 𝑎 = 𝑎, if I is an identity
matrix and A is a square matrix of the same dimensions, 𝐼𝐴 = 𝐴
Square Matrix
• Each size of square matrix has its own identity matrix
• For 2 × 2 matrices, the identity matrix is
1 0
0 1
• For 3 × 3 matrices, the identity matrix is
1 0 0
0 1 0
0 0 1
• For any other square matrix, the identity matrix has 1’s in the left-toright diagonal and 0’s everywhere else
Square Matrix
1
Multiply 0
4
3
2
2
−1
1 0
5 ⋅ 0 1
0 0
1
0
0 =
1
1 ⋅ 1 + 3 ⋅ 0 + −1 ⋅ 0 1 ⋅ 0 + 3 ⋅ 1 + −1 ⋅ 0
0⋅1+2⋅0+5⋅0
0⋅0+2⋅1+5⋅0
4⋅1+2⋅0+1⋅0
4⋅0+2⋅1+1⋅0
1 ⋅ 0 + 3 ⋅ 0 + −1 ⋅ 1
0⋅0+2⋅0+5⋅1 =
4⋅0+2⋅0+1⋅1
1 + 0 + 0 0 + 3 + 0 0 + 0 + (−1)
1
0+0+0 0+2+0
0+0+5 = 0
4+0+0 0+2+0
0 + 0 + _1
4
3 −1
2 5
2 1
Inverse Matrix
• Recall that, among the real numbers, for every non-zero number a
1
1
there is another number so that 𝑎 ⋅ = 1
𝑎
𝑎
• Suppose that A and B represent 𝑛 × 𝑛 matrices and that I is the 𝑛 × 𝑛
identity matrix
• If 𝐴𝐵 = 𝐼, then B is the inverse of A (or A is the inverse of B)
Example
3 8
−5 8
• Suppose that 𝐴 =
and 𝐵 =
2 5
2 −3
• The product 𝐴𝐵 is
3 8 −5 8
3 ⋅ −5 + 8 ⋅ 2 3 ⋅ 8 + 8 ⋅ −3
•
⋅
=
=
2 5
2 −3
2 ⋅ −5 + 5 ⋅ 2 2 ⋅ 8 + 5 ⋅ −3
1 0
−15 + 16 24 − 24
•
=
−10 + 10 16 − 15
0 1
• The result is the 2 × 2 identity matrix, so that 𝐴 is the inverse of 𝐵
• We write the inverse of a matrix A as 𝐴−1 , so 𝐵 = 𝐴−1
Inverse Matrix
• Recall that we use the inverse of a number to solve equations
1
2
1
2
• If 2𝑥 = 10, then multiplying both sides ⋅ 2𝑥 = ⋅ 10
1⋅𝑥 =5
𝑥=5
Inverse Matrix
• In the same way, the inverse of a matrix can be used to solve a matrix
equation
𝐴𝑋 = 𝐵
𝐴−1 𝐴𝑋 = 𝐴−1 𝐵
𝐼𝑋 = 𝐴−1 𝐵
𝑋 = 𝐴−1 𝐵
• Note that we must pay attention to the order of the multiplication
• Since we multiplied to the left of 𝐴𝑋, we must also multiply to the left
of 𝐵
Calculating the Inverse of a Matrix
𝑎
• Given 𝐴 =
𝑐
of A is
𝑏
, where 𝑎, 𝑏, 𝑐, and 𝑑 are real numbers, the inverse
𝑑
𝑑 −𝑏
•
=
−𝑐 𝑎
• Note that his can only make sense if 𝑎𝑑 − 𝑏𝑐 ≠ 0
• What this means is that not every square matrix has an inverse
𝐴−1
1
𝑎𝑑−𝑏𝑐
Example
3 8
Find the inverse of 𝐴 =
.
2 5
Follow the pattern of
2, and 𝑑 = 5.
𝐴−1
1
3⋅5−8⋅2
5
−2
The inverse is
=
1
𝑎𝑑−𝑏𝑐
𝑑
−𝑐
−𝑏
. Then, 𝑎 = 3, 𝑏 = 8, 𝑐 =
𝑎
1
−8
5 −8
−5
=
=
−1
3
−2 3
2
8
−3
Guided Practice
Find the inverse of the matrix.
−3 −4
1.
−1 −2
2.
−1
−4
3.
6 1
2 4
5
8
Guided Practice
Find the inverse of the matrix.
1.
2.
3.
−3 −4 −3 −4
;
−1 −2 −1 −2
−1 5 −1
;
−4 8 −4
5
8
6 1 6 1
;
2 4 2 4
−1
−1
=
=
−1
=
1
−3⋅−2− −4 ⋅−1
1
−1⋅8−5⋅−4
1
6⋅4−2⋅1
4
−2
−1 2
1 −2
−2 4
4
=
= 1 −3
2
1 −3
1 −3
2
2
1 8
8 −5
=
12 4
4 −1
−5
=
−1
1
−1
4 −1
=
=
22 −2
6
6
2
3
1
3
2
11
−1
11
−5
12
−1
12
−1
22
3
11
Solving a Matrix Equation
• To solve a matrix equation, we must find an inverse that will cancel
the coefficient matrix
• If A, B, and X are matrices such that 𝐴𝑋 = 𝐵, then A is the coefficient
matrix and to solve for X we use the inverse of A
• Note that this is similar to finding the inverse (or reciprocal) of real
number a in order to solve the equation 𝑎𝑥 = 𝑏
• Once the inverse is found, multiply
𝐴−1 ⋅ 𝐴𝑋 = 𝐴−1 𝐵
𝐼𝑋 = 𝐴−1 𝐵
𝑋 = 𝐴−1 𝐵
Solving a Matrix Equation
Example
Solve the matrix equation 𝐴𝑋 = 𝐵 for the 2 × 2 matrix X.
2 −7
−21 3
𝐴=
,𝐵 =
−1 4
12 −2
2 −7
−21 3
𝑋=
−1 4
12 −2
How do we know that X must be a 2 × 2 matrix?
Solving a Matrix Equation
First, find the inverse of A
𝐴−1
2 −7
=
−1 4
−1
1 4
=
1 1
1
4
=
2 ⋅ 4 − −7 ⋅ −1 1
7
4
=
2
1
7
2
7
2
Solving a Matrix Equation
Next, multiply both sides of the matrix equation on the left
4
1
𝐴−1 𝐴𝑥 = 𝐴−1 𝐵
7 2 −7
4 7 −21
𝑋=
2 −1 4
1 2 12
4 ⋅ −21 + 7 ⋅ 12
1 0
𝑋=
1 ⋅ −21 + 2 ⋅ 12
0 1
𝑋=
0
3
−2
−1
3
−2
4 ⋅ 3 + 7 ⋅ −2
1 ⋅ 3 + 2 ⋅ −2
Solving a Matrix Equation
Check your answer.
2 −7 0 −2
=
−1 4 3 −1
2 ⋅ 0 + −7 ⋅ 3
−1 ⋅ 0 + 4 ⋅ 3
−21
12
2 ⋅ −2 + −7 ⋅ −1
=
−1 ⋅ −2 + 4 ⋅ −1
3
=𝐵
−2
Guided Practice
Solve the matrix equation
−4
0
1
8
𝑋=
6
24
9
6
Guided Practice
Solve the matrix equation
−4
0
1
8
𝑋=
6
24
The inverse is
1
−1 6
6 −1
=
−4 ⋅ 6 + 1 ⋅ 0 0 −4
24 0
9
6
1
−
−1
4
=
−4
0
1
24
1
6
Guided Practice
Multiply both sides of the equation by the inverse (on the left)
1 1
1 1
−
−
−4
1
4 24
4 24 8 9
𝑋=
1
1 24 6
0 6
0
0
6
6
Since the multiplication on the left is I, we have
1 1
−
4 24 8 9
𝑋=
1 24 6
0
6
Guided Practice
You may use you calculator to perform the final multiplication
1
−
4
𝑋=
0
1
24 8
1 24
6
−2
𝑋=
4
−2
1
9
6
Inverse of a 3 × 3 Matrix
• The inverse of a 3 × 3 matrix can be calculated by hand, but the
calculation is lengthy, so you will use your calculator
• To do this, enter the element values, then raise the matrix to the −1
power
• Use the Math button to display the result with fraction values
Guided Practice
Use your calculator
to find the inverse of the matrix. Check your result by
−1
multiplying 𝐴 ⋅ 𝐴 .
1.
2 −2 0
2
0 −2
12 −4 −6
2.
−3
1
5
3.
2
5
4
4 5
5 0
2 2
1 −2
3 0
3 8
Guided Practice
Use your calculator to find the inverse of the matrix. Check your result
by multiplying 𝐴 ⋅ 𝐴−1 .
1.
2
2
12
−2 0
2
0 −2 ; 2
−4 −6 12
−2 0
0 −2
−4 −6
3
−
2
3
−
2
3
−
2
−1
−2
−1
−1
=
1
2
1
2
1
2
Guided Practice
Use your calculator to find the inverse of the matrix. Check your result
by multiplying 𝐴 ⋅ 𝐴−1 .
2.
−3
1
5
4 5 −3
5 0; 1
2 2
5
4 5
5 0
2 2
−1
=
10
−
153
2
153
23
153
2
−
153
31
153
26
−
153
25
153
5
−
153
19
153
Guided Practice
Use your calculator to find the inverse of the matrix. Check your result
by multiplying 𝐴 ⋅ 𝐴−1 .
3.
2
5
4
1 −2 2
3 0 ; 5
3 8
4
1 −2
3 0
3 8
−1
12
= −20
3
2
−7
12
−1
3
−5
1
2
Solve Linear Systems With a Matrix Equation
• Multiply the matrices on the left
2
1
−3 𝑥
19
⋅ 𝑦 =
−7
4
Solve Linear Systems With a Matrix Equation
• Multiply the matrices on the left
2 −3 𝑥
19
⋅ 𝑦 =
−7
1 4
The result is a system of equations in two variables:
2𝑥 − 3𝑦 = 19
𝑥 + 4𝑦 = −7
Solve Linear Systems With a Matrix Equation
• We can solve systems of equations by converting the system into a
matrix equation
• We can then solve the system in one step by multiplying both sides of
the equation by the inverse of the square matrix
• Example: Use an inverse matrix to solve the linear system.
2𝑥 − 3𝑦 = 19
𝑥 + 4𝑦 = −7
Solve Linear Systems With a Matrix Equation
2𝑥 − 3𝑦 = 19
𝑥 + 4𝑦 = −7
Use the coefficients of the equations to form a matrix
2 −3
1 4
𝑥
Multiply this by the 2 × 1 matrix 𝑦
2
1
−3 𝑥
4 𝑦
Solve Linear Systems With a Matrix Equation
𝑥
Multiply this by the 2 × 1 matrix 𝑦
2 −3 𝑥
1 4 𝑦
Set this equal to the 2 × 1 matrix made using the constants
2
1
−3 𝑥
19
=
−7
4 𝑦
Solve Linear Systems With a Matrix Equation
2 −3 𝑥
19
=
−7
1 4 𝑦
Solve the equation by multiplying both sides by the inverse of the
square matrix. You may use your calculator to find the inverse.
4
11
1
−
11
3
11 2
2 1
11
4
3
−3 𝑥
19
11
11
=
𝑦
1
2 −7
4
−
11 11
Solve Linear Systems With a Matrix Equation
4
11
1
−
11
3
11 2
2 1
11
4
3
−3 𝑥
11 11 19
=
1
2 −7
4 𝑦
−
11 11
4
𝑥
11
=
𝑦
1
−
11
3
11 19 = 5
2 −7
−3
11
Guided Practice
Use an inverse matrix to solve the linear system.
4𝑥 + 𝑦 = 10
1.
3𝑥 + 5𝑦 = −1
2.
2𝑥 − 𝑦 = −6
6𝑥 − 3𝑦 = −18
3.
3𝑥 − 𝑦 = −5
−4𝑥 + 2𝑦 = 8
Guided Practice
Use an inverse matrix to solve the linear system.
4𝑥 + 𝑦 = 10
1.
; 𝑥 = 3, 𝑦 = −2
3𝑥 + 5𝑦 = −1
2.
2𝑥 − 𝑦 = −6
; infinite solutions
6𝑥 − 3𝑦 = −18
3.
3𝑥 − 𝑦 = −5
; 𝑥 = −1, 𝑦 = 2
−4𝑥 + 2𝑦 = 8
Guided Practice
Use an inverse matrix to solve the linear system.
3𝑥 − 7𝑦 + 5𝑧 = −32
2𝑥 + 3𝑦 + 4𝑧 = −8
1.
−3𝑥 + 4𝑦 + 2𝑧 = 5
2.
2𝑥 + 3𝑦 = −2
5𝑥 − 3𝑦 + 2𝑧 = 10
5𝑦 + 3𝑧 = −19
Exercise 3.8
• Handout