Inverses of Matrices

The Inverse of a Matrix

The inverse of matrix [latex][A][/latex] is [latex][A]^{-1}[/latex], and is defined by the property: [latex][A][A]^{-1}=[A]^{-1}[A]=[I][/latex].

Learning Objectives

Practice finding the inverse of a matrix and describe its properties

Key Takeaways

Key Points

  • The definition of a matrix inverse requires commutativity—the multiplication must work the same in either order.
  • To be invertible, a matrix must be square, because the identity matrix must be square as well.
  • To determine the inverse of the matrix [latex]\begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix}[/latex], set [latex]\begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}[/latex]. Then solve for [latex]a[/latex], [latex]b[/latex], [latex]c[/latex], and [latex]d[/latex].

Key Terms

  • singular matrix: A matrix that has no inverse.
  • matrix: A rectangular arrangement of numbers or terms having various uses such as transforming coordinates in geometry, solving systems of linear equations in linear algebra and representing graphs in graph theory.
  • inverse matrix: A square matrix [latex][A][/latex] with an associated matrix [latex][B][/latex] such that [latex][A][/latex] multiplied by [latex][B][/latex] and [latex][B][/latex] multiplied by [latex][A][/latex] both equal the identity matrix.
  • identity matrix: A diagonal matrix all of the diagonal elements of which are equal to [latex]1[/latex], the rest being equal to [latex]0[/latex].

Having seen that the number [latex]1[/latex] plays a special role in multiplication, [latex]1x=x[/latex], the inverse of a number is defined as a number that multiplies by that number to give [latex]1[/latex]. The matrix [latex]B[/latex] is the inverse of the matrix [latex]A[/latex] if when multiplied together, [latex]A\cdot B[/latex] or [latex]B\cdot A[/latex] gives the identity matrix.

Definition of Inverse Matrix

The inverse of matrix [latex][A][/latex], designated as [latex][A]^{-1}[/latex], is defined by the property:

[latex]\displaystyle [A][A]^{-1}=[A]^{-1}[A]=[I][/latex]

Where [latex][I][/latex] is the identity matrix.

Note that, just as in the definition of the identity matrix, this definition requires commutativity—the multiplication must work the same in either order.

Note also that only square matrices can have an inverse. The definition of an inverse matrix is based on the identity matrix [latex][I][/latex], and it has already been established that only square matrices have an associated identity matrix.

The method for finding an inverse matrix comes directly from the definition, along with a little algebra.

Finding an Inverse Matrix

Example: Find the inverse of: [latex]\begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix}[/latex]

First, let the following be true:

[latex]\displaystyle \begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}[/latex]

This is the key step. It establishes [latex]\begin{pmatrix} a & b \\ c & d \end{pmatrix}[/latex] as the inverse that is being looked for, by asserting that it fills the definition of an inverse matrix. When multiplying this mystery matrix by our original matrix, the result is [latex][I][/latex]. When solving for the four variables [latex]a[/latex], [latex]b[/latex], [latex]c[/latex], and [latex]d[/latex], then the inverse of the matrix will be found.

Next, do the multiplication:

[latex]\displaystyle \begin{pmatrix} 3a+4c & 3b+4d \\ 5a+6c & 5b+6d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}[/latex]

For two matrices to be equal, every element in the left must equal its corresponding element on the right. So, for these two matrices to equal each other:

[latex]\displaystyle 3a+4c=1[/latex]

[latex]\displaystyle 3b+4d=0[/latex]

[latex]\displaystyle 5a+6c=0[/latex]

[latex]\displaystyle 5b+6d=1[/latex]

Solve the first two equations for [latex]a[/latex] and [latex]c[/latex] and the second two equations for [latex]b[/latex] and [latex]d[/latex] by using either elimination or substitution.

The results are:

[latex]\displaystyle a=-3[/latex]

[latex]\displaystyle b=2[/latex]

[latex]\displaystyle c=2 \frac12[/latex]

[latex]\displaystyle d= -1 \frac 12[/latex]

Having solved for the four variables, the result is the inverse [latex]\begin{pmatrix} -3 & 2 \\ 2 \frac 12 & -1 \frac 12 \end{pmatrix}[/latex].

If an inverse has been found, then a quick check to be sure it is correct is to multiply it by the original matrix and see if the identify matrix results:

[latex]\displaystyle \begin{pmatrix} -3 & 2 \\ 2 \frac 12 & -1 \frac 12 \end{pmatrix} \begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix} [/latex]

After multiplying, the result is:

[latex]\displaystyle \begin{pmatrix} (-3)(3)+(2)(5) & (-3)(4)+(2)(6) \\ (2 \frac 12)(3)_+(-1 \frac 12)(5) & (2 \frac 12)(4)+(-1 \frac 12)(6) \end{pmatrix} [/latex]

Simplifying the problem gives:

[latex]\displaystyle \begin{pmatrix} -9+10 & -12+12 \\ 7 \frac 12 -7 \frac 12 & 10-9 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}[/latex]

Therefore, the inverse works. But wait!

Note that, to fully test it, one has to try the multiplication in both orders, because, in general, changing the order of a matrix multiplication changes the answer. The definition of an inverse matrix specifies that it must work both ways.

[latex]\displaystyle \begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix} \begin{pmatrix} -3 & 2 \\ 2 \frac 12 & -1 \frac 12 \end{pmatrix} [/latex]

After multiplying, one gets:

[latex]\displaystyle \begin{pmatrix} (3)(-3)+(4)(2\frac 12) & (3)(2)+(4)(-1 \frac 12) \\ (5)(-3)_+(6)(2\frac 12) & (5)(2)+(6)(-1\frac 12) \end{pmatrix}[/latex]

Simplifying the problem gives:

[latex]\displaystyle \begin{pmatrix} -9+10 & 6-6 \\ -15+15 & 10-9 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}[/latex]

So the inverse matrix works both ways.

In some cases, the inverse of a square matrix does not exist. This is called a singular matrix.

Solving Systems of Equations Using Matrix Inverses

A system of equations can be readily solved using the concept of the inverse matrix and matrix multiplication.

Learning Objectives

Practice using inverse matrices to solve a system of linear equations

Key Takeaways

Key Points

  • Using matrices to solve systems of equations can drastically reduce the workload on you. For example, consider the following three equations: [latex]x+2y-z=11[/latex], [latex]2x-y+3z=7[/latex] and [latex]7x-3y-2z = 2[/latex].
  • To solve these equations using matrices, we first define a [latex]3 \times 3 [/latex] matrix [latex][A][/latex], which  are the coefficients of all the variables on the left side of the equal signs:[latex][A] = \begin{bmatrix} 1&2&-1\\2&-1&3\\7&-3&-2 \end{bmatrix}[/latex] Then define a [latex]3 \times 1[/latex] matrix [latex][B][/latex], which are the numbers on the right side of the equal signs: [latex][B] = \begin{bmatrix} 11\\7\\2 \end{bmatrix}.[/latex]
  • In order to determine the values of the variables, we simply multiply the inverse of [latex][A][/latex] times [latex][B][/latex]. This is most readily done using a calculator. The calculator responds with a [latex]3 \times 1[/latex] matrix, which includes the solution to the system as: [latex]x=3[/latex], [latex]y=5[/latex], and [latex]z=2[/latex].

Key Terms

  • inverse matrix: For a matrix [latex][A][/latex], if a matrix [latex][B][/latex] exist such that [latex][A][/latex] multiplied by [latex][B][/latex] and [latex][B][/latex] multiplied by [latex][A][/latex] both equal the identity matrix, then [latex][B][/latex] is the inverse of [latex][A][/latex].
  • linear equation: A polynomial equation of the first degree (such as [latex]x=2y-7[/latex]).

A system of equations can be readily solved using the concepts of the inverse matrix and matrix multiplication.  We have seen, in the chapter on simultaneous equations, how to solve two equations with two unknowns. But suppose we have three equations with three unknowns? Or four, or five?  Such situations are more common than you might suppose in the real world.  And even if you are allowed to use a calculator, it is not at all obvious how to solve such a problem in a reasonable amount of time.

Example 1:  Solve the following system of linear equations:

[latex]\displaystyle x+2y-z = 11 \\ 2x-y+3z = 7 \\ 7x-3y-2z = 2 [/latex]

Step 1:  Set up a matrix [latex]A[/latex], which is comprised of the coefficients of all the variables from each equation from the left hand side of the equal sign:

[latex]\displaystyle [A] = \begin{bmatrix} 1&2&-1\\2&-1&3\\7&-3&-2 \end{bmatrix}[/latex]

Step 2:  Step up a matrix [latex]B[/latex], which consists of the constants on the right hand side of the equal sign:

[latex]\displaystyle [B] = \begin{bmatrix} 11\\7\\2 \end{bmatrix}[/latex]

Now, in order to determine the values of [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex], we simply multiply the inverse of [latex][A][/latex] times [latex][B][/latex]. This can be done by hand, finding the inverse matrix of [latex][A][/latex], then performing the appropriate matrix multiplication with [latex][B][/latex].

However, if you have a graphing calculator, the situation is much easier.  Using the matrix function on the calculator, first enter both matrices.  Then calculate [latex][A^{-1}][B][/latex], that is, the inverse of matrix [latex][A][/latex], multiplied by matrix [latex][B][/latex].

The calculator responds with a [latex]3 \times 1[/latex] matrix, which includes the solution to the system as: [latex]x=3[/latex], [latex]y=5[/latex], and [latex]z=2[/latex]. Solving linear equations in this way is fast and easy.