Lesson Explainer: Properties of Inverse Matrices | Nagwa Lesson Explainer: Properties of Inverse Matrices | Nagwa

Lesson Explainer: Properties of Inverse Matrices Mathematics • Third Year of Secondary School

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In this explainer, we will learn how to use some properties of matrix inverse.

An 𝑛×𝑛 matrix 𝐴 is said to be invertible if there exists an 𝑛×𝑛 matrix 𝐡 such that the product of 𝐴 and 𝐡 is 𝐼, where 𝐼 is the 𝑛×𝑛 identity matrix: Ifthen𝐴𝐡=𝐼,𝐡=𝐴.

If 𝐡 exists, we say that it is the inverse of 𝐴, denoted 𝐴.

Note that, as implied in this definition, a matrix must be square to be invertible, but being square does not guarantee that the inverse exists.

To find the inverse of a 2Γ—2 matrix 𝐴 such that 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ, we apply the formula 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡,det where det𝐴=π‘Žπ‘‘βˆ’π‘π‘. Notice that if the determinant of matrix 𝐴 is equal to zero, the inverse cannot exist. If the determinant is not zero, matrix 𝐴 will have an inverse. We then call matrix 𝐴 invertible or nonsingular. The properties of inverse matrices we will consider in this lesson will apply to all invertible matrices.

Let’s use the definition for an inverse matrix to derive some of the key properties of inverse matrices.

Example 1: Identifying an Equivalent Expression for Matrices Using the Properties of Inverse Matrices

If 𝐴 is a matrix, which of the following is equal to ο€Ήπ΄ο…οŠ±οŠ§οŠ¨?

  1. ο€Ήπ΄ο…οŠ±οŠ§οŽ οŽ‘
  2. 𝐴
  3. 𝐴
  4. ο€Ήπ΄ο…οŠ¨οŠ±οŠ§

Answer

Since 𝐴 exists, 𝐴 must be a square matrix. Let’s imagine 𝐴 is a 2Γ—2 matrix such that 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ.

Based on the definition of the inverse for a 2Γ—2 matrix, 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.det

If we square the inverse of 𝐴, we have 𝐴=1(𝐴)ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡=1(𝐴)𝑑+π‘π‘βˆ’π‘π‘‘βˆ’π‘Žπ‘βˆ’π‘π‘‘βˆ’π‘Žπ‘π‘π‘+π‘Žο‰.detdet

Next, we want to calculate 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆο€Όπ‘Žπ‘π‘π‘‘οˆ=ο€½π‘Ž+π‘π‘π‘Žπ‘+π‘π‘‘π‘Žπ‘+𝑐𝑑𝑏𝑐+𝑑.

Taking the inverse of the square of matrix 𝐴, we have 𝐴=1(𝐴)𝑑+π‘π‘βˆ’π‘π‘‘βˆ’π‘Žπ‘βˆ’π‘π‘‘βˆ’π‘Žπ‘π‘π‘+π‘Žο‰.det

Note that the property of determinants that detdetdet(𝐴𝐡)=(𝐴)(𝐡) allows us to calculate the determinant of 𝐴 to be (𝐴)det.

Since the expression for the inverse of the square of matrix 𝐴 is the same as the expression for the square of the inverse of matrix 𝐴, we have shown that, for an invertible 2Γ—2 matrix 𝐴, 𝐴=𝐴.

In the previous example, we demonstrated that 𝐴=ο€Ήπ΄ο…οŠ±οŠ§οŠ¨οŠ¨οŠ±οŠ§. This can be generalized for higher powers of the inverse matrix such that, for any invertible matrix 𝐴 where π‘›βˆˆπ‘, 𝐴=(𝐴).

In our next example, we will consider the relationship between the transpose of an inverse and the inverse of a transpose.

Example 2: Identifying an Equivalent Expression for Matrices Using the Properties of Inverse Matrices

If 𝐴 is a matrix, which of the following is equal to ο€Ήπ΄ο…οŠ±οŠ§οŒ³?

  1. 𝐴
  2. 𝐴
  3. 𝐴
  4. ο€Ήπ΄ο…οŠ±οŠ§οŽ ο¬
  5. ο€Ήπ΄ο…οŒ³οŠ±οŠ§

Answer

Recall that the capital 𝑇 written in superscript text is the notation for the transpose of a matrix. This means that we switch the rows with the columns. When we transpose a matrix, the values along the diagonal do not change.

Since 𝐴 exists, 𝐴 must be a square matrix. Let’s imagine 𝐴 is a 2Γ—2 matrix such that 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ.

Based on the definition of the inverse for a 2Γ—2 matrix, 𝐴=1π‘Žπ‘‘βˆ’π‘π‘ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.

If we take the transpose of the inverse of matrix 𝐴, we have 𝐴=1π‘Žπ‘‘βˆ’π‘π‘ο€Όπ‘‘βˆ’π‘βˆ’π‘π‘Žοˆ.

Note that since the fraction can be distributed across the matrix, taking the transpose does not affect it and so we can leave it on the outside of the matrix.

Next, we want to calculate the transpose of matrix 𝐴, which gives 𝐴=ο€»π‘Žπ‘π‘π‘‘ο‡.

Taking the inverse of the transpose of matrix 𝐴 would then yield 𝐴=1π‘Žπ‘‘βˆ’π‘π‘ο€Όπ‘‘βˆ’π‘βˆ’π‘π‘Žοˆ.

We have just demonstrated that, for a 2Γ—2 invertible matrix, 𝐴=𝐴.

In the previous example, we demonstrated that 𝐴=ο€Ήπ΄ο…οŒ³οŠ±οŠ§οŠ±οŠ§οŒ³ for a 2Γ—2 matrix. This can be generalized for all invertible matrices 𝐴: 𝐴=𝐴.

In the next example, we will explore what happens when we find the inverse of the inverse of a matrix.

Example 3: Using Properties of Inverse Matrices to Solve a Problem

Consider the matrix 𝐴=ο€Όβˆ’31βˆ’25. Find ο€Ήπ΄ο…οŠ±οŠ§οŠ±οŠ§.

Answer

To find the inverse of the inverse of matrix 𝐴, we first want to find the inverse of matrix 𝐴. For a matrix of the form 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ, the inverse will be equal to 𝐴=1π‘Žπ‘‘βˆ’π‘π‘ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.

Therefore, in this example, the inverse will be 𝐴=1(βˆ’3)(5)βˆ’(1)(βˆ’2)ο€Ό5βˆ’12βˆ’3=βˆ’113ο€Ό5βˆ’12βˆ’3=βŽ›βŽœβŽœβŽβˆ’513113βˆ’213313⎞⎟⎟⎠.

If we then found the inverse of 𝐴, we would take 1ο€»ο‡ο€»ο‡βˆ’ο€»ο‡ο€»ο‡βŽ›βŽœβŽœβŽ313βˆ’113213βˆ’513⎞⎟⎟⎠, which simplifies to βˆ’13βŽ›βŽœβŽœβŽ313βˆ’113213βˆ’513⎞⎟⎟⎠.

Then, we multiply by the scalar βˆ’13: ο€Όβˆ’31βˆ’25.

We have just shown that the inverse of the inverse of a matrix is the original matrix. However, it is not necessary to do all of these calculations to solve this problem because this property is true for all invertible matrices: 𝐴=𝐴.

Once we show that matrix 𝐴 has a nonzero determinant, the properties of inverse matrices tell us that the inverse of the inverse will be the original matrix, ο€Όβˆ’31βˆ’25.

In the previous example, we demonstrated that 𝐴=𝐴 for a 2Γ—2 matrix. This can be generalized for all invertible matrices: 𝐴=𝐴.

In the next example, we will see how we can use the properties of inverse matrices and the identity matrix to simplify problem solving.

Example 4: Using Properties of Inverse Matrices to Solve a Problem

  1. Given the matrices 𝐴 and 𝐡, where 𝐴=1βˆ’230βˆ’14001 and 𝐡=1βˆ’250βˆ’14001, find 𝐴𝐡.
  2. Without doing further calculations, find 𝐴.

Answer

Part 1

The first part of this question has asked us to multiply matrix 𝐴 by matrix 𝐡. As both of these matrices are square 3Γ—3 matrices, they can be multiplied together, remembering that matrix multiplication is not commutative. To multiply matrices, we find the dot product of the rows in the first matrix and the columns in the second: 𝐴𝐡=(1)(1)+(βˆ’2)(0)+(3)(0)(1)(βˆ’2)+(βˆ’2)(βˆ’1)+(3)(0)(1)(5)+(βˆ’2)(4)+(3)(1)(0)(1)+(βˆ’1)(0)+(4)(0)(0)(βˆ’2)+(βˆ’1)(βˆ’1)+(4)(0)(0)(5)+(βˆ’1)(4)+(4)(1)(0)(1)+(0)(0)+(1)(0)(0)(βˆ’2)+(0)(βˆ’1)+(1)(0)(0)(5)+(0)(4)+(1)(1).

When we evaluate each of these expressions, we find 𝐴𝐡=100010001.

Part 2

The second part of the question asked us to find the of inverse of matrix 𝐴 without doing any further calculations. When we multiplied the matrices 𝐴 and 𝐡, the resultant matrix was the identity matrix. We recall that any square matrix (with a nonzero determinant) has an inverse such that 𝐴𝐴=𝐼. Since the product of matrix 𝐴 and matrix 𝐡 was the identity matrix, matrix 𝐡 must be the inverse of matrix 𝐴.

Therefore, 𝐴=1βˆ’250βˆ’14001.

There is one further property that we can use: the relationship between the product of two matrices and their inverse.

For a pair of invertible matrices 𝐴 and 𝐡, (𝐴𝐡)=𝐡𝐴.

In our final example, we will look at how to apply this connection between the inverse of the product of two matrices and the product of their respective inverses.

Example 5: Using Properties of Inverse Matrices to Solve a Problem

Given that (𝐴𝐡)=16ο€Ό5βˆ’3βˆ’3321,𝐴=ο€Όβˆ’2βˆ’1βˆ’3βˆ’2, determine 𝐡.

Answer

We recall, given a pair of invertible matrices 𝐴 and 𝐡, (𝐴𝐡)=𝐡𝐴. This means that we can rewrite our first equation as 𝐡𝐴=16ο€Ό5βˆ’3βˆ’3321.

We now need some strategy to β€œcancel out” the 𝐴 from both sides of this equation. To do that, we use a key property of the identity matrix. We know that 𝐴𝐴=𝐼=𝐴𝐴. This means that we can multiply both sides of this equation by matrix 𝐴 that we were given, remembering that matrix multiplication is not commutative; we must multiply in the correct order: 𝐡𝐴𝐴=16ο€Ό5βˆ’3βˆ’3321οˆβ‹…ο€Όβˆ’2βˆ’1βˆ’3βˆ’2.

From there, we follow the rules for multiplying 2Γ—2 matrices and find that 𝐡𝐴𝐴=16ο€Ύ(5)(βˆ’2)+(βˆ’3)(βˆ’3)(5)(βˆ’1)+(βˆ’3)(βˆ’2)(βˆ’33)(βˆ’2)+(21)(βˆ’3)(βˆ’33)(βˆ’1)+(21)(βˆ’2)=16ο€Όβˆ’113βˆ’9.

On the left side of this equation, we see that we have multiplied the inverse of matrix 𝐴 by matrix 𝐴. We know that this equals the identity matrix. So, we can rewrite the left side of the equation to be 𝐡𝐼=16ο€Όβˆ’113βˆ’9.

Remembering that multiplying any matrix by the identity matrix produces itself, we can say 𝐡=16ο€Όβˆ’113βˆ’9.

The final step will be to multiply through by the scalar 16 to produce 𝐡=βŽ›βŽœβŽœβŽβˆ’161612βˆ’32⎞⎟⎟⎠.

Let’s finish by recapping the key properties we use when working with inverse matrices.

Key Points

  • The product of a matrix and its inverse is the identity matrix: 𝐴𝐴=𝐼.
  • The inverse of the inverse of a matrix is the matrix itself: 𝐴=𝐴.
  • The inverse of a matrix to the 𝑛th power is equal to the 𝑛th power of the inverse of the matrix: 𝐴=(𝐴).
  • The inverse of the product of matrix 𝐴 and matrix 𝐡 is equal to the product of the inverse of matrix 𝐡 and the inverse of matrix 𝐴: (𝐴𝐡)=𝐡𝐴.
  • The transpose of the inverse of a matrix is equal to the inverse of the transpose of a matrix: 𝐴=𝐴.

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