Lesson Explainer: Inverse of a 2 Γ— 2 Matrix Mathematics

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In this explainer, we will learn how to check whether a 2Γ—2 matrix has an inverse and then find its inverse, if possible.

When working with real number systems, we know that multiplying any nonzero number by its reciprocal results in 1. 1 is a special real number because multiplying any number by 1 does not change the number. In this sense, 1 is called the multiplicative identity. The reciprocal of a nonzero number is called the multiplicative inverse.

We can see the usefulness of the multiplicative inverse when we solve a linear equation to find an unknown constant. For instance, consider the problem of finding the value of π‘₯ that satisfies the equation 3π‘₯=6.

We can solve this equation simply by dividing both sides of the equation by 3. Dividing by 3 is the same as multiplying by 13, which is the reciprocal of 3. What is really happening here is that we are multiplying both sides of the equation by the multiplicative inverse of 3, which is the coefficient of the unknown constant π‘₯. This leads to ο€Ό13Γ—3οˆΓ—π‘₯=13Γ—6.

Multiplying 3 by its reciprocal results in 1, which is the multiplicative identity. So, the left-hand side of the equation becomes 1Γ—π‘₯, which is the same as π‘₯. This leads to π‘₯=2, which gives the value of the unknown constant.

Now, let us think of an analogous type of problem using matrix operations. Say that we know two matrices 𝐴 and 𝐡 and we want to find an unknown matrix 𝑋 that satisfies the matrix equation 𝐴𝑋=𝐡.

Thinking back on the solution process in context of real number operations, we would like to β€œdivide” both sides by 𝐴. But we do not know a notion of division in matrix operations. Instead of division, we can think of multiplying both sides of the equation by the β€œreciprocal” of 𝐴, only if such an object as the reciprocal, or a multiplicative inverse, of a matrix can be defined.

The aim of this explainer is precisely to understand when it is possible to find a multiplicative inverse of a 2Γ—2 matrix and to learn how to compute the inverse when possible. In order to discuss the multiplicative inverse of a matrix, we need to first understand the multiplicative identity.

We recall that the identity matrix, denoted 𝐼 or 𝐼, is a diagonal matrix whose diagonal entries are equal to 1. We know that multiplying any matrix by an identity matrix of compatible order does not change the matrix. Hence, the identity matrix is the multiplicative identity for matrix multiplication. This means that we can define the (multiplicative) inverse of a matrix so that multiplying a matrix by its inverse results in the identity matrix. In particular, we will need the 2Γ—2 identity matrix 𝐼=ο€Ό1001 in order to define the inverse of a 2Γ—2 matrix.

Definition: Inverse of a Matrix

Let 𝐴 be a 2Γ—2 matrix. The inverse of 𝐴, denoted 𝐴, is a matrix that satisfies 𝐴𝐴=𝐼,𝐴𝐴=𝐼.

If such a matrix exists, we say that matrix 𝐴 is invertible.

If such a matrix does not exist, we say that matrix 𝐴 is not invertible.

We recall that the multiplication of an π‘šΓ—π‘› matrix by an π‘›Γ—π‘˜ matrix results in a π‘šΓ—π‘˜ matrix. Since both 𝐴 and 𝐼 are 2Γ—2 matrices, this means that the inverse 𝐴, if it exists, must also be a 2Γ—2 matrix. In fact, this definition of the matrix inverse holds for larger matrices, as long as they are square matrices, by replacing 2 by 𝑛 in the statement. We will focus on the case 𝑛=2 in this explainer.

We can see that the inverse, or the multiplicative inverse, of a matrix is defined in a way such that multiplying any matrix by its inverse results in the identity matrix. In this way, the inverse of a matrix is analogous to the reciprocal of a nonzero real number.

We know that matrix multiplication is not commutative, which means that the order of multiplication matters. Thus, we need both 𝐴𝐴 and 𝐴𝐴 to be equal to the identity.

This definition tells us that the inverse of a 2Γ—2 matrix is another 2Γ—2 matrix such that multiplying the two matrices in either order results in the identity matrix 𝐼. If such a matrix 𝐴 exists, we can immediately see that the original matrix 𝐴 satisfies the definition of the inverse of 𝐴. This leads to the following property.

Property: Inverse of the Matrix Inverse

Let 𝐴 be an invertible matrix. Then, 𝐴 is invertible and 𝐴=𝐴.

Before discussing how to find an inverse of a matrix, we need to understand when a matrix is not invertible. In the context of real number operations, the only noninvertible real number is 0. The analog of the real number 0 is the zero matrix, denoted 𝑂, whose entries are all equal to 0. While it is true that the zero matrix is not invertible, there are larger families of matrices that are not invertible.

Property: Singular Matrices

A square matrix is not invertible if and only if its determinant is equal to zero. Matrices with this property are known as singular matrices.

To understand this statement, we recall the property of determinants of a matrix. Given square matrices 𝐡 and 𝐢 of the same order, detdetdet(𝐡𝐢)=(𝐡)Γ—(𝐢).

We can use this property on the first equation in the definition of the matrix inverse: If 𝐴 is an invertible matrix, we have 𝐴𝐴=𝐼.

Taking the determinant of both sides of this equation, detdet𝐴𝐴=(𝐼).

We know that the determinant of an identity matrix is equal to 1. Using the property of the determinant, this equation can be written as detdet(𝐴)×𝐴=1.

Since det(𝐴) is one of the factors on the left-hand side of the equation, this equation cannot be satisfied if det(𝐴)=0. In other words, the condition 𝐴𝐴=𝐼 cannot be true for any matrix 𝐴 if det(𝐴)=0. This proves that 𝐴 is not invertible if det(𝐴)=0.

This argument only shows that a matrix with zero determinant is not invertible. We will not prove the converse statement here, which states that a matrix with nonzero determinant is invertible. Instead, we will develop a formula for finding the inverse of nonsingular matrices, which will serve as proof for the existence of an inverse.

Before going further into this discussion, let us consider a few examples where we will determine whether or not a given matrix is invertible.

Example 1: Determinants and Invertibility

Is the matrix ο€Ό31βˆ’31 invertible?

Answer

In this example, we need to determine whether a given matrix is invertible. We recall that a square matrix is invertible if and only if its determinant is not equal to zero. We can see that the order of the given matrix is 2Γ—2, which means that it is a square matrix. So, we need to check its determinant to see whether it is equal to zero.

Recall that the determinant of a 2Γ—2 matrix is given by detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘ŽΓ—π‘‘βˆ’π‘Γ—π‘.

This leads to detο€Ό31βˆ’31=3Γ—1βˆ’1Γ—(βˆ’3)=6.

We can see that the determinant of the given square matrix is nonzero. This tells us that this matrix is invertible.

Hence, the correct answer to the question in this example is yes.

Let us consider another example where we determine whether a given matrix is invertible.

Example 2: Determinants and Invertibility

Is the matrix ο€Ό31βˆ’3βˆ’1 invertible?

Answer

In this example, we need to determine whether a given matrix is invertible. We recall that a square matrix is invertible if and only if its determinant is not equal to zero. We can see that the order of the given matrix is 2Γ—2, which means that it is a square matrix. So, we need to check its determinant to see whether it is equal to zero.

Recall that the determinant of a 2Γ—2 matrix is given by detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘ŽΓ—π‘‘βˆ’π‘Γ—π‘.

This leads to detο€Ό31βˆ’3βˆ’1=3Γ—(βˆ’1)βˆ’1Γ—(βˆ’3)=0.

We can see that the determinant of the given square matrix is equal to zero. This tells us that this matrix is not invertible.

Hence, the correct answer to the question in this example is no.

In the next example, we will find a condition for an unknown constant in a matrix, given that a matrix is invertible.

Example 3: Determinants and Invertibility

Given that the matrix ο€Ό71βˆ’7π‘Žοˆ is invertible, what must be true of π‘Ž.

Answer

In this example, we are given that the matrix is invertible. We recall that a square matrix is invertible if and only if its determinant is not equal to zero. We can see that the order of the given matrix is 2Γ—2, which means that it is a square matrix. So, we need to make sure that its determinant is not equal to zero to ensure that it is invertible.

Recall that the determinant of a 2Γ—2 matrix is given by detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘ŽΓ—π‘‘βˆ’π‘Γ—π‘.

This leads to detο€Ό71βˆ’7π‘Žοˆ=7Γ—π‘Žβˆ’1Γ—(βˆ’7)=7π‘Ž+7.

Hence, the determinant of the given matrix is 7π‘Ž+7. Since the matrix is invertible, this value must not equal zero. This leads to 7π‘Ž+7β‰ 07π‘Žβ‰ βˆ’7π‘Žβ‰ βˆ’1.

Therefore, it must be true that π‘Žβ‰ βˆ’1.

In the next example, we will find all possible values of an unknown in a matrix when we are given that the matrix is singular.

Example 4: Finding the Unknown Elements of a Singular Matrix

Find the set of real values of (π‘₯) that make the matrix ο€Όπ‘₯βˆ’382π‘₯+3 singular.

Answer

In this example, we are given that the matrix is singular. We recall that a singular matrix is a square matrix with zero determinant. We can see that the order of the given matrix is 2Γ—2, which means that it is a square matrix. So, we need to make sure that its determinant is equal to zero.

Recall that the determinant of a 2Γ—2 matrix is given by detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘ŽΓ—π‘‘βˆ’π‘Γ—π‘.

This leads to detο€Όπ‘₯βˆ’382π‘₯+3=(π‘₯βˆ’3)Γ—(π‘₯+3)βˆ’8Γ—2.

We can expand through the parentheses in this expression by using the difference of squares formula, (π‘Ž+𝑏)(π‘Žβˆ’π‘)=π‘Žβˆ’π‘οŠ¨οŠ¨. Using this formula, we can write the determinant as π‘₯βˆ’3βˆ’16=π‘₯βˆ’25.

Hence, the determinant of the given matrix is π‘₯βˆ’25. Since this is a singular matrix, its determinant must equal to zero. This leads to π‘₯βˆ’25=0π‘₯=25π‘₯=Β±5.

Hence, we must have either π‘₯=5 or π‘₯=βˆ’5. The set of real values of (π‘₯) for which the given matrix is singular is {βˆ’5,5}.

So far, we have considered a few examples concerning the existence of the matrix inverse. Let us turn our attention to finding the inverse of a 2Γ—2 matrix.

In the next example, we will determine whether two given matrices are inverses of each other.

Example 5: Verifying Whether a Given Matrix Is a Multiplicative Inverse of Another Given Matrix

Are the matrices ο€Ό1234,βŽ›βŽœβŽœβŽ1121314⎞⎟⎟⎠ multiplicative inverses of each other?

Answer

In this example, we need to determine whether two given matrices are inverses of each other. We can see that both of these matrices are 2Γ—2 matrices. Recall that, given a 2Γ—2 matrix 𝐴, its multiplicative inverse, denoted 𝐴, is the matrix that satisfies 𝐴𝐴=𝐴𝐴=𝐼, if such a matrix exists. We also recall that 𝐼 is the 2Γ—2 identity matrix ο€Ό1001. If we can obtain the identity matrix by multiplying the matrices in either order, then we will know that they are multiplicative inverses of each other. Let us compute the matrix multiplication in the order provided. ο€Ό1234οˆβŽ›βŽœβŽœβŽ1121314⎞⎟⎟⎠=βŽ›βŽœβŽœβŽ1Γ—1+2Γ—131Γ—12+2Γ—143Γ—1+4Γ—133Γ—12+4Γ—14⎞⎟⎟⎠=βŽ›βŽœβŽœβŽ53113352⎞⎟⎟⎠.

We can see that the resulting matrix is not the identity matrix. Hence, the two matrices are not multiplicative inverses of each other.

The answer to the question in this example is no.

In the previous example, we checked whether two given matrices are inverses of each other. Using this method, we can always check whether two given matrices are inverses of each other. However, this is not a good way to find the inverse of a matrix since we would need to check this for every possible matrix. We introduce the formula for the inverse of a 2Γ—2 matrix.

Formula: Inverse of a 2 Γ— 2 Matrix

Let 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ such that det𝐴≠0. Then, 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.det

Note that we can obtain the inverse from 𝐴 by

  1. swapping the positions of π‘Ž and 𝑑 around,
  2. switching the signs of 𝑏 and 𝑐,
  3. and dividing by the determinant.

Let us prove the validity of this formula by computing the matrix multiplication 𝐴𝐴. The matrix multiplication 𝐴𝐴 is similar and is omitted. We have 𝐴𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆο€Ό1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡οˆ=1π΄ο€Όο€Όπ‘Žπ‘π‘π‘‘οˆο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡οˆ=1π΄ο€Ύπ‘ŽΓ—π‘‘+𝑏×(βˆ’π‘)π‘ŽΓ—(βˆ’π‘)+π‘Γ—π‘Žπ‘Γ—π‘‘+𝑑×(βˆ’π‘)𝑐×(βˆ’π‘)+π‘‘Γ—π‘ŽοŠ=1π΄ο€Όπ‘Žπ‘‘βˆ’π‘π‘00π‘Žπ‘‘βˆ’π‘π‘οˆ=βŽ›βŽœβŽœβŽπ‘Žπ‘‘βˆ’π‘π‘π΄00π‘Žπ‘‘βˆ’π‘π‘π΄βŽžβŽŸβŽŸβŽ =ο€Ό1001.detdetdetdetdetdet

Indeed, multiplying matrix 𝐴 by matrix 𝐴 as defined above results in the identity matrix 𝐼. Along with the verification of the other equation 𝐴𝐴=𝐼, which is omitted here, we have proved that this matrix inverse formula is valid.

Let us pause here for a moment and ask an important philosophical question. Is this the only possible matrix inverse of 𝐴? In other words, is it possible for a matrix to have an inverse that is not given by this formula? This is a question that relates to the uniqueness of the matrix inverse.

Let us rephrase this question a little. We already know that, given a nonsingular matrix, there is a formula defining an inverse of the matrix. So, our question is essentially asking whether two different matrices can satisfy the equations 𝐴𝐴=𝐴𝐴=𝐼, for a given invertible matrix 𝐴. We will show that this is not possible. Say that matrices 𝐡 and 𝐢 are inverses of 𝐴. Then, we must have 𝐴𝐡=𝐡𝐴=𝐼,𝐴𝐢=𝐢𝐴=𝐼.

In particular, this means that 𝐴𝐡=𝐴𝐢.

We can multiply both sides of the equation from the left by 𝐡 to write 𝐡𝐴𝐡=𝐡𝐴𝐢.

Since 𝐡 is an inverse of 𝐴, we know that 𝐡𝐴=𝐼, which gives us 𝐼𝐡=𝐼𝐢.

𝐼 is the multiplicative identity, so 𝐡=𝐢.

This tells us that 𝐡 and 𝐢 are exactly the same matrix. This tells us that, if we have any two matrices that are inverses of a given matrix, they must be equal. In other words, there is only one inverse of an invertible matrix. Since we have a formula for an inverse of a 2Γ—2 matrix, this must be the unique matrix inverse.

In the next example, we will use this formula to compute the inverse of a 2Γ—2 matrix.

Example 6: Finding the Inverse of a Matrix

Find the multiplicative inverse of the matrix 𝐴=ο€Όβˆ’4βˆ’1035, if possible.

Answer

In this example, we need to find the inverse of a given 2Γ—2 matrix, if possible. Recall that the inverse of a square matrix exists (i.e., the matrix is invertible) if and only if its determinant is not equal to zero. We can see that the order of the given matrix is 2Γ—2, which means that it is a square matrix. So, we need to make sure that its determinant is not equal to zero to ensure that it is invertible.

Recall that the determinant of a 2Γ—2 matrix is given by detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘ŽΓ—π‘‘βˆ’π‘Γ—π‘.

This leads to det𝐴=(βˆ’4)Γ—5βˆ’(βˆ’10)Γ—3=10.

Hence, the determinant of the given matrix is 10. Since the determinant is nonzero, we know that the inverse of a matrix exists. Let us find the inverse.

We recall that the inverse of matrix 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ with det𝐴≠0 is given by 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.det

We can see from 𝐴=ο€Όβˆ’4βˆ’1035 that π‘Ž=βˆ’4,𝑏=βˆ’10,𝑐=3,𝑑=5.

We have also computed det𝐴=10. Substituting these values into the formula for the matrix inverse, we obtain 𝐴=110ο€Ό510βˆ’3βˆ’4.

Computing the scalar multiplication, we get 𝐴=βŽ›βŽœβŽœβŽ5101010βˆ’310βˆ’410⎞⎟⎟⎠=βŽ›βŽœβŽœβŽ121βˆ’310βˆ’25⎞⎟⎟⎠.

Now that we know how to find the inverse of a 2Γ—2 matrix, let us return to the problem that motivated this development. If we know the inverse of a matrix, we can solve a matrix equation 𝐴𝑋=𝐡 to find the unknown matrix 𝑋 where 𝐴 and 𝐡 are known matrices. To solve this equation for the unknown matrix 𝑋, we need to multiply on the left of 𝐴𝑋 by the inverse matrix 𝐴, if it exists. Since this is an equation, we must do the same on the right-hand side of the equation. This leads to 𝐴𝐴𝑋=𝐴𝐡.

We know 𝐴𝐴=𝐼, which is the identity matrix. This leads to 𝑋=𝐴𝐡.

How To: Solving Matrix Equations

Let 𝐴 be an invertible matrix, and let both 𝐴 and 𝐡 be 2Γ—2 matrices. Matrix 𝑋 satisfying the equation 𝐴𝑋=𝐡 is 𝑋=𝐴𝐡.

Similarly, matrix 𝑋 satisfying the equation 𝑋𝐴=𝐡 is 𝑋=𝐡𝐴.

We can see that the solution for the equation 𝐴𝑋=𝐡 is different from the solution of 𝑋𝐴=𝐡 because the matrix multiplication is noncommutative. In either case, we need to multiply both sides of the equation by the inverse of 𝐴, but it matters which side we are multiplying from.

In our final example, we will apply our knowledge of the matrix inverse to solve a matrix equation involving 2Γ—2 matrices.

Example 7: Solving Matrix Equations

Using matrix inverses, solve the following for 𝑋: π‘‹ο€Όβˆ’324βˆ’3=ο€Ό0βˆ’230.

Answer

In this example, we need to solve a matrix equation. Recall that we can solve a matrix equation of the form 𝐴𝑋=𝐡 or 𝑋𝐴=𝐡 by using the inverse of matrix 𝐴, if it exists. In this example, matrix 𝐴 is the matrix on the left-hand side of the equation 𝐴=ο€Όβˆ’324βˆ’3.

Let us first check whether the inverse of this matrix exists. Recall that the inverse of a square matrix exists (i.e. the matrix is invertible) if and only if its determinant is not equal to zero. We can see that the order of matrix 𝐴 is 2Γ—2, which means that it is a square matrix. So, we need to make sure that its determinant is not equal to zero to ensure that it is invertible.

Recall that the determinant of a 2Γ—2 matrix is given by detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘ŽΓ—π‘‘βˆ’π‘Γ—π‘.

This leads to det𝐴=(βˆ’3)Γ—(βˆ’3)βˆ’2Γ—4=1.

Hence, the determinant of the given matrix is 1. Since the determinant is nonzero, we know that the inverse of a matrix exists. Let us find the inverse.

We recall that the inverse of matrix 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ with det𝐴≠0 is given by 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.det

We can see from 𝐴=ο€Όβˆ’324βˆ’3 that π‘Ž=βˆ’3,𝑏=2,𝑐=4,𝑑=βˆ’3.

We also know that det𝐴=1, which means that the scalar factor 1𝐴det in the formula can be disregarded. Substituting these values into the formula of the matrix inverse, we obtain 𝐴=ο€Όβˆ’3βˆ’2βˆ’4βˆ’3.

Now that we have computed the matrix inverse, let us consider how to solve the equation. Denoting the matrix on the right-hand side of the equation as 𝐡, our equation is in the form 𝑋𝐴=𝐡.

We can multiply 𝐴 from the right on both sides of the equation to write 𝑋𝐴𝐴=𝐡𝐴.

We know that 𝐴𝐴=𝐼, which is the identity matrix. Since multiplying by the identity matrix does not change the matrix, we have 𝑋=𝐡𝐴.

Hence, we need to find the matrix multiplication 𝐡𝐴: 𝐡𝐴=ο€Ό0βˆ’230οˆο€Όβˆ’3βˆ’2βˆ’4βˆ’3=ο€Ύ0Γ—(βˆ’3)+(βˆ’2)Γ—(βˆ’4)0Γ—(βˆ’2)+(βˆ’2)Γ—(βˆ’3)3Γ—(βˆ’3)+0Γ—(βˆ’4)3Γ—(βˆ’2)+0Γ—(βˆ’3)=ο€Ό86βˆ’9βˆ’6.

Therefore, the answer is 𝑋=ο€Ό86βˆ’9βˆ’6.

Let us finish by recapping a few important concepts from the explainer.

Key Points

  • Let 𝐴 be a 2Γ—2 matrix. The inverse of 𝐴, denoted 𝐴, is a matrix that satisfies 𝐴𝐴=𝐼,𝐴𝐴=𝐼. If such a matrix exists, we say that matrix 𝐴 is invertible. If such a matrix does not exist, we say that matrix 𝐴 is not invertible.
  • Let 𝐴 be an invertible matrix. Then, 𝐴 is invertible and 𝐴=𝐴.
  • A singular matrix is a square matrix whose determinant is equal to zero. A square matrix is not invertible if and only if it is singular.
  • Let 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ such that det𝐴≠0. Then, 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.det
  • Let 𝐴 be an invertible matrix and both 𝐴 and 𝐡 be 2Γ—2 matrices. Matrix 𝑋 satisfying the equation 𝐴𝑋=𝐡 is 𝑋=𝐴𝐡. Similarly, matrix 𝑋 satisfying the equation 𝑋𝐴=𝐡 is 𝑋=𝐡𝐴.

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