Explainer: Inverse of a 2 Γ— 2 Matrix

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

Recall that a matrix that has the same number of columns and rows is called a square matrix. By the definition of matrix products, any two 𝑛×𝑛 matrices can be multiplied together, and we can make the following definition.

Definition: The Inverse of an 𝑛 Γ— 𝑛 Matrix

The inverse of an 𝑛×𝑛 matrix 𝐴 is a matrix 𝐡 such that 𝐴𝐡=𝐼𝐡𝐴=𝐼,and where 𝐼 is the 𝑛×𝑛 identity matrix.

Notice that:

  1. We call 𝐡 the inverse of 𝐴 because it is unique and we denote it by 𝐴. In particular, if we know that 𝐴𝐡=𝐼, then 𝐡 is just 𝐴.
  2. The definition says what is expected of an inverse, but it does not say that such an inverse exists. For example, consider 𝐴=0000, the 2Γ—2 zero matrix. Then for any 2Γ—2 matrix 𝐡=ο”π‘Žπ‘π‘π‘‘ο , we have 𝐴𝐡=0000ο ο”π‘Žπ‘π‘π‘‘ο =0(π‘Ž)+0(𝑐)0(𝑏)+0(𝑑)0(π‘Ž)+0(𝑐)0(𝑏)+0(𝑑)=0000. Therefore, we can never get 𝐼=1001 as the product with any 𝐡 because the zero matrix has no inverse.
  3. On the other hand, the identity matrix is its own inverse because 𝐼𝐼=𝐼.

One approach to finding the inverse of 𝐴=ο”π‘Žπ‘π‘π‘‘ο  is to try and solve some equations. Suppose that 𝐴=π‘₯π‘¦π‘§π‘€οŸοŠ±οŠ§. Then we must solve for π‘₯,𝑦,𝑧,𝑀 in the system ο”π‘Žπ‘π‘π‘‘ο ο“π‘₯π‘¦π‘§π‘€οŸ=1001.

In other words, ο–π‘Žπ‘₯+π‘π‘§π‘Žπ‘¦+𝑏𝑀𝑐π‘₯+𝑑𝑧𝑐𝑦+𝑑𝑀=1001.

This is a system of 4 equations in the 4 unknowns, not as straightforward as we would like.

A simplification comes from the observation that given 𝐴, if we look at the matrix 𝐴=ο“π‘‘βˆ’π‘βˆ’π‘π‘ŽοŸ,βˆ— then we find products 𝐴𝐴=𝐴𝐴=ο“π‘‘βˆ’π‘βˆ’π‘π‘ŽοŸο”π‘Žπ‘π‘π‘‘ο =ο”π‘Žπ‘‘βˆ’π‘π‘00π‘Žπ‘‘βˆ’π‘π‘ο =(𝐴)1001.βˆ—βˆ—det

This gives a diagonal matrix with the determinant of 𝐴—the number det(𝐴)β€”on the main diagonal.

We conclude two things from this fact:

  1. If det(𝐴)=0, then we have found a matrix whose product with 𝐴 is the zero matrix. It is not hard to see that it is, therefore, impossible for 𝐴 to have an inverse.
  2. If det(𝐴)β‰ 0, then the matrix 1(𝐴)𝐴=βŽ‘βŽ’βŽ’βŽ’βŽ£π‘‘(𝐴)βˆ’π‘(𝐴)βˆ’π‘(𝐴)π‘Ž(𝐴)⎀βŽ₯βŽ₯βŽ₯⎦detdetdetdetdetβˆ— is the inverse of 𝐴.

The matrix π΄βˆ— is called the adjugate of 𝐴=ο”π‘Žπ‘π‘π‘‘ο  and we get it by two steps:

  • Swap the main diagonal entries: π‘Žβ†”π‘‘.
  • Negate the off-diagonal entries: π‘β†’βˆ’π‘, π‘β†’βˆ’π‘.

For example, with 𝐴=12βˆ’34, we have det(𝐴)=(1)(4)βˆ’(2)(βˆ’3)=4βˆ’(βˆ’6)=10 and adjugate 𝐴=4βˆ’231ο βˆ— and, therefore, inverse 𝐴=1104βˆ’231=⎑⎒⎒⎣410βˆ’210310110⎀βŽ₯βŽ₯⎦=⎑⎒⎒⎣25βˆ’15310110⎀βŽ₯βŽ₯⎦.

To summarize,

Inverse Matrices

  1. The inverse of an 𝑛×𝑛 matrix 𝐴 is an 𝑛×𝑛 matrix 𝐡 such that 𝐴𝐡=𝐼=𝐡𝐴, where 𝐼 is the 𝑛×𝑛 identity matrix. We call this matrix 𝐴.
  2. For 2Γ—2 matrices, the matrix 𝐴=ο”π‘Žπ‘π‘π‘‘ο  has an inverse exactly when its determinant π‘Žπ‘‘βˆ’π‘π‘ is not zero.
  3. When the 2Γ—2 matrix 𝐴 is invertible (has an inverse), then it is given as 1(𝐴)det times the adjugate matrix ο“π‘‘βˆ’π‘βˆ’π‘π‘ŽοŸ.

Example 1: Identifying Whether a Given Matrix Is Invertible Using the Determinant

Is the following matrix invertible? 31βˆ’3βˆ’1

Answer

We know that this matrix, which we will call 𝐴, is invertible if, and only if, it has a nonzero determinant. We compute det(𝐴)=(3)(βˆ’1)βˆ’(1)(βˆ’3)=βˆ’3βˆ’(βˆ’3)=0.

Therefore, this matrix is not invertible.

Example 2: Identifying Whether Two Matrices Are Inverses of One Another

Are the matrices 1234,⎑⎒⎒⎣1121314⎀βŽ₯βŽ₯⎦ multiplicative inverses of each other?

Answer

Of course, β€œmultiplicative” is for emphasis. Usually, the word β€œinverse” is enough. To decide whether they are inverses or not, we multiply the matrices together (in either order). We will find 1234⎑⎒⎒⎣1121314⎀βŽ₯βŽ₯⎦=⎑⎒⎒⎣53113352⎀βŽ₯βŽ₯⎦, which is not the identity matrix.

These matrices are not inverses of each other.

Example 3: Finding the Inverse of a Matrix

Find the multiplicative inverse of the matrix 𝐴=ο”βˆ’4βˆ’1035, if possible.

Answer

We determine whether an inverse exists at all by calculating the determinant: det(𝐴)=(βˆ’4)(5)βˆ’(βˆ’10)(3)=βˆ’20+30=10.

It is not zero, so we can proceed to find the inverse: 𝐴=110510βˆ’3βˆ’4=⎑⎒⎒⎣5101010βˆ’310βˆ’410⎀βŽ₯βŽ₯⎦=⎑⎒⎒⎣121βˆ’310βˆ’25⎀βŽ₯βŽ₯⎦.

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