Question Video: Proving Properties of the Adjoint Matrix of a Square Matrix of Order 𝑛. Mathematics

True or False: If 𝐴 is any given square matrix of order 𝑛, then 𝐴 Γ— adj (𝐴) = adj (𝐴) Γ— 𝐴 = |𝐴|𝐼, where 𝐼 is the identity matrix of order 𝑛

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Video Transcript

True or False: If 𝐴 is any given square matrix of order 𝑛, then 𝐴 multiplied by the adjoint matrix of 𝐴 is equal to the adjoint of 𝐴 multiplied by 𝐴 is equal to the determinant of 𝐴 multiplied by 𝐼, where 𝐼 is the identity matrix of order 𝑛.

We’ll begin by noting that if the matrix is singular, that is, the determinant of 𝐴 is equal to zero, then the inverse of 𝐴 does not exist. And we’ll come back to this case a little later on. If, on the other hand, an 𝑛-by-𝑛 matrix is nonsingular, that means its inverse does exist, then to show whether the given statement is true or false, we can use the theorem that for a nonsingular 𝑛-by-𝑛 matrix 𝐴, 𝐴 inverse β€” that’s 𝐴 negative one β€” is the adjoint of 𝐴 divided by the determinant of 𝐴. And recall that the adjoint of matrix 𝐴 is the transpose of the matrix of its cofactors.

So let’s assume that the inverse of our matrix exists. This then means that the determinant is nonzero. To show whether this statement is true or false, let’s take our definition of the inverse of 𝐴 and multiply through on the left by the matrix 𝐴. Now we know that for any nonsingular 𝑛-by-𝑛 matrix, 𝐴 multiplied by its inverse is the identity matrix. So now on the left-hand side, we have the 𝑛-by-𝑛 identity matrix. And now since the determinant of the matrix 𝐴 is a scalar, we can multiply through by this. And this gives us the determinant of 𝐴 multiplied by the identity matrix is 𝐴 multiplied by its adjoint. And this shows that our first statement 𝐴 adjoint 𝐴 is indeed equal to the determinant of 𝐴 multiplied by 𝐼 for a nonsingular matrix 𝐴.

Now, if we begin again with our formula for the inverse of 𝐴 and multiply through this time on the right by our matrix 𝐴, we again have the identity matrix on the left since 𝐴 inverse 𝐴 is the identity matrix. And again multiplying through by the determinant of 𝐴, which we can do since it’s a scalar, we have the determinant of 𝐴 multiplied by the identity matrix is equal to the adjoint of 𝐴 multiplied by 𝐴. And so we can say that our second expression is also true for a nonsingular matrix. The whole of the given expression is therefore true for a nonsingular matrix 𝐴.

We still don’t know whether the given statement is true or false if our matrix 𝐴 is singular. That’s when the determinant is equal to zero. So writing out our statement again, using slightly different notation for the determinant just for clarity, and now suppose we take the determinant of each of our expressions, we can use the theorem that for 𝑛-by-𝑛 matrices 𝐴 and 𝐡, the determinant of the product 𝐴𝐡 is equal to the determinant of matrix 𝐴 multiplied by determinant of matrix 𝐡. Our first expression is then the determinant of 𝐴 multiplied by the determinant of the adjoint of 𝐴. And similarly, for our second expression, and on the right we leave our expression as it is.

Now, if 𝐴 is a singular matrix, then the inverse doesn’t exist and the determinant is equal to zero. In our first two expressions then, we’re multiplying by zero. And anything multiplied by zero is equal to zero. On our right-hand side, we’re multiplying the identity matrix by zero, which gives us the zero matrix. And the determinant of the zero matrix is zero. And so the left-hand, the middle term, and the right-hand term are all equal to zero. This means then that for a singular matrix where the determinant is equal to zero, then the given statement is also true. Hence, 𝐴 multiplied by adjoint 𝐴 is equal to adjoint 𝐴 multiplied by 𝐴 is equal to the determinant of 𝐴 multiplied by 𝐼 is true for any square matrix 𝐴 of order 𝑛.

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