Proof about adjoint matrix

Question

Prove: If $A_{n\times n}$ is not invertible then $Adjoint(A)$ is not invertible also.

I have made the following:

By contradiction.

Suppose that $Adj(A)^{-1}$ exists

Then write $Adj(A)*B=I$

\begin{align} det(Adj(A)\cdot B) &= det(I) \\ det(Adj(A)) \cdot det(B) &= 1 \\ det(B) &= \frac{1}{det(Adj(A))} \\ &= \frac{1}{det(A)^{n-1}} \\ &= \frac{1}{0} \text{Contradiction} \end{align}

Im not sure all about the proof. It is correct to use that $det(Adj(A))=det(A)^{n-1}$ i know this is a fact. However,(im not sure) in the proof is used that $A^{-1}$ exists

Answer 1

Just getting this off the unanswered list.

As @EuYu noted, your proof is a little awkward but could be cleaned up to be correct. I would argue that part of the reason it's awkward is that $\det(\mathop{\mathrm{adj}}(A))=\det(A)^{n-1}$ is more than you need to establish your result -- see the more direct proof here for example.

I'll add that $\det(\mathop{\mathrm{adj}}(A))=\det(A)^{n-1}$ can be proved for a "generic matrix" $A$ with indeterminate entries, which shows that it's a polynomial identity and therefore holds for any, not necessarily invertible, "scalar matrix" $A$ by substitution -- see here.

Finally I'll add that you can prove much more: the rank of $\mathop{\mathrm{adj}}(A)$ drops off quickly as the rank of $A$ drops -- see here for example.

Answer 2

Restatement of the Problem to Make it Easier to View in the Answer

Prove: If $A_{n\times n}$ is not invertible then $Adjoint(A)$ is not invertible also. I have made the following: By contradiction. Suppose that $Adj(A)^{-1}$ exists Then write $Adj(A)*B=I$ $$ \begin{align} det(Adj(A)\cdot B) &= det(I) \\ det(Adj(A)) \cdot det(B) &= 1 \\ det(B) &= \frac{1}{det(Adj(A))} \\ &= \frac{1}{det(A)^{n-1}} \\ &= \frac{1}{0} \text{Contradiction} \end{align} $$ I am not sure all about the proof. It is correct to use that $det(Adj(A))=det(A)^{n-1}$ i know this is a fact. However,(I am not sure) in the proof is used that $A^{-1}$ exists

Answer Solution Steps

First, division by zero is not defined. So the proof needs to deal with this issue.

From the quote in "Cramer's Rule":

Let A be an n × n matrix with entries in a field F. Then $${\displaystyle A\,\operatorname {adj} (A)=\operatorname {adj} (A)\,A=\det(A)I}$$ ... If $\det(A)$ is nonzero, then the inverse matrix of A is $${\displaystyle A^{-1}={\frac {1}{\det(A)}}\operatorname {adj} (A).}$$

This can be rewritten as: $${\det(A)}\,{\displaystyle A^{-1}=\operatorname {adj} \left(A\right).}$$ The "adjoint" (also known as the Adjugate matrix) $\operatorname {adj} \left(A\right)=C^T$ (the transpose of the matrix C) *is always defined * as follows:

$${\displaystyle \mathbf {C} ={\begin{bmatrix}C_{11}&C_{12}&\cdots &C_{1n}\\C_{21}&C_{22}&\cdots &C_{2n}\\\vdots &\vdots &\ddots &\vdots \\C_{n1}&C_{n2}&\cdots &C_{nn}\end{bmatrix}}}$$ One can write down the inverse of an invertible matrix by computing its cofactors by using Cramer's rule, as follows. The matrix formed by all of the cofactors of a square matrix A is called the cofactor matrix (also called the matrix of cofactors or, sometimes, comatrix):

The case with $A_{x\times x}$ not being invertable occurs when $A_{x\times x}$ is not uniquely defined.

From: $${\det(A)}\,{\displaystyle A^{-1}=\operatorname {adj} \left(A\right).}$$ From here, only if ${\det(A)}=0$, then the matrix $A$ can take on any value with the product $\left({\det(A)}=0\right)\,{\displaystyle A^{-1}=0}$. So only if ${\det(A)}=0$, since $A_{n\times n}$ is not unique, it is not invert-able. Also in this case:

$$0\,A^{-1}=\operatorname {adj} \left(A\right)=0$$

Then since ${\operatorname {adj} \left(A\right)=0}$ is a zero matrix, it is not invertable since matrix division by a zero matrix is not defined. Technically there is also the case where $Adj(A)$ is defined, but a zero matrix. In that case either $Det(A)=0$ which results in $A$ not being invertable since it is a zero $nxn$ matrix. Or $det(A)=0$ resulting in $A$ not being invertable since zero times any matrix has the same result as a zero matrix, and therefore within the limitation that the determinant is zero, $A$ is not unique. For $A$ to be invert-able, it must be unique. Distinct examples resulting in a zero determinant are, for example, for any two rows to be the same value.

Since $A_{n\times n}$ is not invertible then $Adjoint(A)$ is not invertible also.