Chapter 5.2, Problem 12PS

To determine

The cofactor matrix $C$ of $A$ and multiply $A$ times $C^T$. Compare $A{C}^T$ with $A^{-1}$.

The cofactor matrix is $C=\left[\begin{array}{ccc}3& 2& 1\\ 2& 4& 2\\ 1& 2& 3\end{array}\right]$

   $A{C}^T=\left[\begin{array}{ccc}4& 0& 0\\ 0& 4& 0\\ 0& 0& 4\end{array}\right]$

The product of $A{C}^T$ and $A^{-1}$ is the cofactor matrix $C$.

Given:

Given information:

The matrix $A=\left[\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right]$ and $A^{-1}=\frac{1}{4}\left[\begin{array}{ccc}3& 2& 1\\ 2& 4& 2\\ 1& 2& 3\end{array}\right]$

Concept Used:

A matrix B is invertible if B is non-singular and its inverse is given by $B^{-1}=\frac{adj\left(B\right)}{\left|B\right|}$ where,

   $\begin{array}{l}{B}^{-1}=\frac{adj\left(B\right)}{\left|B\right|}\\ adj\left(B\right)=\left[\begin{array}{ccc}{C}_{11}& C_{21}& C_{31}\\ C_{12}& C_{22}& C_{32}\\ C_{13}& C_{23}& C_{33}\end{array}\right]\\ C_{ij}=\text{Cofactor of }{a}_{ij}={\left(-1\right)}^{i+j}{M}_{ij}\\ M_{ij}=\text{Minor of }{a}_{ij}\end{array}$

Calculation:

Here, we have

   $A=\left[\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right]$

Let us find all the cofactors of $A$

Therefore, the cofactor matrix is

   $C=\left[\begin{array}{ccc}3& 2& 1\\ 2& 4& 2\\ 1& 2& 3\end{array}\right]$

And, the transpose of $C$ is

   $C^T=\left[\begin{array}{ccc}3& 2& 1\\ 2& 4& 2\\ 1& 2& 3\end{array}\right]$

Multiply $A$ times $C$.

   $\begin{array}{l}A{C}^T=\left[\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right]\left[\begin{array}{ccc}3& 2& 1\\ 2& 4& 2\\ 1& 2& 3\end{array}\right]\\ \Rightarrow A{C}^T=\left[\begin{array}{ccc}2\times 3+(-1)\times 2+0\times 1& 2\times 2+(-1)\times 4+0\times 2& 2\times 1+(-1)\times 2+0\times 3\\ -1\times 3+2\times 2+(-1)\times 1& -1\times 2+2\times 4+(-1)\times 2& -1\times 1+2\times 2+(-1)\times 3\\ 0\times 3+(-1)\times 2+2\times 1& 0\times 2+(-1)\times 4+2\times 2& 0\times 1+(-1)\times 2+2\times 3\end{array}\right]\\ \Rightarrow A{C}^T=\left[\begin{array}{ccc}6-2+0& 4-4+0& 2-2+0\\ -3+4-1& -2+8-2& -1+4-3\\ 0-2+2& 0-4+4& 0-2+6\end{array}\right]\\ \Rightarrow A{C}^T=\left[\begin{array}{ccc}4& 0& 0\\ 0& 4& 0\\ 0& 0& 4\end{array}\right]\end{array}$

From the above result we can write

   $A{C}^T=4I$

Since $A^{-1}=\frac{1}{4}\left[\begin{array}{ccc}3& 2& 1\\ 2& 4& 2\\ 1& 2& 3\end{array}\right]$ and $C^T=\left[\begin{array}{ccc}3& 2& 1\\ 2& 4& 2\\ 1& 2& 3\end{array}\right]$ we can write

   $A^{-1}=\frac{1}{4}{C}^T$

Now, multiply the above equation both sides by $A{C}^T$

   $(A{C}^T)A^{-1}=\frac{1}{4}(A{C}^T)C^T$

Substitute $A{C}^T=4I$ on the right side

   $(A{C}^T)A^{-1}=\frac{1}{4}(4I)C^T$

   $\begin{array}{l}(A{C}^T)A^{-1}=\frac{1}{4}.4(I{C}^T)\\ (A{C}^T)A^{-1}=C^T\end{array}$

Since $C^T=C$ we get

   $(A{C}^T)A^{-1}=C$

Thus, we can say that the product of $A{C}^T$ and $A^{-1}$ is the cofactor matrix $C$.