Answer to Question #101851 in Linear Algebra for kelbesa Gemechu

Question

Let A = (-1 0 4)
6 3 2
2 5 6 3x3 then find
A/ Det(A)
B/ Cofactor of A
C/ adjoint of A
D/ Inverse of A

Answer 1

We have matrix A = $\left(\begin{matrix}-1 & 0 & 4\\6 & 3 & 2\\2 & 5 & 6\end{matrix}\right)$;

Solution:

A) We compute determinant of A by using Laplace expansion (in this case for 3x3 matrix) $Det(A) = \sum_{j=1}^3 (-1)^{1+j}a_{1j}M_{1j},$ where $M_{1j}$ is a minor of A.

So,

$Det(A) = -1 * \left[\begin{matrix}3 & 2\\5 & 6\end{matrix}\right] - 0 * \left[\begin{matrix}6 & 2\\2 & 6\end{matrix}\right] + 4*\left[\begin{matrix}6 & 3\\2 & 5\end{matrix}\right]$ $= -1*(18 - 10)+4*(30-6) = 88$

As a result

$Det(A) = 88$.

B) Let's call cofactor matrix of A as B. The B can be computed with next expression $b_{ij} = (-1)^{i+j}a_{ij}M_{ij}.$ Following the given formula, we obtain: $b_{11} = (-1)^{1+1} \left[\begin{matrix}3 & 2\\5 & 6\end{matrix}\right] = 8;$ $b_{12} = (-1)^{1+2} \left[\begin{matrix}6 & 2\\2 & 6\end{matrix}\right] =-32;$

$b_{13} = (-1)^{1+3} \left[\begin{matrix}6 & 3\\2 & 5\end{matrix}\right] =24;$ $b_{21} = (-1)^{2+1} \left[\begin{matrix}0 & 4\\5 & 6\end{matrix}\right] =20;$

$b_{22} = (-1)^{2+2} \left[\begin{matrix}-1 & 4\\2 & 6\end{matrix}\right] =-14;$ $b_{23} = (-1)^{2+3} \left[\begin{matrix}-1 & 0\\2 & 5\end{matrix}\right] = 5;$

$b_{31} = (-1)^{3+1} \left[\begin{matrix}0 & 4\\3 & 2\end{matrix}\right] =-12;$ $b_{32} = (-1)^{3+2} \left[\begin{matrix}-1 & 4\\6 & 2\end{matrix}\right] =26;$

$b_{33} = (-1)^{3+3} \left[\begin{matrix}-1 & 0\\6 & 3\end{matrix}\right] =-3;$

Finally, we have

$B = \left(\begin{matrix} 8 & -32 & 24\\20 & -14 & 5\\-12 & 26 & -3\end{matrix}\right)$.

C) By definition adjoint (also called "adjugate") is just a transpose matrix of cofactor matrix A

$adj(A) = B^T.$

In this case

$adj(A) = \left(\begin{matrix} 8 & 20 & -12\\-32 & -14 & 26\\24 & 5 & -3\end{matrix}\right).$

D) Using the expression $A^{-1} = \frac{1}{det(A)}adj(A)$, we obtain

$A^{-1}=\frac{1}{88}\left(\begin{matrix} 8 & 20 & -12\\-32 & -14 & 26\\24 & 5 & -3\end{matrix}\right) =$ $\left[\begin{matrix}\frac{1}{11} & \frac{5}{22} & - \frac{3}{22}\\- \frac{4}{11} & - \frac{7}{44} & \frac{13}{44}\\\frac{3}{11} & \frac{5}{88} & - \frac{3}{88}\end{matrix}\right]$