Q.Show that the matrices $A = \begin{bmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} \dfrac{-4}{35} & \dfrac{11}{35} & \dfrac{-5}{35} \\[6pt] \dfrac{-1}{35} & \dfrac{-6}{35} & \dfrac{25}{35} \\[6pt] \dfrac{6}{35} & \dfrac{1}{35} & \dfrac{-10}{35} \end{bmatrix}$ are inverses of each other.

Use the property $|A^{-1}|=\dfrac{1}{|A|}$, giving $\dfrac{1}{5}$.

From $A\,A^{-1}=I$, taking determinants and using $|AB|=|A||B|$:

$|A|\cdot|A^{-1}|=|I|=1.$

Therefore

$|A^{-1}|=\frac{1}{|A|}.$

Since $|A|=5\neq 0$, $A$ is invertible and

$|A^{-1}|=\frac{1}{5}.$

…

Compute $|A|=\dfrac{7}{30}$ and $\text{adj}(A)$, then $A^{-1}=\dfrac{1}{|A|}\text{adj}(A)=\dfrac{30}{7}\begin{pmatrix}1/2 & 5/12\\ 2/5 & 4/5\end{pmatrix}$.

Let $A=\begin{pmatrix} \dfrac{4}{5} & \dfrac{-5}{12} \\[4pt] \dfrac{-2}{5} & \dfrac{1}{2} \end{pmatrix}$.

Step 1 — Determinant.

$|A|=\frac{4}{5}\cdot\frac{1}{2}-\left(\frac{-5}{12}\right)\left(\frac{-2}{5}\right)=\frac{4}{10}-\frac{10}{60}=\frac{2}{5}-\frac{1}{6}.$

$=\frac{12}{30}-\frac{5}{30}=\frac{7}{30}.$

Step 2 — Adjoint. For a $2\times 2$ matrix $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$, $\text{adj}=\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}$:

$\text{adj}(A)=\begin{pmatrix} \dfrac{1}{2} & \dfrac{5}{12} \\[4pt] \dfrac{2}{5} & \dfrac{4}{5} \end{pmatrix}.$

Step 3 — Inverse. …

Input-output analysis, a matrix-based technique for modelling the interdependence of industries in an economy, was invented by Prof. Wassily W. Leontief.

In the Tamil Nadu HSC Class-11 Business Mathematics syllabus, the Matrices and Determinants chapter introduces input-output analysis as an application of matrices. In this model the inter-industry demand of an economy is written as a matrix equation $X = AX + D$, whose solution $X = (I - A)^{-1}D$ needs the matrix inverse.

…

The Hawkins-Simon conditions for the viability of an input-output model are two in number, so the answer is option (c) 2.

In Leontief's input-output analysis the technology matrix $B$ (or the matrix $I - A$, where $A$ is the coefficient matrix) must satisfy conditions that guarantee the system produces non-negative output levels able to meet the final demand.

The Hawkins-Simon conditions state that:

1. The main diagonal elements of $(I - A)$ must be positive, and …