If $A = \begin{bmatrix} 5 & -7 \\ 3 & -4 \end{bmatrix}$ then prove that $(A^{-1})^{-1} = A$.

If A = 5 -7 \3 -4 then prove that (A⁻¹)⁻¹ = A.

Question 23 of 47

A = \begin{bmatrix} 5 & -7 \\ 3 & -4 \end{bmatrix}

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2020Subjective· 3mImportance★★★★★

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$\det A=1$ so $A^{-1}=\operatorname{adj}A=\begin{bmatrix}-4&7\\-3&5\end{bmatrix}$ ; inverting that again gives back $A$ , proving $(A^{-1})^{-1}=A$ .

An inverse-property proof from the Matrices and Determinants unit of the Tamil Nadu HSC Class-11 Business Mathematics syllabus.

Step 1 — Determinant of $A$ . $A=\begin{bmatrix}5&-7\\3&-4\end{bmatrix}$ , so $\det A=(5)(-4)-(-7)(3)=-20+21=1.$

Step 2 — Inverse of $A$ . For a $2\times2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ , $A^{-1}=\dfrac{1}{\det A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ . Since $\det A=1$ ,

$A^{-1}=\begin{bmatrix}-4&7\\-3&5\end{bmatrix}.$

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set MARCH1 markMCQ
$|A|=5$
(a) $1$
(b) $5$
(c) does not exist
(d) $\dfrac{1}{5}$
›Reveal solutionSolution

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

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

…

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CBSE 2025Set MARCH1 markMCQ
$\begin{pmatrix} \dfrac{4}{5} & \dfrac{-5}{12} \\ \dfrac{-2}{5} & \dfrac{1}{2} \end{pmatrix}$
(a) $\dfrac{30}{7}\begin{pmatrix} \dfrac{1}{2} & \dfrac{5}{12} \\ \dfrac{2}{5} & \dfrac{4}{5} \end{pmatrix}$
(b) $\dfrac{7}{30}\begin{pmatrix} \dfrac{1}{2} & \dfrac{5}{12} \\ \dfrac{2}{5} & \dfrac{4}{5} \end{pmatrix}$
(c) $\dfrac{30}{7}\begin{pmatrix} \dfrac{1}{2} & \dfrac{-5}{12} \\ -2 & \dfrac{4}{5} \end{pmatrix}$
(d) $\dfrac{7}{30}\begin{pmatrix} \dfrac{1}{2} & \dfrac{-5}{12} \\ \dfrac{-2}{5} & \dfrac{1}{5} \end{pmatrix}$
›Reveal solutionSolution

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. …

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CBSE 2024Set MARCH1 markMCQ
$Q. The inventor of input-output analysis is :$
(a) Prof. Wassily W. Leontief
(b) Sir Francis Galton
(c) Arthur Caylay
(d) Fisher
›Reveal solutionSolution

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.

…

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CBSE 2023Set MARCH1 markMCQ
$Q. The number of Hawkin's-Simon conditions for the viability of an input-output analysis is :$
(a) 4
(b) 1
(c) 2
(d) 3
›Reveal solutionSolution
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:

The main diagonal elements of $(I - A)$ must be positive, and …
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Q.If $A = \begin{bmatrix} 5 & -7 \\ 3 & -4 \end{bmatrix}$ then prove that $(A^{-1})^{-1} = A$ .

Concept understanding — Adjoint, Inverse and Solving Linear Equations

Q.If A=[5−73−4]A = \begin{bmatrix} 5 & -7 \\ 3 & -4 \end{bmatrix}A=[53​−7−4​] then prove that (A−1)−1=A(A^{-1})^{-1} = A(A−1)−1=A.

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Q.If $A = \begin{bmatrix} 5 & -7 \\ 3 & -4 \end{bmatrix}$ then prove that $(A^{-1})^{-1} = A$ .

Unlock everything free for 14 days