(a) If $A = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 0 & -1 \\ 1 & 2 \end{bmatrix}$ then, show that $(AB)^{-1} = B^{-1}A^{-1}$. OR (b) Prove that : $\dfrac{\sin(180^\circ + A)\,\cos(90^\circ - A)\,\tan(270^\circ - A)}{\sec(540^\circ - A)\,\cos(360^\circ + A)\,\operatorname{cosec}(270^\circ + A)} = -\sin A\,\cos^{2} A$

(a) If A = 1 2 \1 1 , B = 0 -1 \1 2 then, show that (AB)⁻¹ = B⁻¹A⁻¹. OR (b) Prove that : dfracsin(180^° + A)co

Question 30 of 47

A = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}

(OR)

(b) Prove that :

\dfrac{\sin(180^\circ + A)\,\cos(90^\circ - A)\,\tan(270^\circ - A)}{\sec(540^\circ - A)\,\cos(360^\circ + A)\,\operatorname{cosec}(270^\circ + A)} = -\sin A\,\cos^{2} A

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2022Subjective· 5mImportance★★★★★

64% · 30/47 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

(a) Both $(AB)^{-1}$ and $B^{-1}A^{-1}$ equal $\begin{bmatrix}-1&3\\1&-2\end{bmatrix}$ . (b) Using allied-angle reductions the ratio simplifies to $-\sin A\cos^{2}A$ .

Part (a) — Show $(AB)^{-1}=B^{-1}A^{-1}$ .

Step 1 — Compute $AB$ .

$AB=\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}0&-1\\1&2\end{bmatrix}=\begin{bmatrix}2&3\\1&1\end{bmatrix}.$

Step 2 — $(AB)^{-1}$ . $\det(AB)=2(1)-3(1)=-1$ .

$(AB)^{-1}=\dfrac{1}{-1}\begin{bmatrix}1&-3\\-1&2\end{bmatrix}=\begin{bmatrix}-1&3\\1&-2\end{bmatrix}.$

Step 3 — $A^{-1}$ and $B^{-1}$ . $\det A=1(1)-2(1)=-1$ , $\det B=0(2)-(-1)(1)=1$ .

$A^{-1}=\dfrac{1}{-1}\begin{bmatrix}1&-2\\-1&1\end{bmatrix}=\begin{bmatrix}-1&2\\1&-1\end{bmatrix},\qquad B^{-1}=\begin{bmatrix}2&1\\-1&0\end{bmatrix}.$

Step 4 — Compute $B^{-1}A^{-1}$ .

$B^{-1}A^{-1}=\begin{bmatrix}2&1\\-1&0\end{bmatrix}\begin{bmatrix}-1&2\\1&-1\end{bmatrix}=\begin{bmatrix}-1&3\\1&-2\end{bmatrix}.$

Step 5 — Compare. $(AB)^{-1}=B^{-1}A^{-1}=\begin{bmatrix}-1&3\\1&-2\end{bmatrix}$ . Proved.

Part (b) — Prove the trigonometric identity.

Reduce each factor using allied-angle (compound-angle) rules:

$\sin(180^\circ+A)=-\sin A,\qquad \cos(90^\circ-A)=\sin A,\qquad \tan(270^\circ-A)=\cot A,$ …

Start your 14-day free trial to unlock the full solution →

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

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

…

Start your 14-day free trial to see the full solution →
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. …

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →

←Previous questionQ11 — The technology matrix of an economic system of two industries is 0.6 0…→Next questionQ13 — The number of Hawkin's-Simon conditions for the viability of an input-…

Q.(a) If $A = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}$ , $B = \begin{bmatrix} 0 & -1 \\ 1 & 2 \end{bmatrix}$ then, show that $(AB)^{-1} = B^{-1}A^{-1}$ .

Concept understanding — Adjoint, Inverse and Solving Linear Equations

Q.(a) If A=[1211]A = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}A=[11​21​], B=[0−112]B = \begin{bmatrix} 0 & -1 \\ 1 & 2 \end{bmatrix}B=[01​−12​] then, show that (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}(AB)−1=B−1A−1.

🎓Unlock everything free for 14 days

Q.(a) If $A = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}$ , $B = \begin{bmatrix} 0 & -1 \\ 1 & 2 \end{bmatrix}$ then, show that $(AB)^{-1} = B^{-1}A^{-1}$ .

Unlock everything free for 14 days