(a) Show that the matrices $A = \begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} \frac{4}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-2}{5} & \frac{4}{5} \end{pmatrix}$ are inverse of each other. OR (b) If $\tan\alpha = \dfrac{1}{3}$ and $\tan\beta = \dfrac{1}{7}$ then prove that $(2\alpha + \beta) = \dfrac{\pi}{4}$.

> > [!TLDR] > **(a)** Multiplying $A$ by $B$ gives the identity matrix, proving they are inverses. **(b)** $\tan 2\alpha = \tfrac34$; then $\tan(2\alpha+\beta) = 1$, so $2\alpha+\beta = \tfrac{\pi}{4}$. > > **Part (a): Show $A$ and $B$ are inverses.** > Two square matrices are inverses iff their product is the identity $I$. Compute $AB$ (with $B = \tfrac15\begin{pmatrix}4&-2&-1\\-1&3&-1\\-1&-2&4\end{pmatrix}$). > > - Row 1 $\times$ B: $\tfrac15(2\cdot4 + 2\cdot(-1) + 1\cdot(-1)) = \tfrac{5}{5}=1$; $\tfrac15(2\cdot(-2)+2\cdot3+1\cdot(-2)) = \tfrac{0}{5}=0$; $\tfrac15(2\cdot(-1)+2\cdot(-1)+1\cdot4) = \tfrac{0}{5}=0$. > - Row 2 $\times$ B: $\tfrac15(1\cdot4+3\cdot(-1)+1\cdot(-1)) = 0$; $\tfrac15(1\cdot(-2)+3\cdot3+1\cdot(-2)) = \tfrac{5}{5}=1$; $\tfrac15(1\cdot(-1)+3\cdot(-1)+1\cdot4)=0$. > - Row 3 $\times$ B: $\tfrac15(1\cdot4+2\cdot(-1)+2\cdot(-1)) = 0$; $\tfrac15(1\cdot(-2)+2\cdot3+2\cdot(-2)) = 0$; $\tfrac15(1\cdot(-1)+2\cdot(-1)+2\cdot4)=\tfrac{5}{5}=1$. > > Thus $AB = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} = I$. Since $AB = I$, $B = A^{-1}$; the matrices are inverses of each other. > > **Part (b): Prove $2\alpha+\beta = \dfrac{\pi}{4}$, given $\tan\alpha=\tfrac13,\ \tan\beta=\tfrac17$.** > > *Step 1 — Find $\tan 2\alpha$* using $\tan 2\alpha = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$: …

(a) Show that the matrices A = 2 2 1 \1 3 1 \1 2 2 and B = (4)/(5) (-2)/(5) (-1)/(5) ; frac-15 (3)/(5) (-1)/(5

Question 35 of 47

A = \begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix}

(OR)

(b) If

\tan\alpha = \dfrac{1}{3}

and

\tan\beta = \dfrac{1}{7}

then prove that

(2\alpha + \beta) = \dfrac{\pi}{4}

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(a) Multiplying $A$ by $B$ gives the identity matrix, proving they are inverses. (b) $\tan 2\alpha = \tfrac34$ ; then $\tan(2\alpha+\beta) = 1$ , so $2\alpha+\beta = \tfrac{\pi}{4}$ .

Part (a): Show $A$ and $B$ are inverses.

Two square matrices are inverses iff their product is the identity $I$ . Compute $AB$ (with $B = \tfrac15\begin{pmatrix}4&-2&-1\\-1&3&-1\\-1&-2&4\end{pmatrix}$ ).

Row 1 $\times$ B: $\tfrac15(2\cdot4 + 2\cdot(-1) + 1\cdot(-1)) = \tfrac{5}{5}=1$ ; $\tfrac15(2\cdot(-2)+2\cdot3+1\cdot(-2)) = \tfrac{0}{5}=0$ ; $\tfrac15(2\cdot(-1)+2\cdot(-1)+1\cdot4) = \tfrac{0}{5}=0$ .
Row 2 $\times$ B: $\tfrac15(1\cdot4+3\cdot(-1)+1\cdot(-1)) = 0$ ; $\tfrac15(1\cdot(-2)+3\cdot3+1\cdot(-2)) = \tfrac{5}{5}=1$ ; $\tfrac15(1\cdot(-1)+3\cdot(-1)+1\cdot4)=0$ .
Row 3 $\times$ B: $\tfrac15(1\cdot4+2\cdot(-1)+2\cdot(-1)) = 0$ ; $\tfrac15(1\cdot(-2)+2\cdot3+2\cdot(-2)) = 0$ ; $\tfrac15(1\cdot(-1)+2\cdot(-1)+2\cdot4)=\tfrac{5}{5}=1$ .

Thus $AB = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} = I$ . Since $AB = I$ , $B = A^{-1}$ ; the matrices are inverses of each other.

Part (b): Prove $2\alpha+\beta = \dfrac{\pi}{4}$ , given $\tan\alpha=\tfrac13,\ \tan\beta=\tfrac17$ .

Step 1 — Find $\tan 2\alpha$ using $\tan 2\alpha = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$ : …

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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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←Previous questionQ16 — Find λ, if the matrix 1 1 3 \2 λ 4 \9 7 11 has no inverse.→Next questionQ18 — The inventor of input-output analysis is : (a) Prof. Wassily W. Leonti…

Q.(a) Show that the matrices $A = \begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} \frac{4}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-2}{5} & \frac{4}{5} \end{pmatrix}$ are inverse of each other.

Concept understanding — Adjoint, Inverse and Solving Linear Equations

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