Let X = [ \beginmatrix 0 & 1 & 0 0 & 0 & 1 0 & 0 & 0 \endmatrix ],\,Y = \alpha I + \beta X + \gamma X^2 and Z = \alpha ^2I - \alpha \beta X + (\beta ^2 - \alpha \gamma )X^2 , \alpha , \beta , \gamma\in R. If Y^ - 1 = [ \beginmatrix 1 \over 5 & - 2 \over 5 & 1 \over 5 0 & 1 \over 5 & - 2 \over 5 0 & 0 & 1 \over 5 \endmatrix ] , then ( \alpha-\beta + \gamma )2 is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2022 · 26 Jun · Shift 2 · Q42 with Solution

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Correct answer: 100

Given $X=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$ we have $X^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix},\qquad X^3=0.$

So any matrix of the form $aI+bX+cX^2$ is upper triangular Toeplitz.

Since $Y=\alpha I+\beta X+\gamma X^2,$ therefore

\alpha & \beta & \gamma\\ 0 & \alpha & \beta\\ 0 & 0 & \alpha \end{bmatrix}.$$ We are given $$Y^{-1}=\begin{bmatrix} \frac15 & -\frac25 & \frac15\\ 0 & \frac15 & -\frac25\\ 0 & 0 & \frac15 \end{bmatrix}.$$ Notice this is also of the form $$Y^{-1}=aI+bX+cX^2$$ with $$a=\frac15,\quad b=-\frac25,\quad c=\frac15.$$ --- 3. Use the standard inverse formula. For $$Y=\alpha I+\beta X+\gamma X^2,$$ because $X^3=0$, its inverse is $$Y^{-1}=\frac1\alpha I-\frac\beta{\alpha^2}X+\left(\frac{\beta^2}{\alpha^3}-\frac\gamma{\alpha^2}\right)X^2.$$ Comparing with the given $Y^{-1}$: - Coefficient of $I$: $$\frac1\alpha=\frac15 \implies \alpha=5.$$ - Coefficient of $X$: $$-\frac\beta{\alpha^2}=-\frac25.$$ With $\alpha=5$, $$-\frac\beta{25}=-\frac25 \implies \beta=10.$$ - Coefficient of $X^2$: $$\frac{\beta^2}{\alpha^3}-\frac\gamma{\alpha^2}=\frac15.$$ Substitute $\alpha=5,\beta=10$: $$\frac{100}{125}-\frac\gamma{25}=\frac15$$ $$\frac45-\frac\gamma{25}=\frac15$$ $$-\frac\gamma{25}=-\frac35$$ $$\gamma=15.$$ So, $$\alpha=5,\quad \beta=10,\quad \gamma=15.$$ --- 4. Compute the required quantity. The expression is interpreted as $$\bigl(\alpha-\beta+\gamma\bigr)^2.$$ Thus, $$\alpha-\beta+\gamma=5-10+15=10,$$ so $$\bigl(\alpha-\beta+\gamma\bigr)^2=10^2=100.$$ --- 5. Final answer. $$\boxed{100}$$ This matches the stored correct answer.

View written solutionFree

Correct answer: 100

Given $X=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$ we have $X^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix},\qquad X^3=0.$

So any matrix of the form $aI+bX+cX^2$ is upper triangular Toeplitz.

Since $Y=\alpha I+\beta X+\gamma X^2,$ therefore

\alpha & \beta & \gamma\\ 0 & \alpha & \beta\\ 0 & 0 & \alpha \end{bmatrix}.$$ We are given $$Y^{-1}=\begin{bmatrix} \frac15 & -\frac25 & \frac15\\ 0 & \frac15 & -\frac25\\ 0 & 0 & \frac15 \end{bmatrix}.$$ Notice this is also of the form $$Y^{-1}=aI+bX+cX^2$$ with $$a=\frac15,\quad b=-\frac25,\quad c=\frac15.$$ --- 3. Use the standard inverse formula. For $$Y=\alpha I+\beta X+\gamma X^2,$$ because $X^3=0$, its inverse is $$Y^{-1}=\frac1\alpha I-\frac\beta{\alpha^2}X+\left(\frac{\beta^2}{\alpha^3}-\frac\gamma{\alpha^2}\right)X^2.$$ Comparing with the given $Y^{-1}$: - Coefficient of $I$: $$\frac1\alpha=\frac15 \implies \alpha=5.$$ - Coefficient of $X$: $$-\frac\beta{\alpha^2}=-\frac25.$$ With $\alpha=5$, $$-\frac\beta{25}=-\frac25 \implies \beta=10.$$ - Coefficient of $X^2$: $$\frac{\beta^2}{\alpha^3}-\frac\gamma{\alpha^2}=\frac15.$$ Substitute $\alpha=5,\beta=10$: $$\frac{100}{125}-\frac\gamma{25}=\frac15$$ $$\frac45-\frac\gamma{25}=\frac15$$ $$-\frac\gamma{25}=-\frac35$$ $$\gamma=15.$$ So, $$\alpha=5,\quad \beta=10,\quad \gamma=15.$$ --- 4. Compute the required quantity. The expression is interpreted as $$\bigl(\alpha-\beta+\gamma\bigr)^2.$$ Thus, $$\alpha-\beta+\gamma=5-10+15=10,$$ so $$\bigl(\alpha-\beta+\gamma\bigr)^2=10^2=100.$$ --- 5. Final answer. $$\boxed{100}$$ This matches the stored correct answer.