Let A be a symmetric matrix such that |A|=2 and [ \beginmatrix 2 & 1 3 & 3 \over 2 \endmatrix ]A = [ \beginmatrix 1 & 2 \alpha & \beta \endmatrix ] . If the sum of the diagonal elements of A is s , then \frac\beta s\alpha^2 is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2023 · 29 Jan · Shift 2 · Q44 with Solution

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Correct answer: 5

Let $M=\begin{bmatrix}2&1\\[2pt]3&\frac32\end{bmatrix},\qquad A=\begin{bmatrix}a&b\\ b&d\end{bmatrix}$ since $A$ is symmetric.

We are given $MA=\begin{bmatrix}1&2\\ \alpha&\beta\end{bmatrix}.$ Also, $|A|=ad-b^2=2.$

1. Use the first row of the matrix equation

From $\begin{bmatrix}2&1\end{bmatrix}A=\begin{bmatrix}1&2\end{bmatrix},$ we get $[2a+b,\;2b+d]=[1,\;2].$ So, $2a+b=1 \quad ...(1)$ $2b+d=2 \quad ...(2)$

Hence,

\qquad d=2-2b=2-2(1-2a)=4a.$$ Thus, $$A=\begin{bmatrix}a&1-2a\\ 1-2a&4a\end{bmatrix}.$$ ## 2. Use the determinant condition Since $|A|=2$, $$a(4a)-(1-2a)^2=2.$$ Now, $$(1-2a)^2=1-4a+4a^2,$$ so $$4a^2-(1-4a+4a^2)=2.$$ This gives $$4a-1=2 \Rightarrow 4a=3 \Rightarrow a=\frac34.$$ Then, $$b=1-2\cdot \frac34=1-\frac32=-\frac12,$$ $$d=4a=3.$$ Therefore, $$A=\begin{bmatrix}\frac34&-\frac12\\[2pt]-\frac12&3\end{bmatrix}.$$ ## 3. Find $s$, the sum of diagonal elements $$s=a+d=\frac34+3=\frac{15}{4}.$$ ## 4. Compute $\alpha$ and $\beta$ The second row of $MA$ is $$\begin{bmatrix}3&\frac32\end{bmatrix}A=[\alpha,\beta].$$ So, $$\alpha=3a+\frac32 b, \qquad \beta=3b+\frac32 d.$$ Substitute $a=\frac34,\; b=-\frac12,\; d=3$: ### For $\alpha$ $$\alpha=3\cdot \frac34+\frac32\cdot \left(-\frac12\right) =\frac94-\frac34=\frac64=\frac32.$$ ### For $\beta$ $$\beta=3\left(-\frac12\right)+\frac32\cdot 3 =-\frac32+\frac92=3.$$ ## 5. Evaluate $\dfrac{\beta s}{\alpha^2}$ $$\frac{\beta s}{\alpha^2} =\frac{3\cdot \frac{15}{4}}{\left(\frac32\right)^2} =\frac{\frac{45}{4}}{\frac94} =\frac{45}{4}\cdot \frac{4}{9}=5.$$ Thus, the required integer is $$\boxed{5}.$$

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1. Use the first row of the matrix equation

From

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

we get

[2a+b,\;2b+d]=[1,\;2].

So,

2a+b=1 \quad ...(1)

2b+d=2 \quad ...(2)

Hence,

\qquad d=2-2b=2-2(1-2a)=4a.$$ Thus, $$A=\begin{bmatrix}a&1-2a\\ 1-2a&4a\end{bmatrix}.$$ ## 2. Use the determinant condition Since $|A|=2$, $$a(4a)-(1-2a)^2=2.$$ Now, $$(1-2a)^2=1-4a+4a^2,$$ so $$4a^2-(1-4a+4a^2)=2.$$ This gives $$4a-1=2 \Rightarrow 4a=3 \Rightarrow a=\frac34.$$ Then, $$b=1-2\cdot \frac34=1-\frac32=-\frac12,$$ $$d=4a=3.$$ Therefore, $$A=\begin{bmatrix}\frac34&-\frac12\\[2pt]-\frac12&3\end{bmatrix}.$$ ## 3. Find $s$, the sum of diagonal elements $$s=a+d=\frac34+3=\frac{15}{4}.$$ ## 4. Compute $\alpha$ and $\beta$ The second row of $MA$ is $$\begin{bmatrix}3&\frac32\end{bmatrix}A=[\alpha,\beta].$$ So, $$\alpha=3a+\frac32 b, \qquad \beta=3b+\frac32 d.$$ Substitute $a=\frac34,\; b=-\frac12,\; d=3$: ### For $\alpha$ $$\alpha=3\cdot \frac34+\frac32\cdot \left(-\frac12\right) =\frac94-\frac34=\frac64=\frac32.$$ ### For $\beta$ $$\beta=3\left(-\frac12\right)+\frac32\cdot 3 =-\frac32+\frac92=3.$$ ## 5. Evaluate $\dfrac{\beta s}{\alpha^2}$ $$\frac{\beta s}{\alpha^2} =\frac{3\cdot \frac{15}{4}}{\left(\frac32\right)^2} =\frac{\frac{45}{4}}{\frac94} =\frac{45}{4}\cdot \frac{4}{9}=5.$$ Thus, the required integer is $$\boxed{5}.$$

Matrices and Determinants question

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