Let A be a 2 \times 2 symmetric matrix such that A[\beginarrayl1 1\endarray]=[\beginarrayl3 7\endarray] and the determinant of A be 1 . If A^-1=\alpha A+\beta I , where I is an identity matrix of order 2 \times 2 , then \alpha+\beta equals \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q53 with Solution

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Correct answer: 5

Let $A=\begin{pmatrix}a&b\\ b&d\end{pmatrix}$ since $A$ is a $2\times 2$ symmetric matrix.
Given $A\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}3\\7\end{pmatrix}$ so, $\begin{pmatrix}a+b\\ b+d\end{pmatrix}=\begin{pmatrix}3\\7\end{pmatrix}.$ Hence, $a+b=3, \qquad b+d=7.$ Thus, $a=3-b, \qquad d=7-b.$
Also, $\det(A)=1$ . Therefore, $ad-b^2=1.$ Substitute $a=3-b$ and $d=7-b$ : $(3-b)(7-b)-b^2=1.$ Expand: $21-10b+b^2-b^2=1$ $21-10b=1$ $10b=20$ $b=2.$ Then, $a=3-2=1, \qquad d=7-2=5.$ So, $A=\begin{pmatrix}1&2\\2&5\end{pmatrix}.$
Since $\det(A)=1$ , for a $2\times 2$ matrix, $A^{-1}=\begin{pmatrix}5&-2\\-2&1\end{pmatrix}.$
Now use $A^{-1}=\alpha A+\beta I.$ That is,

= \alpha\begin{pmatrix}1&2\\2&5\end{pmatrix} +\beta\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}\alpha+\beta&2\alpha\\2\alpha&5\alpha+\beta\end{pmatrix}.$$ 6. Compare corresponding entries: $$2\alpha=-2 \Rightarrow \alpha=-1.$$ Then, $$\alpha+\beta=5\quad \text{(from the top-left entry)}.$$ So the required value is $$\boxed{5}.$$ 7. Verification using Cayley-Hamilton: For any $2\times 2$ matrix, $$A^2-(\operatorname{tr}A)A+(\det A)I=0.$$ Since $\det A=1$, $$A^{-1}=(\operatorname{tr}A)I-A.$$ Here, $$\operatorname{tr}A=1+5=6,$$ so $$A^{-1}=-A+6I.$$ Thus, $$\alpha=-1,\ \beta=6 \Rightarrow \alpha+\beta=5,$$ which confirms the answer.

View written solutionFree

Correct answer: 5

Let $A=\begin{pmatrix}a&b\\ b&d\end{pmatrix}$ since $A$ is a $2\times 2$ symmetric matrix.
Given $A\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}3\\7\end{pmatrix}$ so, $\begin{pmatrix}a+b\\ b+d\end{pmatrix}=\begin{pmatrix}3\\7\end{pmatrix}.$ Hence, $a+b=3, \qquad b+d=7.$ Thus, $a=3-b, \qquad d=7-b.$
Also, $\det(A)=1$ . Therefore, $ad-b^2=1.$ Substitute $a=3-b$ and $d=7-b$ : $(3-b)(7-b)-b^2=1.$ Expand: $21-10b+b^2-b^2=1$ $21-10b=1$ $10b=20$ $b=2.$ Then, $a=3-2=1, \qquad d=7-2=5.$ So, $A=\begin{pmatrix}1&2\\2&5\end{pmatrix}.$
Since $\det(A)=1$ , for a $2\times 2$ matrix, $A^{-1}=\begin{pmatrix}5&-2\\-2&1\end{pmatrix}.$
Now use $A^{-1}=\alpha A+\beta I.$ That is,

= \alpha\begin{pmatrix}1&2\\2&5\end{pmatrix} +\beta\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}\alpha+\beta&2\alpha\\2\alpha&5\alpha+\beta\end{pmatrix}.$$ 6. Compare corresponding entries: $$2\alpha=-2 \Rightarrow \alpha=-1.$$ Then, $$\alpha+\beta=5\quad \text{(from the top-left entry)}.$$ So the required value is $$\boxed{5}.$$ 7. Verification using Cayley-Hamilton: For any $2\times 2$ matrix, $$A^2-(\operatorname{tr}A)A+(\det A)I=0.$$ Since $\det A=1$, $$A^{-1}=(\operatorname{tr}A)I-A.$$ Here, $$\operatorname{tr}A=1+5=6,$$ so $$A^{-1}=-A+6I.$$ Thus, $$\alpha=-1,\ \beta=6 \Rightarrow \alpha+\beta=5,$$ which confirms the answer.