2×2 Matrix A With A’=αA+I and det(A²−A)=4, Sum of All Possible α

JEE Main 2023. Previous Year Question — exact date and shift not confidently identified.

Problem

Let $A$ be a $2\times2$ matrix with real entries such that $A’ = \alpha A + I$, where $\alpha \in \mathbb{R} – \{-1, 1\}$. If $\det(A^2 – A) = 4$, then the sum of all possible values of $\alpha$ is equal to:

Key insight. Writing out $A’ = \alpha A + I$ entry-by-entry shows the off-diagonal entries must vanish (since $\alpha \neq \pm1$) — meaning $A$ is forced to be a scalar multiple of the identity matrix. Once $A = aI$, the determinant condition becomes a plain one-variable equation.

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Approach

Write $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$, so $A’ = \begin{bmatrix}a&c\\b&d\end{bmatrix}$. Matching this against $\alpha A + I$ entry-by-entry gives four equations. Two of them force $b$ and $c$ to be zero (using $\alpha \neq \pm1$), collapsing $A$ down to a simple diagonal matrix — after which the determinant condition is just algebra in one variable.

Solution

Step 1 — Match entries of A’ and αA + I

$$\begin{bmatrix} a & c \\ b & d \end{bmatrix} = \alpha\begin{bmatrix} a & b \\ c & d \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \alpha a + 1 & \alpha b \\ \alpha c & \alpha d + 1 \end{bmatrix}$$

Comparing entries: $a = \alpha a + 1$, $\;c = \alpha b$, $\;b = \alpha c$, $\;d = \alpha d + 1$.

Step 2 — Show A must be diagonal

From $b = \alpha c$ and $c = \alpha b$: substituting one into the other gives $b = \alpha(\alpha b) = \alpha^2 b$, so $b(1-\alpha^2) = 0$. Since $\alpha \neq \pm1$, $1-\alpha^2 \neq 0$, forcing $b = 0$ — and then $c = \alpha b = 0$ too.

From $a = \alpha a + 1$: $a(1-\alpha) = 1 \implies a = \dfrac{1}{1-\alpha}$. The same equation gives $d = \dfrac{1}{1-\alpha}$, so $a = d$.

So $A = aI$ where $a = \dfrac{1}{1-\alpha}$.

Step 3 — Apply the determinant condition

$$A^2 – A = a^2I – aI = a(a-1)I$$

For a scalar multiple $kI$ of the $2\times2$ identity, $\det(kI) = k^2$, so:

$$\det(A^2-A) = \big[a(a-1)\big]^2 = 4 \implies a(a-1) = \pm2 \implies a^2 – a \mp 2 = 0$$

Step 4 — Solve both cases

Case $a^2 – a – 2 = 0$: factorises as $(a-2)(a+1) = 0$, giving $a = 2$ or $a = -1$ — both real.

Case $a^2 – a + 2 = 0$: discriminant $= 1 – 8 = -7 < 0$ — no real solutions.

So only $a = 2$ and $a = -1$ are valid.

Step 5 — Convert back to α

Since $a = \dfrac{1}{1-\alpha}$, we get $\alpha = 1 – \dfrac{1}{a}$:

$$a = 2 \implies \alpha = 1 – \frac{1}{2} = \frac{1}{2}, \qquad a = -1 \implies \alpha = 1 – \frac{1}{-1} = 2$$

Sum of all possible values of $\alpha$:

$$\frac{1}{2} + 2 = \frac{5}{2}$$

Answer

Common mistakes

• Not checking that $\alpha \neq \pm1$ is what forces $b=c=0$. Without invoking this restriction explicitly, it’s tempting to assume $A$ could be a general (non-diagonal) matrix, which makes the determinant condition far messier to handle.
• Forgetting to discard the case with negative discriminant. Both $a^2-a-2=0$ and $a^2-a+2=0$ come from squaring $a(a-1)=\pm2$, but only one of the two actually has real roots — skipping this check risks including invalid, non-real values of $\alpha$.

Practise next

• Let $A$ be a $2\times2$ real matrix with $A’ = \alpha A + 2I$, $\alpha \in \mathbb{R}-\{-1,1\}$. If $\det(A^2-A) = 9$, find the sum of all possible values of $\alpha$.