JEE Main 2026 — Mathematics Algebra

Given $A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}$A=[24​−2−2​], we find its determinant:

$$|A| = (2)(-2) - (-2)(4) = 4$$∣A∣=(2)(−2)−(−2)(4)=4

Since $|A| \neq 0$∣A∣=0, $A$A is invertible. The inverse of $A$A is:

$$A^{-1} = \dfrac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$$A−1=41​[−2−4​22​]=[−1/2−1​1/21/2​]

We are given $PA = B \Rightarrow P = BA^{-1}$PA=B⇒P=BA−1 and $AQ = B \Rightarrow Q = A^{-1}B$AQ=B⇒Q=A−1B.

Using the cyclic property of trace $\text{Tr}(XY) = \text{Tr}(YX)$Tr(XY)=Tr(YX), we have:

$$\text{Tr}(P) = \text{Tr}(BA^{-1}) = \text{Tr}(A^{-1}B) = \text{Tr}(Q)$$Tr(P)=Tr(BA−1)=Tr(A−1B)=Tr(Q)

Let us calculate $\text{Tr}(P)$Tr(P) by finding the diagonal elements of $P = BA^{-1}$P=BA−1:

$$P = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$$P=[31​93​][−1/2−1​1/21/2​]

The diagonal elements of $P$P are:

$$P_{11} = 3\left(-\dfrac{1}{2}\right) + 9(-1) = -\dfrac{21}{2}$$P11​=3(−21​)+9(−1)=−221​

$$P_{22} = 1\left(\dfrac{1}{2}\right) + 3\left(\dfrac{1}{2}\right) = 2$$P22​=1(21​)+3(21​)=2

$$\text{Tr}(P) = P_{11} + P_{22} = -\dfrac{21}{2} + 2 = -\dfrac{17}{2}$$Tr(P)=P11​+P22​=−221​+2=−217​

Since $\text{Tr}(P) = \text{Tr}(Q) = -\dfrac{17}{2}$Tr(P)=Tr(Q)=−217​, the sum of the diagonal elements of $2(P + Q)$2(P+Q) is:

$$\text{Tr}(2(P + Q)) = 2(\text{Tr}(P) + \text{Tr}(Q)) = 2\left(-\dfrac{17}{2} - \dfrac{17}{2}\right) = 2(-17) = -34$$Tr(2(P+Q))=2(Tr(P)+Tr(Q))=2(−217​−217​)=2(−17)=−34

The absolute value is $|-34| = 34$∣−34∣=34.

Answer: $34$34