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Matrix Equations PA=B, AQ=B: Diagonal Sum |−34| = 34 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Consider the matrices A=[2−24−2]A=\begin{bmatrix}2&-2\\4&-2\end{bmatrix} and B=[3913]B=\begin{bmatrix}3&9\\1&3\end{bmatrix}. If matrices PP and QQ are such that PA=BPA=B and AQ=BAQ=B, then the absolute value of the sum of the diagonal elements of 2(P+Q)2(P+Q) is
Solution
Answer: 34 (± 0.01)
Consider the matrices A=[2−24−2]A=\begin{bmatrix}2&-2\\4&-2\end{bmatrix} and B=[3913]B=\begin{bmatrix}3&9\\1&3\end{bmatrix}. If matrices PP and QQ are such that PA=BPA=B and AQ=BAQ=B, then the absolute value of the sum of the diagonal elements of 2(P+Q)2(P+Q) is ______. Step 1: Compute A−1A^{-1}. det⁡A=2(−2)−(−2)(4)=−4+8=4\det A=2(-2)-(-2)(4)=-4+8=4, so
A−1=14[−22−42]=12[−11−21].A^{-1}=\frac{1}{4}\begin{bmatrix}-2&2\\-4&2\end{bmatrix}=\frac12\begin{bmatrix}-1&1\\-2&1\end{bmatrix}.
 Step 2: From PA=BPA=B, P=BA−1P=BA^{-1}:
P=[3913]⋅12[−11−21]=12[−2112−74].P=\begin{bmatrix}3&9\\1&3\end{bmatrix}\cdot\frac12\begin{bmatrix}-1&1\\-2&1\end{bmatrix}=\frac12\begin{bmatrix}-21&12\\-7&4\end{bmatrix}.
 Step 3: From AQ=BAQ=B, Q=A−1BQ=A^{-1}B:
Q=12[−11−21][3913]=12[−2−6−5−15].Q=\frac12\begin{bmatrix}-1&1\\-2&1\end{bmatrix}\begin{bmatrix}3&9\\1&3\end{bmatrix}=\frac12\begin{bmatrix}-2&-6\\-5&-15\end{bmatrix}.
 Step 4: 2(P+Q)=[−2112−74]+[−2−6−5−15]=[−236−12−11].2(P+Q)=\begin{bmatrix}-21&12\\-7&4\end{bmatrix}+\begin{bmatrix}-2&-6\\-5&-15\end{bmatrix}=\begin{bmatrix}-23&6\\-12&-11\end{bmatrix}. Step 5: Sum of diagonal elements =−23+(−11)=−34=-23+(-11)=-34; its absolute value is
∣−34∣=34.|-34|=34.
 Correct answer: 34
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