Q47 / 75 · JEE Main 2026 (02 April Shift 2) · Mathematics
Consider the matrices $A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}$. If matrices P and Q are such that $PA = B$ and $AQ = B$, then the absolute value of the sum of the diagonal elements of $2(P + Q)$ is _______.
Correct answer: 34
Solution
Given $A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}$, we find its determinant:
$|A| = (2)(-2) - (-2)(4) = 4$
Since $|A| \neq 0$, $A$ is invertible. The inverse of $A$ is:
$A^{-1} = \dfrac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$
We are given $PA = B \Rightarrow P = BA^{-1}$ and $AQ = B \Rightarrow Q = A^{-1}B$.
Using the cyclic property of trace $\text{Tr}(XY) = \text{Tr}(YX)$, we have:
$\text{Tr}(P) = \text{Tr}(BA^{-1}) = \text{Tr}(A^{-1}B) = \text{Tr}(Q)$
Let us calculate $\text{Tr}(P)$ by finding the diagonal elements of $P = BA^{-1}$:
$P = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$
The diagonal elements of $P$ are:
$P_{11} = 3\left(-\dfrac{1}{2}\right) + 9(-1) = -\dfrac{21}{2}$
$P_{22} = 1\left(\dfrac{1}{2}\right) + 3\left(\dfrac{1}{2}\right) = 2$
$\text{Tr}(P) = P_{11} + P_{22} = -\dfrac{21}{2} + 2 = -\dfrac{17}{2}$
Since $\text{Tr}(P) = \text{Tr}(Q) = -\dfrac{17}{2}$, the sum of the diagonal elements of $2(P + Q)$ is:
$\text{Tr}(2(P + Q)) = 2(\text{Tr}(P) + \text{Tr}(Q)) = 2\left(-\dfrac{17}{2} - \dfrac{17}{2}\right) = 2(-17) = -34$
The absolute value is $|-34| = 34$.
Answer: $34$
$|A| = (2)(-2) - (-2)(4) = 4$
Since $|A| \neq 0$, $A$ is invertible. The inverse of $A$ is:
$A^{-1} = \dfrac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$
We are given $PA = B \Rightarrow P = BA^{-1}$ and $AQ = B \Rightarrow Q = A^{-1}B$.
Using the cyclic property of trace $\text{Tr}(XY) = \text{Tr}(YX)$, we have:
$\text{Tr}(P) = \text{Tr}(BA^{-1}) = \text{Tr}(A^{-1}B) = \text{Tr}(Q)$
Let us calculate $\text{Tr}(P)$ by finding the diagonal elements of $P = BA^{-1}$:
$P = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$
The diagonal elements of $P$ are:
$P_{11} = 3\left(-\dfrac{1}{2}\right) + 9(-1) = -\dfrac{21}{2}$
$P_{22} = 1\left(\dfrac{1}{2}\right) + 3\left(\dfrac{1}{2}\right) = 2$
$\text{Tr}(P) = P_{11} + P_{22} = -\dfrac{21}{2} + 2 = -\dfrac{17}{2}$
Since $\text{Tr}(P) = \text{Tr}(Q) = -\dfrac{17}{2}$, the sum of the diagonal elements of $2(P + Q)$ is:
$\text{Tr}(2(P + Q)) = 2(\text{Tr}(P) + \text{Tr}(Q)) = 2\left(-\dfrac{17}{2} - \dfrac{17}{2}\right) = 2(-17) = -34$
The absolute value is $|-34| = 34$.
Answer: $34$
Hence, the correct answer is 34.