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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Mathematics
Chapter
Matrices

JEE Advanced 2011 · Paper 2 · Mathematics · Matrices

Let M be a 3 × 3 matrix satisfying M[ array l 0 1 0 array ]= [...

 Let \text{ Let }M  be a \text{ be a }3 ×\times3  matrix satisfying \text{ matrix satisfying }M\left[010\begin{array}{l}0 \\ 1 \\ 0 \end{array}\right]=\left[−123\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right], M\left[1−10\begin{array}{c}1 \\ -1 \\ 0 \end{array}\right]=\left[11−1\begin{array}{c}1 \\ 1 \\ -1 \end{array}\right]\text{ and }M\left[111\begin{array}{l}1 \\ 1 \\ 1 \end{array}\right]=\left[0012\begin{array}{c}0 \\ 0 \\ 12 \end{array}\right] Then, the sum of the diagonal entries of MM is

Correct answer

9

Step-by-step solution

Let M=\left[a1a2a3b1b2b3c1c2c3\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right]\therefore M\left[010\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[−123\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right], M\left[1−10\begin{array}{r}1 \\ -1 \\ 0\end{array}\right]=\left[11−1\begin{array}{r}1 \\ 1 \\ -1\end{array}\right], M\left[111\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[0012\begin{array}{r}0 \\ 0 \\ 12\end{array}\right], ⇒[a2b2c2]=[−123],[a1−a2b1−b2c1−c2]=[11−1],[a1+a2+a3b1+b2+b3c1+c2+c3]=[0012]⇒a2=−1,b2=2,c2=3,a1−a2=1,⇒b1−b2=1,c1−c2=−1⇒a1+a2+a3=0,b1+b2+b3=0,∴c1+c2+c3=12∴a1=0,b2=2 and c3=7\begin{aligned} \Rightarrow& {\left[\begin{array}{l}a_2 \\ b_2 \\ c_2 \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ 3 \end{array}\right],\left[\begin{array}{l}a_1-a_2 \\ b_1-b_2 \\ c_1-c_2 \end{array}\right]=\left[\begin{array}{r}1 \\ 1 \\ -1 \end{array}\right], } \\ & {\left[\begin{array}{l}a_1+a_2+a_3 \\ b_1+b_2+b_3 \\ c_1+c_2+c_3 \end{array}\right]=\left[\begin{array}{r}0 \\ 0 \\ 12 \end{array}\right] } \\ \Rightarrow & a_2=-1, b_2=2, c_2=3, a_1-a_2=1, \\ \Rightarrow & b_1-b_2=1, c_1-c_2=-1 \\ \Rightarrow \quad & a_1+a_2+a_3=0, b_1+b_2+b_3=0, \\ \therefore & c_1+c_2+c_3=12 \\ \therefore & a_1=0, b_2=2 \text{ and }c_3=7 \end{aligned}Hence, sum of diagonal elements =0+2+7=9
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