Let A = |111| and B = 92−102112122132−142−152162172 , then the value of A'BA is:

A'BA=[1 1 1][92−102112122132−142−152162172][111]= [92+122−152−102+132+162 112−142+172][111] =[92+122−152−102+132+162+112−142+172]=[539]

Let A = $| \begin{matrix} 1 & 1 & 1 \end{matrix} |$ and B = $\begin{matrix} 9^{2} & - 1 0^{2} & 1 1^{2} \\ 1 2^{2} & 1 3^{2} & - 1 4^{2} \\ - 1 5^{2} & 1 6^{2} & 1 7^{2} \end{matrix}$ , then the value of A'BA is:

Option 1 - 1224

Option 2 - 1042

Option 3 -  540

Option 4 -  539

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Answered by

Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 4

Detailed Solution:

$A' B A = [1 1 1] [\begin{matrix} 9^{2} & - 1 0^{2} & 1 1^{2} \\ 1 2^{2} & 1 3^{2} & - 1 4^{2} \\ - 1 5^{2} & 1 6^{2} & 1 7^{2} \end{matrix}] [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}] =$

$[9^{2} + 1 2^{2} - 1 5^{2} - 1 0^{2} + 1 3^{2} + 1 6^{2} 1 1^{2} - 1 4^{2} + 1 7^{2}] [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$

$= [9^{2} + 1 2^{2} - 1 5^{2} - 1 0^{2} + 1 3^{2} + 1 6^{2} + 1 1^{2} - 1 4^{2} + 1 7^{2}] = [539]$

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Kindly consider the following figure

B = (I – adjA)⁵

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Kindly consider the following figure

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