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Matrices and Determinants question

2022 · 26 Jul · Shift 2 · Q22
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Matrices and Determinants question

2022 · 26 Jul · Shift 2 · Q22

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
 Let A=[111] and B=[92−102112122132−142−152162172], then the value of A′BA is: \text { Let } A=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] \text { and } B=\left[\begin{array}{ccc} 9^{2} & -10^{2} & 11^{2} \\ 12^{2} & 13^{2} & -14^{2} \\ -15^{2} & 16^{2} & 17^{2} \end{array}\right] \text {, then the value of } A^{\prime} B A \text { is: } Let A=​111​​ and B=​92122−152​−102132162​112−142172​​, then the value of A′BA is: 
  1. A
    1224
  2. B
    1042
  3. C
    540
  4. D
    539
View written solutionFree

Correct answer: D

  1. We are given A=[111]A=\begin{bmatrix}1\\1\\1\end{bmatrix}A=​111​​ and B=[92−102112122132−142−152162172].B=\begin{bmatrix}9^2 & -10^2 & 11^2\\ 12^2 & 13^2 & -14^2\\ -15^2 & 16^2 & 17^2\end{bmatrix}.B=​92122−152​−102132162​112−142172​​.

Here A′A'A′ means the transpose of AAA, so A′=[111].A' = \begin{bmatrix}1&1&1\end{bmatrix}.A′=[1​1​1​].

We need to find A′BA.A'BA.A′BA.

  1. First compute the numerical entries of BBB: 92=81,102=100,112=121,9^2=81,\quad 10^2=100,\quad 11^2=121,92=81,102=100,112=121, 122=144,132=169,142=196,12^2=144,\quad 13^2=169,\quad 14^2=196,122=144,132=169,142=196, 152=225,162=256,172=289.15^2=225,\quad 16^2=256,\quad 17^2=289.152=225,162=256,172=289.

So, B=[81−100121144169−196−225256289].B=\begin{bmatrix}81 & -100 & 121\\ 144 & 169 & -196\\ -225 & 256 & 289\end{bmatrix}.B=​81144−225​−100169256​121−196289​​.

  1. Now compute BABABA: BA=[81−100121144169−196−225256289][111].BA = \begin{bmatrix}81 & -100 & 121\\ 144 & 169 & -196\\ -225 & 256 & 289\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}.BA=​81144−225​−100169256​121−196289​​​111​​.

This gives row sums:

  • First row: 81−100+121=10281-100+121=10281−100+121=102
  • Second row: 144+169−196=117144+169-196=117144+169−196=117
  • Third row: −225+256+289=320-225+256+289=320−225+256+289=320

Hence, BA=[102117320].BA=\begin{bmatrix}102\\117\\320\end{bmatrix}.BA=​102117320​​.

  1. Now multiply by A′A'A′: A′BA=[111][102117320]=102+117+320.A'BA = \begin{bmatrix}1&1&1\end{bmatrix}\begin{bmatrix}102\\117\\320\end{bmatrix} = 102+117+320.A′BA=[1​1​1​]​102117320​​=102+117+320.

So, A′BA=539.A'BA = 539.A′BA=539.

  1. Comparing with the options:
  • A: 122412241224
  • B: 104210421042
  • C: 540540540
  • D: 539539539

Therefore, the correct option is D.\boxed{D}.D​.

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