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UPSC NDA Math Model Paper 9

Question : 115 of 120

Marks: +1, -0

\begin{bmatrix} 1 & a & b \\ 0 & x & 0 \\ 0 & 0 & 1 \end{bmatrix}\;\text{is}

\begin{bmatrix} 1 & -a & -b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

then x is

Solution:

\text{Let }A = \begin{bmatrix} 1 & a & b \\ 0 & x & 0 \\ 0 & 0 & 1 \end{bmatrix}

So,

A^{-1} = \begin{bmatrix} 1 & -a & -b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

We know that,

A A^{-1} = I

Taking Determinant both sides,

\det(A A^{-1}) = \det(I)

We know that Determinant of an Identity matrix is 1.

\Rightarrow |A||A^{-1}| = 1

\Rightarrow \begin{vmatrix} 1 & a & b \\ 0 & x & 0 \\ 0 & 0 & 1 \end{vmatrix}

\begin{vmatrix} 1 & -a & -b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = \;1

Now, Expanding along

C_3

⇒ [0 – 0 + 1(x – 0)] [0 – 0 + 1(1 – 0)] = 1

⇒ x = 1

∴ Option 3 is correct.

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