Let A=≤ft[{array}{cc}x & 1 \\ 1 & 0{array}], x {R}^{+}and A^{4}=≤ft[a_{i j}]₂. If a₁1}=109, then ≤ft(A^{4})^{-1}=

Correct Answer is : [10−33−33109]≤ft[{array}{cc}10 & -33 \\ -33 & 109{array}][10−33−33109]

≤ft[{array}{rr}109 & 33 \\ 33 & 10{array}]

≤ft[{array}{ll}10 & 33 \\ 33 & 10{array}]

≤ft[{array}{cc}10 & 33 \\ 33 & 109{array}]

≤ft[{array}{cc}10 & -33 \\ -33 & 109{array}]

Correct Answer is : [10−33−33109]≤ft[{array}{cc}10 & -33 \\ -33 & 109{array}][10−33−33109]

Test Index

MHTCET 16 May 2024 Shift 1 Paper

Question : 101 of 150

Marks: +1, -0

Let

A=\left[\begin{array}{cc}x & 1 \\ 1 & 0\end{array}\right], x \in \mathbb{R}^{+}

A^{4}=\left[a_{i j}\right]_{2}

a_{11}=109

\left(A^{4}\right)^{-1}=

Solution:

Concept:

Matrix multiplication and inverse of a

2 \times 2

matrix.

Explanation:

First, compute

A^2 = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} x^2+1 & x \\ x & 1 \end{bmatrix}.

Now,

A^4 = A^2 A^2 = \begin{bmatrix} (x^2+1)^2+x^2 & x(x^2+2) \\ x(x^2+2) & x^2+1 \end{bmatrix}.

Given

a_{11}=109

, so

(x^2+1)^2+x^2=109.

Let

x^2=y

. Then

y^2+3y+1=109

, giving

y^2+3y-108=0

Thus

y=9

y=-12

. Since

x \in \mathbb{R}^+

x^2=9

, so

x=3

Therefore,

A^4 = \begin{bmatrix} 109 & 33 \\ 33 & 10 \end{bmatrix}.

Its determinant is

109 \times 10 - 33 \times 33 = 1090 - 1089 = 1

For a

2 \times 2

matrix,

A^{-1} = \frac{1}{\det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

Hence,

(A^4)^{-1} = \begin{bmatrix} 10 & -33 \\ -33 & 109 \end{bmatrix}.

Answer:

\begin{bmatrix} 10 & -33 \\ -33 & 109 \end{bmatrix}

Option D.

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