If A≤ft[{array}{ll}2 & 1 \\ 7 & 4{array}] then ≤ft(A²-5 A)^{-1} is

Correct Answer is : (14)[−317−1]≤ft(1/4)≤ft[{array}{cc}-3 & 1 \\ 7 & -1{array}](41)[−371−1]

≤ft(-1/4)≤ft[{array}{cc}-3 & 1 \\ 7 & -1{array}]

≤ft(1/4)≤ft[{array}{cc}-3 & 1 \\ 7 & -1{array}]

≤ft(1/4)≤ft[{array}{ll}3 & 1 \\ 7 & 1{array}]

≤ft(1/-4)≤ft[{array}{ll}3 & -1 \\ 7 & -1{array}]

Correct Answer is : (14)[−317−1]≤ft(1/4)≤ft[{array}{cc}-3 & 1 \\ 7 & -1{array}](41)[−371−1]

Test Index

MHTCET 11 May 2024 Shift 2 Paper

Question : 141 of 150

Marks: +1, -0

A\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]

\left(A^{2}-5 A\right)^{-1}

$\left(-\frac{1}{4}\right)\left[\begin{array}{cc}-3 & 1 \\ 7 & -1\end{array}\right]$
$\left(\frac{1}{4}\right)\left[\begin{array}{cc}-3 & 1 \\ 7 & -1\end{array}\right]$
$\left(\frac{1}{4}\right)\left[\begin{array}{ll}3 & 1 \\ 7 & 1\end{array}\right]$
$\left(\frac{1}{-4}\right)\left[\begin{array}{ll}3 & -1 \\ 7 & -1\end{array}\right]$

Solution:

Concept:

For a

2 \times 2

matrix,

A^{-1} = \frac{1}{|A|} \text{adj}(A)

First compute

A^2 - 5A

, then find its inverse.

Explanation:

Given

A = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}

Compute

A^2

A^2 = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix} = \begin{bmatrix} 11 & 6 \\ 42 & 23 \end{bmatrix}

Then

5A = \begin{bmatrix} 10 & 5 \\ 35 & 20 \end{bmatrix}

A^2 - 5A = \begin{bmatrix} 1 & 1 \\ 7 & 3 \end{bmatrix}

Its determinant is

|A^2 - 5A| = (1)(3) - (1)(7) = -4

The inverse is:

\left(A^2 - 5A\right)^{-1} = \frac{1}{-4} \begin{bmatrix} 3 & -1 \\ -7 & 1 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -3 & 1 \\ 7 & -1 \end{bmatrix}

This matches option B.

Answer:

\left(\frac{1}{4}\right)\begin{bmatrix} -3 & 1 \\ 7 & -1 \end{bmatrix}

Option B.

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