If A=≤ft[{array}{cc}1 & -5 \\ -2 & 4{array}], then A^{-1}=

Correct Answer is : −16[4521]-1/6≤ft[{array}{ll}4 & 5 \\ 2 & 1{array}]−61[4251]

-1/6≤ft[{array}{cc}-4 & 5 \\ 2 & -1{array}]

1/14≤ft[{array}{cc}-1 & 5 \\ 2 & -4{array}]

1/14≤ft[{array}{ll}4 & 5 \\ 2 & 1{array}]

-1/6≤ft[{array}{ll}4 & 5 \\ 2 & 1{array}]

Correct Answer is : −16[4521]-1/6≤ft[{array}{ll}4 & 5 \\ 2 & 1{array}]−61[4251]

Test Index

MHTCET 20 Apr 2026 Shift 2 Paper

Question : 114 of 150

Marks: +1, -0

A=\left[\begin{array}{cc}1 & -5 \\ -2 & 4\end{array}\right]

A^{-1}=

Solution:

Concept:

For a

2 \times 2

matrix, the inverse is found using the determinant and the adjugate matrix.

Explanation:

Let

A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}

Then

A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

For

A = \begin{bmatrix} 1 & -5 \\ -2 & 4 \end{bmatrix}

, we have

a=1

b=-5

c=-2

d=4

Compute

|A| = ad - bc = (1)(4) - (-5)(-2) = 4 - 10 = -6

Substitute into the inverse formula:

A^{-1} = \frac{1}{-6} \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix} = -\frac{1}{6} \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}

This matches option D.

Answer:

-\frac{1}{6}\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]

(Option D).

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