Inverse of ≤ft[{array}{cc}3 & -2 \\ 1 & 4{array}] is

Correct Answer is : [2717−114314]≤ft[{array}{cc}2/7 & 1/7 \\ -1/14 & 3/14{array}][72−14171143]

≤ft[{array}{cc}2/7 & 1/7 \\ -1/14 & 3/14{array}]

≤ft[{array}{cc}3/14 & -1/7 \\ 1/14 & 2/7{array}]

≤ft[{array}{cc}2/7 & -1/7 \\ 1/14 & 3/14{array}]

≤ft[{array}{cc}-3/14 & 1/7 \\ 1/14 & 2/7{array}]

Correct Answer is : [2717−114314]≤ft[{array}{cc}2/7 & 1/7 \\ -1/14 & 3/14{array}][72−14171143]

Test Index

MHTCET 16 Apr 2026 Shift 1 Paper

Question : 114 of 150

Marks: +1, -0

Inverse of

\left[\begin{array}{cc}3 & -2 \\ 1 & 4\end{array}\right]

$\left[\begin{array}{cc}\frac{2}{7} & \frac{1}{7} \\ -\frac{1}{14} & \frac{3}{14}\end{array}\right]$
$\left[\begin{array}{cc}\frac{3}{14} & -\frac{1}{7} \\ \frac{1}{14} & \frac{2}{7}\end{array}\right]$
$\left[\begin{array}{cc}\frac{2}{7} & -\frac{1}{7} \\ \frac{1}{14} & \frac{3}{14}\end{array}\right]$
$\left[\begin{array}{cc}-\frac{3}{14} & \frac{1}{7} \\ \frac{1}{14} & \frac{2}{7}\end{array}\right]$

Solution:

Concept:

The inverse of a

2 \times 2

matrix is found using its determinant and the adjugate matrix.

Explanation:

For a matrix

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

, the inverse is given by:

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

Here,

A = \begin{bmatrix} 3 & -2 \\ 1 & 4 \end{bmatrix}

So,

a = 3

b = -2

c = 1

, and

d = 4

First, compute the determinant:

|A| = (3)(4) - (-2)(1) = 12 + 2 = 14

Now apply the formula:

A^{-1} = \frac{1}{14} \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix}

Multiply each element by

\frac{1}{14}

A^{-1} = \begin{bmatrix} \frac{4}{14} & \frac{2}{14} \\ -\frac{1}{14} & \frac{3}{14} \end{bmatrix} = \begin{bmatrix} \frac{2}{7} & \frac{1}{7} \\ -\frac{1}{14} & \frac{3}{14} \end{bmatrix}

This matches Option A.

Answer:

The correct option is

\left[\begin{array}{cc}\frac{2}{7} & \frac{1}{7} \\ -\frac{1}{14} & \frac{3}{14}\end{array}\right]

, i.e., Option A.

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