Inverse of the matrix ≤ft[{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8{array}] is

Correct Answer is : [0.80.6−0.60.8]≤ft[{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8{array}][0.8−0.60.60.8]

≤ft[{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8{array}]

≤ft[{array}{ll}-0.8 & 0.6 \\ -0.6 & 0.8{array}]

≤ft[{array}{cc}-0.8 & -0.6 \\ 0.6 & 0.8{array}]

≤ft[{array}{cc}8 & -6 \\ 6 & 8{array}]

Correct Answer is : [0.80.6−0.60.8]≤ft[{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8{array}][0.8−0.60.60.8]

Test Index

MHTCET 10 May 2024 Shift 2 Paper

Question : 127 of 150

Marks: +1, -0

Inverse of the matrix

\left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]

Solution:

Concept:

For a

2 \times 2

matrix, the inverse is found by swapping the diagonal elements, changing the signs of the off-diagonal elements, and dividing by the determinant.

Explanation:

Let

A = \begin{bmatrix} 0.8 & -0.6 \\ 0.6 & 0.8 \end{bmatrix}

First, compute the determinant:

|A| = (0.8)(0.8) - (-0.6)(0.6) = 0.64 + 0.36 = 1

Since

|A| = 1 \neq 0

, the inverse exists.

For a

2 \times 2

matrix

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

, the inverse is:

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

Substitute

a = 0.8

b = -0.6

c = 0.6

d = 0.8

A^{-1} = \frac{1}{1} \begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix} = \begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix}

This matches option A.

Answer:

\begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix}

i.e., Option A.

Go to Question:

Prev Question

Next Question