Question

Topic: Matrices



✨Detailed solution please​

Answer 1

Answer:

d= A^2 = I

Step-by-step explanation:

A inverse exists bcz determinant of the given matrix is not equal to 0

A = A cannot be equal to unit matrix even after taking -1 common because matrix which has 1 on its principal diagonal and zeroes at other places is unit matrix. But in given matrix 1 are given on the second diagonal and not on the principal diagonal.

A is of course not a unit matrix

So option left is A^2 = unit matrix. It is true and you can check by solving.

Answer 2

Given:-

$A= \begin{bmatrix} 0& 0 & - 1\\ 0& - 1 & 0 \\ -1&0 & 0 \end{bmatrix}$

To Find:-

Which of the following is correct:-

1. A^-1 Doesn't Exist.
2. A=(-1)I
3. A is a Unit Matrix
4. A²=I

Solution:-

For Part(a)

for checking that inverse of a Matrix is exist or not , we need to Find it's Mod Value

i e,

$⟼|A|$

$|A| = 0( 0 - 0) - 0(0 - 0) - 1(0 - 1)$

$= > |A| = 0 - 0 + 1$

$= > |A| = 1$

So,

$∣A∣ ≠ 0$

$\therefore { ∣A∣}^{ - 1}$

exist.

For Part(b)

$A= \begin{bmatrix} 0& 0 & - 1\\ 0& - 1 & 0 \\ -1&0 & 0 \end{bmatrix}$

The Given Matrix is in the Order of 3 x 3

So,

$I= \begin{bmatrix} 1& 0 & 0\\ 0& 1 & 0 \\ 0&0 & 1 \end{bmatrix}$

We Can Also Write Matrix A as:

$= > A= - 1 \begin{bmatrix} 0& 0 & 1\\ 0& 1 & 0 \\ 1&0 & 0 \end{bmatrix}$

$\therefore \: A≠-I$

For Part(c)

Given Matrix is

$A= \begin{bmatrix} 0& 0 & - 1\\ 0& - 1 & 0 \\ -1&0 & 0 \end{bmatrix}$

The Given Matrix is in 3 x 3 Order

So, Unit Matrix of Order 3 x 3 is

$I= \begin{bmatrix} 1& 0 & 0\\ 0& 1 & 0 \\ 0&0 & 1 \end{bmatrix}$

$\therefore \: A≠I$

For Part(d)

Given Matrix is

$A= \begin{bmatrix} 0& 0 & - 1\\ 0& - 1 & 0 \\ -1&0 & 0 \end{bmatrix}$

$= > A²=A \times A$

$= > {A}^{2} = { \begin{bmatrix} 0& 0 & - 1\\ 0& - 1 & 0 \\ -1&0 & 0 \end{bmatrix}}^{2}$

$= > \begin{bmatrix} 0& 0 & - 1\\ 0& - 1 & 0 \\ -1&0 & 0 \end{bmatrix} \times \begin{bmatrix} 0& 0 & - 1\\ 0& - 1 & 0 \\ -1&0 & 0 \end{bmatrix}$

$= > \begin{bmatrix} 0 + 0 + 1& 0 + 0 + 0 & 0 + 0 + 0\\ 0 + 0 + 0& 0 + 1 + 0& 0 + 0 + 0 \\ 0 + 0 + 0&0 + 0 + 0& 1 + 0 + 0 \end{bmatrix}$

$= > \begin{bmatrix} 1& 0 & 0\\ 0& 1 & 0 \\ 0&0 & 1 \end{bmatrix}$

As it is equal to Unit Matrix

$= > I$

$\therefore \: A²=I$

Therefore, Option (d) A²=I is Correct answer