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Given,

A = \(\begin{bmatrix} ab & b^2 \\[0.3em] -a^2 & -ab \\[0.3em] \end{bmatrix}\)

A² = \(\begin{bmatrix} ab & b^2 \\[0.3em] -a^2 & -ab \\[0.3em] \end{bmatrix}\)\(\begin{bmatrix} ab & b^2 \\[0.3em] -a^2 & -ab \\[0.3em] \end{bmatrix}\)

= \(\begin{bmatrix} a^2b^2-a^2b^2 & ab^3-ab^3 \\[0.3em] -a^3b+a^3b & -a^2b^2+a^2b^2 \\[0.3em] \end{bmatrix}\) 

= \(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}\) 

= 0

Hence,

A² = 0

True

∵If A and B are any two matrices for which AB is defined, then

(AB)’=B’A’.

X has a + b rows and a + 2 columns.

⇒ Order of X = (a + b) × (a + 2)

Y has b + 1 rows and a + 3 columns.

⇒ Order of Y = (b + 1) × (a + 3)

Recall that the product of two matrices A and B is defined only when the number of columns of A is equal to the number of rows of B.

It is given that the matrix XY exists.

⇒ Number of columns of X = Number of rows of Y

⇒ a + 2 = b + 1 

∴ a = b – 1 

The matrix YX also exists.

⇒ Number of columns of Y = Number of rows of X 

⇒ a + 3 = a + b 

∴ b = 3 

We have, 

a = b – 1 

⇒ a = 3 – 1 

∴ a = 2 

Thus,

 a = 2 and b = 3. 

Hence, 

Order of X = 5 × 4 

And,

Order of Y = 4 × 5.

Order of XY = Number of rows of X × Number of columns of Y 

⇒ Order of XY = 5 × 5 

Order of YX = Number of rows of Y × Number of columns of X 

⇒ Order of XY = 4 × 4 

As the orders of the two matrices XY and YX are different, they are not of the same type and thus unequal.

Number of entries = (Number of rows) x (Number of columns) = 18

If order is (a x b) then, Number of entries = a x b

So now a x b = 18 (in this case)

Possible cases are (1 x 18), (2 x 9), (3 x 6), (6 x 3), (9 x 2), (18 x 1)

Conclusion: If a matrix has 18 elements, then possible orders

are (1 x 18), (2 x 9), (3 x 6), (6 x 3), (9 x 2), (18 x 1)

Answer is (B)

X is of order 2 x n

Z is of order 2 x p

2 x n = 2 x p

given n = p

Answer is (A)

[P]p x k [Y]3x k + [W]n x 3[Y]3x k

[P]{y} is possible only when k = 3

[P][Y] + [W][Y] is possible only

when the order of [P][y] = order of [W][Y]

P x K = n x k

∴ P = n

We are given that,

[2 ,1 ,3]\(\begin{bmatrix}-1& 0 & -1 \\[0.3em]-1 &1 &0 \\[0.3em]0 &1 &1\end{bmatrix}\)\(\begin{bmatrix}1 \\[0.3em]0 \\[0.3em]-1\end{bmatrix}\)= A

We need to find the order of the matrix A. 

Let the matrices be,

X = [2 ,1 ,3]

Y = \(\begin{bmatrix}-1& 0 & -1 \\[0.3em]-1 &1 &0 \\[0.3em]0 &1 &1\end{bmatrix}\)

Z = \(\begin{bmatrix}1 \\[0.3em]0 \\[0.3em]-1\end{bmatrix}\)

Let us find the order of X. 

Number of rows of matrix X = 1 

Number of columns of matrix X = 3 

So, 

Order of matrix X = 1 × 3 …(i) 

Now, 

let us find the order of Y. 

Number of rows of matrix Y = 3 

Number of columns of matrix Y = 3 

So, 

Order of matrix Y = 3 × 3 …(ii) 

From (i) and (ii), 

Order of resulting XY = 1 × 3 

[∵ Number of columns of X = Number of rows of Y] …(iii) 

Let us find the order of Z. 

Number of rows of matrix Z = 3 

Number of columns of matrix Z = 1 

So, 

Order of matrix Z = 3 × 1 …(iv)

Order of resulting XYZ = 1 × 1 

[∵ Number of columns of XY = Number of rows of Z] 

Thus, 

The order of matrix A = 1 × 1

We are given with the information that, 

Each element of the 2 × 2 matrix can be filled in 3 ways, either 1, 2 or 3. 

We need to find the number of total 2 × 2 matrices with each entry 1, 2 or 3. 

Let A be 2 × 2 matrix such that,

A =\(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)

Note that, 

There are 4 elements in the matrix.

So, 

If 1 element can be filled in 3 ways, either 1, 2 or 3.

That is, 

Number of ways in which 1 element can be filled = 3¹ 

Then, 

Number of ways in which 4 elements can be filled = 3⁴ 

⇒ Number of ways in which 4 elements can be filled = 81 

Thus, 

Total number of 2 × 2 matrices with each entry 1, 2 or 3 is 81.

We are given that,

A = \(\begin{bmatrix} 0 & 2b & -2 \\[0.3em] 3& 1 &3 \\[0.3em] 3a & 3 &-1 \end{bmatrix}\)is symmetric matrix.

We need to find the values of a and b. 

We must understand what symmetric matrix is. 

A symmetric matrix is a square matrix that is equal to its transpose. 

A symmetric matrix ⇔ A = A^T 

This means, 

We need to find the transpose of matrix A. 

Let us take transpose of the matrix A. 

We know that, 

The transpose of a matrix is a new matrix whose rows are the columns of the original. 

We have, 

1^st row of matrix A = (0 ,2b, -2) 

2^nd row of matrix A = (3,1, 3) 

3^rd row of matrix A = (3a ,3 ,-1) 

For matrix A^T, it will become 

1^st column of A^T = 1^st row of A = (0 ,2b,-2)

2^nd column of A^T = 2^nd row of A = (3,1 ,3) 

3^rd column of A T = 3^rd row of A = (3a,3 ,-1)

∴ A^T = \(\begin{bmatrix} 0 & 3 & 3a \\[0.3em] 2b& 1 &3 \\[0.3em] -2 & 3 &-1 \end{bmatrix}\)

Now, 

as  A = A^T.

Substituting the matrices A and A^T, we get

\(\begin{bmatrix} 0 & 2b & -2 \\[0.3em] 3& 1 &3 \\[0.3em] 3a & 3 &-1 \end{bmatrix}\)= \(\begin{bmatrix} 0 & 3 & 3a \\[0.3em] 2b& 1 &3 \\[0.3em] -2 & 3 &-1 \end{bmatrix}\)

We know by the property of matrices,

 \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies, 

a₁₁ = b₁₁, 

a₁₂ = b₁₂, 

a₂₁ = b₂₁ and

 a₂₂ = b₂₂ 

Applying this property, we can write

2b = 3 …(i) 

-2 = 3a …(ii) 

3 = 2b 

3a = -2 

We can find a and b from equations (i) and (ii).

From equation (i),

2b = 3

⇒ b = \(\frac{ 3}{2}\)

From equation (ii), 

- 2 = 3a

 ⇒ a = \(-\frac{ 2}{3}\)

Thus, we get

 a = \(-\frac{ 2}{3}\) and  b = \(\frac{ 3}{2}\).