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# NCERT Exemplar - Matrices(Part-1) Commerce Notes | EduRev

## Commerce : NCERT Exemplar - Matrices(Part-1) Commerce Notes | EduRev

The document NCERT Exemplar - Matrices(Part-1) Commerce Notes | EduRev is a part of the Commerce Course JEE Revision Notes.
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Q.1. If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?
Ans.
The possible orders that a matrix having 28 elements are {28 Ã— 1, 1 Ã— 28, 2 Ã— 14, 14 Ã— 2, 4 Ã— 7, 7 Ã— 4}. The possible orders of a matrix having 13 elements are {1 Ã— 13, 13 Ã— 1}.

Q.2. In the matrix A = write:
(i) The order of the matrix A
(ii) The number of elements
(iii) Write elements  a23, a31, a12
Ans.
(i) The order of the given matrix A is 3 Ã— 3
(ii) The number of elements in matrix A = 3 Ã— 3 = 9
(iii) aij = the elements of ith row and jth column.
So, a23 = x2 â€“ y, a31 = 0, a12 = 1.

Q.3. Construct a2 Ã— 2 matrix where

(ii) aij = |âˆ’2i + 3j|
Ans.
Let

(i) Given that

Hence, the matrix A =

(ii) Given that

Hence, the matrix A =

Q.4. Construct a 3 Ã— 2 matrix whose elements are given by aij = eixsinjx

Ans. Let

Given that aij = eix sin jx
a11 = ex sin x    a12 = esin 2x
a21 = e2x sin x   a22 = e2x sin 2x
a31 = e3x sin x    a32 = e3x sin 2x
Hence, the matrix

Q.5. Find values of a and b if A = B, where

Ans.
Given that A = B
â‡’
Equating the corresponding elements, we get
a + 4 = 2a + 2, 3b = b2 + 2 and b2 â€“ 5b = â€“ 6
â‡’ 2a - a = 2, b2 - 3b + 2 = 0, b2 - 5b + 6 = 0
âˆ´ a = 2

but here 2 is common.
Hence, the value of a = 2 and b = 2.

Q.6. If possible, find the sum of the matrices A and B, where A = and B =
Ans.
The order of matrix A = 2 Ã— 2 and the order of matrix B = 2 Ã— 3 Addition of matrices is only possible when they have same order. So, A + B is not possible.

Q.7. Iffind
(i) X +Y
(ii) 2X â€“ 3Y
(iii) A matrix Z such that X + Y + Z is a zero matrix.
Ans.
Given that

(i) X + Y  =

(ii) 2X â€“ 3Y =

(iii) X + Y + Z = 0

Equating the corresponding elements, we get
5 + a = 0 â‡’ a = â€“ 5, 2 + b = 0 â‡’ b = â€“ 2, â€“ 2 + c = 0 â‡’ c = 2
12 + d = 0 â‡’ d = â€“ 12, e = 0, 1 + f = 0 â‡’ f = â€“ 1
Hence, the matrix,

Q.8. Find non-zero values of x satisfying the matrix equation:

Ans.
The given equation can be written as

Equating the corresponding elements we get
12x = 48, 3x + 8 = 20, x2 + 8x = 12x
âˆ´ x = 48/12 = 4, 3x = 20 â€“ 8 = 12, â‡’ x2 = 12x â€“ 8x = 4x
âˆ´ x = 4, â‡’ x2 â€“ 4x = 0 x = 0, x = 4
Hence, the non-zero values of x is 4.

Q.9. If show that (A + B) (A â€“ B) â‰  A2 â€“ B2.
Ans.
Given that

A + B =

â‡’ A + B =

â‡’
âˆ´ (A + B) Ã— (A - B) =

Now, R.H.S. = A2 â€“ B2
= A Ã— A - B Ã— B

Hence,

Hence, (A + B) . (A â€“ B) â‰  A2 - B2

Q.10. Find the value of x if

Ans.
Given that,

â‡’
â‡’

â‡’ [2x + 16 + 10x + 12 + x2 + 4x] = 0;  â‡’ x2 + 16x + 28 = 0
â‡’ x2 + 14x + 2x + 28 = 0; â‡’ x(x + 14) + 2(x + 14) = 0
â‡’ (x + 2) (x + 14) = 0;  x + 2 = 0 or x + 14 = 0
âˆ´ x = â€“ 2 or x = â€“ 14
Hence, the values of x are â€“ 2 and â€“ 14.

Q.11. Show thatsatisfies the equation A2 â€“ 3A â€“ 7I = O and hence find Aâ€“1.
Ans.
Given that

A2 = A Ã— A

A2 â€“ 3A â€“ 7I = O

We are given A2 â€“ 3A â€“ 7I = O
â‡’ Aâ€“1 [A2 â€“ 3A â€“ 7I] = Aâ€“1O
[Pre-multiplying both sides by Aâ€“1]
â‡’ Aâ€“1A Ã— A â€“ 3Aâ€“1 Ã— A â€“ 7Aâ€“1 I = O [Aâ€“1O = O]
â‡’ I Ã— A â€“ 3I â€“ 7Aâ€“1 I  = O
â‡’ A â€“ 3I â€“ 7Aâ€“1  = O
â‡’ â€“7Aâ€“1 = 3I â€“ A

Hence,

Q.12. Find the matrix A satisfying the matrix equation:

Ans.
Let

Equating the corresponding elements, we get,
â€“ 6a â€“ 3c + 10b + 5d = 1 ...(1)
â€“9a â€“ 6c + 15b + 10d = 0 ...(2)
4a + 2c â€“ 6b â€“ 3d = 0 ...(3)
6a + 4c â€“ 9b â€“ 6d = 1 ...(4)
Multiplying eq. (1) by 2 and subtracting eq. (2), we get,

â€“ 3a + 5b = 2 ...(5)
Now, multiplying eq. (3) by 2 and subtracting eq. (4), we get

2a â€“ 3b = â€“ 1 ...(6)
Solving eq. (5) and (6) i.e.,
â€“ 3a + 5b = 2
2a â€“ 3b = â€“ 1
2 x (- 3a + 5b = 2) â‡’ 6a + 10b = 4
3 x (2a - 3b = -1) â‡’ 6a - 9b = - 3
Putting the value of b in eq. (6), we get,
2a â€“ 3 x 1 = - 1
â‡’ 2a - 3 = - 1 â‡’ 2a = 3 - 1 â‡’ 2a = 2
âˆ´ a = 1
Now, putting the values of a and b in equations (1) and (3)
â€“6 x 1 - 3c + 10 x 1 + 5d = 1
â‡’ - 6 - 3c + 10 + 5d = 1
â‡’ â€“ 3c + 5d + 4 = 1 â‡’ 3c + 5d = - 3 ...(7)
and from eq. (3)
4 x 1 + 2c - 6 x 1 - 3d = 0; â‡’  4 + 2c - 6 - 3d = 0
â‡’ - 2 + 2c - 3d = 0 â‡’ 2c - 3d = 2 ...(8)
Solving eq. (7) and (8) we get,

Putting the value of d in eq. (8) we get,
2c â€“ 3 x 0 = 2 â‡’ 2c= 2 â‡’ c = 1
Hence,

Q.13. Find A, if

Ans.
Order ofis 3 Ã— 1 and order ofis 3 x 3. So, the  order of matrix A must be 1 x 3.

Equating the corresponding elements to get the values of a, b and c
4a = â€“ 4, 4b = 8, 4c = 4
âˆ´ a = â€“ 1 âˆ´ b = 2 âˆ´ c = 1
Hence, matrix A = [- 1  2  1]

Q.14. If A =and B =then verify (BA)2 â‰  B2A2
Ans.
Here,

âˆ´

Here, number of columns of first i.e., 3 is not equal to the number of rows of second matrix i.e., 2.
So, B2 is not possible. Similarly, A2 is also not possible.
Hence, (BA)2 â‰  B2A2

Q.15. If possible, find BA and AB, where

Ans.

Hence,

Q.16. Show by an example that for  A â‰  O, B â‰  O, AB = O.
Ans.
Let

Hence,

Q.17. Given Is (AB)â€² = Bâ€²Aâ€²?

Ans.
Here,

L.H.S.

Hence, L.H.S. = R.H.S.

Q.18. Solve for x and y:

Ans.
Given that:

Comparing the corresponding elements of both sides, we get,
2x + 3y â€“ 8 = 0 â‡’ 2x + 3y = 8 ...(1)
x + 5y â€“ 11 = 0 â‡’ x + 5y = 11 ...(2)
Multiplying eq. (1) by 1 and eq. (2) by 2, and then on subtracting, we get,

âˆ´ y = 2
Putting y = 2 in eq. (2) we get,
x + 5 x 2 = 11 â‡’ x + 10 = 11
âˆ´ x = 11 â€“ 10 = 1
Hence, the values of x and y are 1 and 2 respectively.

Q.19. If X and Y are 2 Ã— 2 matrices, then solve the following matrix equations for X and Y

Ans.
Given that:

Multiplying eq. (1) by 3 and eq. (2) by 2, we get,

On subtracting eq. (4) from eq. (3) we get

Now, putting the value of Y in equation (1) we get,

Hence,

Q.20. If A = [3 5],  B = [7 3], then find a non-zero matrix C such that AC = BC.
Ans.
Given that:

Let,

AC = BC (Given)
â‡’ [3Î± + 5Î²] = [7Î± + 3Î²]
â‡’ 3Î± + 5Î² = 7Î± + 3Î²
â‡’ 3Î± â€“ 7Î± = 3Î² â€“ 5Î²
â‡’ â€“ 4Î± = â€“ 2Î²
âˆ´
So, let Î± = K and Î² = 2K, K is some real number.
Hence, matrix C =

Q.21. Give an example of matrices A, B and C such that AB = AC, where A is nonzero matrix, but B â‰  C.
Ans.
Let,

Hence, AB = AC for matrix A is non-zero and B â‰  C.

Q.22. If

(i) (AB) C = A (BC)
(ii) A (B + C) = AB + AC.
Ans.
Given that

(i) To verify: (AB)C = A(BC)

L.H.S. = R.H.S.
So, (AB)C = A(BC)
(ii) To verify: A(B + C) = AB + AC
L.H.S. B + C

L.H.S.A(B + C)

R.H.S. AB

AC

R.H.S. AB + AC

L.H.S. = R.H.S.
Hence, A(B + C) = AB + AC

Q.23. If

Ans.
Given that:

Now

Hence, PQ = QP.

Q.24. If :find A.
Ans.
Given that:

L.H.S.

â‡’

Hence, matrix A = [â€“ 4]

Q.25. If A = [ 2   1] ,  B =verify that A (B + C) = (AB + AC).
Ans.
Given that:

L.H.S.

R.H.S.

L.H.S. = R.H.S.
Hence, A(B + C) = (AB + AC) is verified.

Q.26. If then verify that A2 + A  = A (A + I), where I is 3 Ã— 3 unit matrix.
Ans.
Given that:

A2 = A Ã— A

L.H.S. A2 + A =

R.H.S. A(A + I) =

L.H.S. = R.H.S.
A2 + A = A(A + I). Hence verified.

Q.27. If then verify  that :
(i) (Aâ€²)â€² = A
(ii) (AB)â€² = Bâ€²Aâ€²
(iii) (kA)â€² = (kAâ€²).

Ans.
Given that:

(i)

(ii) L.H.S.

R.H.S.

L.H.S. = R.H.S.
Hence,(AB)â€² = Bâ€²Aâ€²is verified.
(iii) L.H.S.

R.H.S.

Hence, L.H.S. = R.H.S.
(kA)â€² = (kAâ€²) is verified.

Q.28. If then  verify that:
(i) (2A + B)â€² = 2Aâ€² + Bâ€²
(ii) (A â€“ B)â€² = Aâ€² â€“ Bâ€².

Ans.
Given that:

(i) To verify that:(2A + B)â€² = 2Aâ€² + Bâ€²
L.H.S. (2A + B)â€² =

R.H.S. 2Aâ€² + Bâ€² =

Hence, L.H.S. = R.H.S.
(2A + B)â€² = 2Aâ€² + Bâ€²is verified.
(ii) To verify that:(A â€“ B)â€² = Aâ€² â€“ Bâ€².
L.H.S. (A â€“ B)â€² =

R.H.S.Aâ€² â€“ Bâ€²=

Hence, L.H.S. = R.H.S.
(A â€“ B)â€² = Aâ€² â€“ Bâ€² is verified.

Q.29. Show that Aâ€²A and AAâ€² are both symmetric matrices for any matrix A.
Ans.
Let P = Aâ€²A
â‡’ Pâ€² = (Aâ€²A)â€²
â‡’ Pâ€² = Aâ€²(Aâ€²)â€² [(AB)â€² = Bâ€²Aâ€²]
â‡’ Pâ€² = Aâ€²A [âˆµ (Aâ€²)â€² = A]
â‡’ Pâ€² = P
Hence, AÂ¢A is a symmetric matrix.
Now, Let Q = AAâ€²
â‡’ Qâ€² = (AAâ€²)â€²
â‡’ Qâ€² = (Aâ€²)â€² Aâ€² [(AB)â€² = Bâ€²Aâ€²]
â‡’ Qâ€² = AAâ€² [âˆµ (Aâ€²)â€² = A]
â‡’ Qâ€² = Q
Hence, AAâ€² is also a symmetric matrix.

Q.30. Let A and B be square matrices of the order 3 Ã— 3. Is (AB)2 = A2 B2 ? Give reasons.
Ans.
Given that A and B are the matrices of the order 3 x 3.
(AB)2 = AB Ã— AB
= AA Ã— BB
= A2 Ã— B2
Hence, (AB)2 = A2B2

Q.31. how that if A and B are square matrices such that AB = BA, then
(A + B)2 = A2 + 2AB + B2.
Ans.
To prove that (A + B)2 = A2 + 2AB + B2
L.H.S. (A + B)2 = (A + B) Ã— (A + B) [âˆµ A2 = A Ã— A]
= A Ã— A + AB + BA + B Ã— B
= A2 + AB + AB + B2 [AB = BA]
= A2 + 2AB + B2 R.H.S.
So, L.H.S. = R.H.S.

Q.32. Let and a = 4, b = â€“2.
Show that:
(a) A + (B + C) = (A + B) + C
(b) A (BC) = (AB) C
(c) (a + b)B = aB + bB
(d) a (Câ€“A) = aC â€“ aA
(e) (AT)= A
(f) (bA)T = b AT
(g) (AB)T = BT AT
(h) (A â€“B)C = AC â€“ BC
(i) (A â€“ B)T = AT â€“ BT
Ans.
(a) To prove that: A + (B + C) = (A + B) + C
L.H.S. A + (B + C) =

R.H.S. (A + B) + C

Hence, L.H.S. = R.H.S.
A + (B + C) = (A + B) + C Hence proved.
(b) To prove that: A(BC) = (AB)C
L.H.S. A(BC) =

R.H.S. (AB)C

Hence, L.H.S. = R.H.S.
A(BC) = (AB)C Hence proved.
(c) To prove that: (a + b)B = aB + bB
Here, a  = 4  and b = â€“ 2
L.H.S. (a + b)B =

R.H.S. aB + bB

Hence, L.H.S. = R.H.S.
(a + b)B = aB + bB Hence proved.
(d) To prove that: a(C â€“ A) =  aC â€“ aA
L.H.S. a(C â€“ A) =

R.H.S. aC â€“ aA =

Hence, L.H.S. = R.H.S.
a(C â€“ A) = aC â€“ aA Hence proved.
(e) To prove that: (AT)T = A
L.H.S.

R.H.S.
Hence, (AT)T = A
(f) To prove that: (bA)T = bAT
L.H.S. (bA)T =

R.H.S. bAT =

Hence, L.H.S. = R.H.S.
(bA)T = bAT Hence proved.
(g) To prove that: (AB)T = BTAT
L.H.S. (AB)T

R.H.S. BTAT=

Hence, L.H.S. = R.H.S.
(AB)T = BTAHence proved.
(h) To prove that: (A â€“ B)C = AC â€“ BC
L.H.S. (A â€“ B)C =

R.H.S. AC â€“ BC =

Hence, L.H.S. = R.H.S.
(A â€“ B)C = AC â€“ BC
(i) To prove that: (A â€“ B)T = AT â€“ BT
L.H.S. (A â€“ B)T =

R.H.S. AT â€“ BT =

Hence, L.H.S. = R.H.S.
(A â€“ B)T = AT â€“ BT Hence proved.

Q.33. If A =then show that A2 =
Ans. Given that

A = A . A=

Hence proved.

Q.34. If A = and x2 = â€“1, then show that (A + B)2 = A2 + B2.
Ans. Given that:

L.H.S. (A + B)2 = (A + B) Ã— (A + B)

Put x2 = â€“1 (given)

R.H.S.
A2 + B2 = A Ã—  A + B Ã— B

Hence, L.H.S. = R.H.S.
(A + B)2 = A2 + B2

Q.35. Verify that A2 = I when A =
Ans. Given that:

L.H.S. A2 = A Ã— A =

Hence, A2 = I is verified.

Q.36. Prove by Mathematical Induction that (Aâ€²)â€³ = (Aâ€³)â€², where n âˆˆ N for any square matrix A.

Ans.
To prove that (Aâ€²)â€³ = (Aâ€³)â€²
Let P (n): (Aâ€²)â€³ = (Aâ€³)â€²
Step 1: Put n = 1, P(1): Aâ€² = Aâ€² which is true for n = 1
Step 2: Put n = K, P(K): (Aâ€²)K = (AK)â€² Let  it be true for n = K
Step 3: Put n = K + 1, P(K + 1): (Aâ€²)K + 1 = (AK + 1)â€²
L.H.S. (Aâ€²)K + 1 = (Aâ€²)K Ã— (Aâ€²)
= (AK)â€² Ã— (Aâ€²) (From step 2)
= (AK Ã— A)â€²
= (AK+1)â€² R.H.S.
The given statement is true for P(K+1) whenever it is true for P(K), where K âˆˆ N.

Q.37. Find inverse, by elementary row operations (if possible), of the following matrices
(i)
(ii)
Ans.
(i) Let

|A| = 1 x 7 - (- 5) x 3 = 7 + 15 = 22 â‰  0
So, A is invertible.
Let A = IA

R2 â†’ R2 + 5R1
â‡’

R1 â†’ R1 - 3R2
â‡’=
So
Hence, inverse of
(ii) Let

|A| = 1 x 6 - (- 3) (- 2) = 6 - 6 = 0
|A| = 0 so A is not invertible.
Hence, inverse ofis not possible.

Q.38. Ifthen find values of x, y, z and w.
Ans.
Given that:

Equating the corresponding elements,
xy = 8, w = 4, z + 6 = 0 â‡’ z = â€“ 6, x + y = 6
Now, solving x + y = 6 ...(i)
and xy = 8 ...(ii)
From eqn. (i), y = 6 â€“ x ...(iii)
Putting the value of y in eqn. (ii) we get,
x(6 â€“ x) = 8 â‡’ 6x â€“ x2 = 8
â‡’ x2 â€“ 6x + 8 = 0 â‡’ x2 â€“ 4x â€“ 2x + 8 = 0
â‡’ x(x â€“ 4) â€“ 2(x â€“ 4) = 0 â‡’ (x - 4) (x - 2) = 0
âˆ´ x = 4, 2
From eqn. (iii), y = 2, 4.
Hence, x = 4 or  2, y = 2 or 4, z = â€“ 6 and w = 4.

Q.39. Iffind a matrix C such that 3A + 5B + 2C is a null matrix.
Ans.
Order of matrices A and B is 2 Ã— 2.
âˆ´ Order of matrix C must be 2 Ã— 2.
Let
âˆ´ 3A + 5B + 2C = 0

Equating the corresponding elements, we get,
48 + 2a = 0 â‡’ 2a = - 48 â‡’ a = - 24
20 + 2b = 0 â‡’ 2b = - 20 â‡’ b = - 10
56 + 2c = 0 â‡’ 2c = - 56 â‡’ c = - 28
76 + 2d = 0 â‡’ 2d = - 76 â‡’ d = - 38
Hence,

Q.40. If  A =then find A2 â€“ 5A â€“ 14I. Hence, obtain A3.
Ans.
Given that:

A2 = A . A =

âˆ´ A2 â€“ 5A â€“ 14I

Hence, A2 â€“ 5A â€“ 14I = O
Now, multiplying both sides by A, we get,
A2 . A â€“ 5A . A â€“ 14IA = OA
â‡’ A3 â€“ 5A2 â€“ 14A = 0
â‡’ A3 = 5A2 + 14A

Hence, A=

Q.41. Find the values of a, b, c and d, if

Ans.
Given that:

Equating the corresponding elements, we get,
3a = a + 4 â‡’ 3a â€“ a = 4 â‡’ 2a = 4 â‡’ a = 2
3b = 6 + a + b â‡’ 3b â€“ b â€“ a = 6 â‡’ 2b â€“ a = 6 â‡’ 2b â€“ 2 = 6
â‡’ 2b = 8
â‡’ b = 4 3c = â€“ 1 + c + d
â‡’ 3c â€“ c â€“ d = â€“ 1 â‡’ 2c â€“ d = â€“ 1
and 3d = 2d + 3 â‡’ 3d â€“ 2d = 3 â‡’ d = 3
Now 2c â€“ d = â€“ 1
â‡’ 2c â€“ 3 = â€“ 1 â‡’ 2c = 3 - 1 â‡’ 2c = 2
âˆ´ c = 1
âˆ´ a = 2, b = 4, c = 1 and d = 3.

Q.42. Find the matrix A such that

Ans.
Order of matrix is 3 Ã— 2 and the matrix
is 3 Ã— 3
âˆ´ Order of matrix A must be 2 Ã— 3
Let A =

So,

Equating the corresponding elements, we get,
2a â€“ d = â€“ 1 and a = 1 â‡’ 2 Ã— 1 â€“ d = â€“ 1 â‡’ d = 2 + 1 â‡’ d = 3
2b â€“ e = â€“ 8 and b = â€“ 2 â‡’ 2(â€“ 2) â€“ e =  â€“ 8 â‡’ â€“ 4 â€“ e = â€“ 8
â‡’ e = 4
2c â€“ f = â€“ 10 and c = â€“ 5 â‡’ 2(â€“ 5)  â€“ f = â€“ 10 â‡’ â€“ 10 â€“ f = â€“ 10
â‡’ f = 0
a = 1, b = â€“ 2, c = â€“ 5, d = 3, e = 4 and f = 0
Hence,

Q.43. If A =find A2 + 2A + 7I.

Ans.
Given that:

Hence, A+ 2A + 7I =

Q.44. If A =and A â€“1 = Aâ€² , find value of Î±.
Ans.
Here,
A=
Given that: Aâ€“ 1 = Aâ€²
Pre-multiplying both sides by
AAâ€“ 1 = AAâ€²
â‡’ I = AAâ€² [âˆµ AA-1 = I]

Hence, it is true for all values of Î±.

Q.45. If the matrix is a skew symmetric matrix, find the values of a, b and c.
Ans.
Let,

For skew symmetric matrix, Aâ€² = - A.

â‡’
Equating the corresponding elements, we get
a = â€“ 2, b = â€“ b â‡’ 2b = 0 â‡’ b = 0 and c = - 3
Hence, a = - 2,  b = 0 and c = - 3.

Q.46. If P (x) =then show that
P(x) . P(y) = P(x + y) = P(y) . P(x).
Ans.
Given that:
[Replacing x by y]
P(x).P(y)

P(x + y)
Now
P(y).P(x)

= P(x + y)
Hence, P(x).P(y) = P(x + y) = P(y).P(x).

Q.47. If A is square matrix such that A2 = A, show that (I + A)3 = 7A + I.
Ans.
To show that: (I + A)3 = 7A + I
L.H.S. (I + A)3 = I3 + A3 + 3I2A + 3IA2
â‡’ I + A2.A + 3IA + 3IA2
â‡’ I + A.A + 3IA + 3IA [âˆµ A2 = A]
â‡’ I + A2 + 3IA + 3IA
â‡’ I + A + 3IA + 3IA [âˆµ A2 = A]
â‡’ I + A + 3A + 3A  â‡’ 7A + I  R.H.S.
L.H.S. = R.H.S. Hence, Proved.

Q.48. If A, B are square matrices of same order and B is a skew-symmetric matrix, show that Aâ€²BA is skew symmetric.
Ans.
Given that B is a skew symmetric matrix
âˆ´ Bâ€² = - B
Let P = Aâ€²BA
â‡’ Pâ€² = (Aâ€²BA)â€²
= Aâ€²Bâ€²(Aâ€²)â€² [(AB)â€² = Bâ€²Aâ€²]
= Aâ€²(- B) A
= â€“ Aâ€²BA = - P
So Pâ€² = â€“ P
Hence, Aâ€²BA is a skew symmetric matrix.

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