You can Download RBSE Solutions for Class 12 Maths Chapter 3 Matrix Miscellaneous Exercise Guide Pdf, help you to revise the complete Syllabus and score more marks in your examinations.
Rajasthan Board RBSE Class 12 Maths Chapter 3 Matrix Miscellaneous Exercise
Question 1.
, then find A.
Solution:
Question 2.
, then find (A – 2I) (A – 3I).
Solution:
Question 3.
, then find AB.
Solution:
Question 4.
, then find BA.
Solution:
Question 5.
, then find matrices A and B.
Solution:
Question 6.
, then find the value of x and y.
Solution:
On comparing,
x + 2 = -2 ∴ x = -4
– y – 2 = 5 ⇒ y = -7
Hence, x = -4, y = -7
Question 7.
Order of matrix A is 3 x 4 and B is a matrix, such that ATB and ABT defined, then write the order of B.
Solution:
∴ Order of A = 3 x 4
∴ Order of AT = 4 x 3
But ATB and ABT is defined.
So, order of B is 3 x 4.
Question 8.
, is a symmetric matrix, then determine x.
Solution:
Given,
On comparing aij = aji
a32 = a23 ⇒ -x = 4
∴ x = -4
Question 9.
Construct a matrix of order 3 x 3, B = [bij], whose elements are bij= (i) (j).
Solution:
B11 = 1 x 1 = 1
B12 = 1 x 2 = 2
B13 = 1 x 3 = 3
B21 = 2 x 1 = 2
B22 = 2 x 2 = 4
B23 = 2 x 3 = 6
B31 = 3 x 1 = 3
B32 = 3 x 2 = 6
B33 = 3 x 3 = 9
Question 10.
Solution:
Question 11.
Express matrix A as the sum of symmetric and skew-symmetric matrices, where
.
Solution:
Given,
Question 12.
then prove that :
(i) (AT)T = A
(ii) A + AT is a symmetric matrix.
(iii) A – AT is a skew-symmetric matrix.
(iv) AAT and ATA are symmetric matrix.
Solution:
(i) (AT)T = A
(ii) A + AT is symmetric
So, A + AT is symmetric matrix. Hence Proved.
So, A – AT is skew symmetric matrix. Proved.
Here,
a21 = a12 = 0
a31 = a13 = 0
a23 = a32 = 6
So, AAT is symmetric matrix.
Here,
a12 = a21 = 0
a13 = a31 = 0
a32 = a23 = 4
So, ATA is symmetric matrix.
Question 13.
, and 3A – 2B + C is a null matrix, then determine matrix ‘C’.
Solution:
Question 14.
Construct a matrix B = [bij] of the order 2 x 3, whose elements are bij = (i +2j)2/2
Solution:
Given, B = [bij] whose elements are
Question 15.
, then find the element of 1st row of ABC.
Solution:
So, element of 1st row is 8.
Question 16.
, then find AAT.
Solution:
Given,
Question 17.
, then find x.
Solution:
Question 18.
, then prove
Solution:
Given,
= (bc – ad)I2 = R.H.S.
Hence Proved.
Question 19.
, then find the matrix form of the following (aA + bB) (aA – bB).
Solution:
Givn,
Question 20.
, then prove that (A – B)2 ≠ A2 – 2AB + B2.
Solution:
Given,
From (i) and (ii),
(A – B)2 + A2 – 2AB + B2
Hence Proved.
Question 21.
, then find k, where A2 = kA – 2IA2 .
Solution:
Given,
On comparing corresponding element
From 3k – 2 = 1
3k = 3 ⇒ k = \(\frac { 3 }{ 3 }\) ⇒ k = 1.
Question 22.
i = √-1 then prove that :
(1) A2 = B2 = C2 = -I2
(ii) AB = – BA = -C
Solution:
Question 23.
and f(A) = A2 – 5A + 7I then find f(A).
Solution:
Question 24.
Prove that
Solution:
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