S = R. . We have excellent notes prepared by Ex-IITian to best match the requirement of the exam. Evaluate the following:Solution:(i) GivenFirst we have to add first two matrix,On simplifying, we get(ii) Given,First we have to multiply first two given matrix,= 82(iii) GivenFirst we have subtract the matrix which is inside the bracket,Solution:GivenWe know that,Again we know that,Now, consider,We have,Now, from equation (1), (2), (3) and (4), it is clear that A2 = B2= C2= I2Solution:GivenConsider,Now we have to find,Solution:GivenConsider,Hence the proof. 60Solution:GivenConsider, … (i) … (ii)From (i) and (ii) we can see thatA skew-symmetric matrix is a square matrix whose transpose equal to its negative, that is,X = – XTSo, A – AT is a skew-symmetric.

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In todays world getting such education free of cost will definitely help the aspirants to crack the exam. 2 SolutionsExercise 5. So, A = [ai j] 3×4R1 = first row of A = [a11, a12, a13, a14]
So, order of matrix R1 = 1 × 4C2 = second column ofTherefore order of C2 = 3 × 14. Get excellent practice papers and Solved examples to grasp the concept and check for speed and make you ready for big day.

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H. H. will make sure you understand the concept well. 1 Page No: 5. Construct a 3×4 matrix A = [ai j] whose elements ai j are given by:
(i) ai j = i + j(ii) ai j = i – j(iii) ai j = 2i(iv) ai j = j(v) ai j = ½ |-3i + j|Solution:(i) Given ai j = i + jLet A = [ai j]2×3So, the elements in a 3 × 4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 1 + 1 = 2a12 = 1 + 2 = 3a13 = 1 + 3 = 4a14 = 1 + 4 = 5a21 = 2 + 1 = 3a22 = 2 + 2 = 4a23 = 2 + 3 = 5a24 = 2 + 4 = 6a31 = 3 + 1 = 4a32 = 3 + 2 = 5a33 = 3 + 3 = 6a34 = 3 + 4 = 7Substituting these values in matrix A we get,A =
(ii) Given ai j = i – jLet A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 1 – 1 = 0a12 = 1 – 2 = – 1a13 = navigate here – 3 = – 2a14 = 1 – 4 = – 3a21 = 2 – 1 = 1a22 = 2 – 2 = 0a23 = 2 – 3 = – 1a24 = 2 – 4 = – 2a31 = 3 – 1 = 2a32 = 3 – 2 = 1a33 = 3 – 3 = 0a34 = 3 – 4 = – 1Substituting these values in matrix A we get,A =
(iii) Given ai j = 2iLet A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 2×1 = 2a12 = 2×1 = 2a13 = 2×1 = 2a14 = 2×1 = 2a21 = 2×2 = 4a22 = 2×2 = 4a23 = 2×2 = 4a24 = 2×2 = 4a31 = 2×3 = 6a32 = 2×3 = 6a33 = 2×3 = 6a34 = 2×3 = 6Substituting these values in matrix A we get,A =
(iv) Given ai j = jLet A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 1a12 = 2a13 = 3a14 = 4a21 = 1a22 = 2a23 = 3a24 = 4a31 = 1a32 = 2a33 = 3a34 = 4Substituting these values in matrix A we get,A =
(vi) Given ai j = ½ |-3i + j|Let A = from this source j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 =
a12 =
a13 =
a14 =
a21 =
a22 =
a23 =
a24 =
a31 =
a32 =
a33 =
a34 =
Substituting these values in matrix A we get,A =
Multiplying by negative sign we get,7. .

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