3. Let A = [aij] be an n x n matrix. Determine whether A is...

Transcript

00:01 High so in this question they have given four equations we have to find which is symmetric or not okay before that what is mean by symmetric symmetric is a transpose equal to a so this is the condition for symmetric if both a transpose and a matrix are equal then the given equation is symmetric so first we take the first equation a i j is equal to i squared plus j squared so where i ranges from 1 to 2 and j ranges from 1 to 2 so we just take a11 a 111a 1 1 1 1 s is the first row first column element so we just in the given question i squared plus j square so 1 square plus 1 is equal to 2 likewise a 1 2 is equal to 1 square plus 2 square is equal to 5 i we proceed like that up to a 2 2 so we just accumulate this in the form of matrix so here i put those 4 values in the form of matrix which is a11 a1 2 a 2 1 and a 2 i just write the corresponding values which is 2 5 5 8 i mark it as equation 1 this is a so now find a transpose what is a transpose if we write the row as a column and if your column as a row what it is so 2 5 and 5 8 so both the matrix of a transpose and a are equal here a transpose is equal to a so the given equation is symmetric so the first equation is symmetric likewise now we find the second equation likewise we find for the second subpart which is a i j is equal to i squared minus j squared so the i value and j values ranges from one to two so likewise we find a11 a12 a 2 1 2 because it is i square minus j squared so i minus the given values so we find it in the matrix form which is a is equal to a one one a two one eight two two two two three so we just put those values in a corresponding matrix which is zero minus three and three zero just the first equation which is a part and we find it is a transpose so we just make row as column we just write row element in the way of column matrix so zero three minus three and zero so in this equation we can able to see that that a transpose is not equal to a because the values of a transpose and a are different in the place of a transpose the three it will be in minus three in a matrix so it is not equal to so it is not symmetric so the given equation is not symmetric for the second equation now the third support is eij is equal to two i plus 2 j likewise we substitute the i value and j values ranges from 1 to 2 so now i substitute the i value and j value in the given equation so a 1 1 1 2 a 2 1 2 so i found the 4 values and i put those values in a matrix form so a equal to 4 6 6 8 so this is the matrix form of a and we found the a transpose a transpose is the same value 4 6 6 6 6 8 so this is the same value 4 6 6 6 6 6 8 so this is the 1.

03:42 So from this what we can say is a transpose is equal to a so the given equation is symmetric so the third equation is symmetric likewise we found for the fourth subpart now the fourth subpart is 2 i squared which is a ij is equal to 2 i squared plus 2 j cube likewise x i and j is ranges from 1 to 2 2 2.

04:16 So a11, a12, a2, a21 and a2.

04:20 Substitute the i value and j value in the given equation.

04:23 So we found the value of a in the matrix form.

04:26 So a value is 4, 18, 10 and 24...

Question

3. Let A = [aij] be an n x n matrix. Determine whether A is symmetric. (a) aij = i^2 + j^2 (b) aij = i^2 - j^2 (c) aij = 2i + 2j (d) aij = 2i^2 + 2j^3

Question

3. Let A = [aij] be an n x n matrix. Determine whether A is symmetric. (a) aij = i^2 + j^2 (b) aij = i^2 - j^2 (c) aij = 2i + 2j (d) aij = 2i^2 + 2j^3

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