Let A=[aij] be a matrix of order 3×3, with aij=(2)i+j. If the sum of all the elements in the third row of A2 is α+β2, α,β∈Z, then α+β is equal to:
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Step-by-Step Solution
Step 1: Define the matrix A
First, we define the elements of the matrix A using the given rule aij=(2)i+j. This means each element is determined by the sum of its row and column indices.
Step 2: Calculate the elements of A
Now, we calculate each element aij by substituting the values of i and j into the formula (2)i+j. For example, a11=(2)1+1=(2)2=2, and a12=(2)1+2=(2)3=22. We do this for all 9 elements.
Step 3: Calculate the third row of A squared
To find the elements of the third row of A2, we use the matrix multiplication rule. The element (A2)3j is the dot product of the third row of A and the j -th column of A.
Step 4: Compute the elements of the third row of A squared
We calculate each element of the third row of A2: (A2)31, (A2)32, and (A2)33. This involves multiplying the elements of the third row of A by the corresponding elements of the first, second, and third columns of A, respectively, and summing the products.
Step 5: Sum the elements of the third row of A squared
Now, we sum the three elements we just calculated for the third row of A2. This gives us the total sum in the form α+β2.
Step 6: Determine α and β and calculate their sum
By comparing the sum 168+562 with α+β2, we identify α=168 and β=56. Finally, we calculate the sum α+β.