Tor K, H, M,IV and P (17) Let A be a 2x 2 matrix and let B be a 3x3 matrix. Cons

Question

Tor K, H, M,IV and P (17) Let A be a 2x 2 matrix and let B be a 3x3 matrix. Consider the following matrices (i.) A' is obtained from A by multiplying one of the rows by 3. (ii.) A" is obtained from A by multiplying one of the columns by 3. (iii.) A",-5A is obtained from multiplying all entries of A by 5. (iv.) B, is obtained fron B by multiplying one of the columns by 3 (v.) B" is obtained from B by multiplying two of the columns by 3. (m) B", = 5B is obtained from multiplying all entries of A by 5. , B compute the determinant, assuming det A and det B7

Explanation / Answer

17. We know that

(i). det (A’) = 3* de4t(A) = 3*7 = 21.

(ii). Since det(AT) = det(A) and since multiplying a column of A by 3 is same as multiplying the corresponding row of AT by 3, hence det(A’’) = 3*det(AT) = 3*det(A) = 3*7 = 21.

(iii). If A’’’ = 5A, then det(A) = 52det(A) = 25*7 = 175.

(iv). Since det(BT) = det(B) and since multiplying a column of B by 3 is same as multiplying the corresponding row of BT by 3, hence det(B’’) = 3*det(AT) = 3*det(A) = 3*7 = 21.

(v). Since det(BT) = det(B) and since multiplying 2 columns of B by 3 is same as multiplying the corresponding 2 rows of BT by 3, hence det(B’’) = 32 det(BT) = 9*7 = 63.

(vi). There is a misprint. B’’’ = 5B ( how can we get B”’ by multiplying all the entries of A by 5; It should be B instead of A). det(B’’’) = 53*det(B) = 125*7 = 875.

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