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Question 178 of 182

Q.If for three matrices 𝐴 = [π‘Žπ‘–π‘—]π‘šΓ—4 , B = [𝑏𝑖𝑗]𝑛×3 π‘Žπ‘›π‘‘ C = [𝑐𝑖𝑗]π‘Γ—π‘ž products 𝐴𝐡 and 𝐴𝐢 both are defined and are square matrices of same order, then value of π‘š, 𝑛, 𝑝 and π‘ž are:
(A) π‘š = π‘ž = 3 π‘Žπ‘›π‘‘ 𝑛 = 𝑝 = 4
(B) π‘š = 2, π‘ž = 3 π‘Žπ‘›π‘‘ 𝑛 = 𝑝 = 4
(C) π‘š = π‘ž = 4 π‘Žπ‘›π‘‘ 𝑛 = 𝑝 = 3
(D) π‘š = 4, 𝑝 = 2 π‘Žπ‘›π‘‘ 𝑛 = π‘ž = 3

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Compatibility forces n=4n=4 and p=4p=4; requiring both products to be square (and of the same order) forces m=3m=3 and q=3q=3. So m=q=3,Β n=p=4m=q=3,\ n=p=4 β€” option (A).

The two rules we need

For a product XYXY to exist, the number of columns of XX must equal the number of rows of YY, and the result takes the outer dimensions. A matrix is square when its number of rows equals its number of columns.

We are told A=[aij]mΓ—4A=[a_{ij}]_{m\times 4}, B=[bij]nΓ—3B=[b_{ij}]_{n\times 3}, C=[cij]pΓ—qC=[c_{ij}]_{p\times q}, and that both ABAB and ACAC are defined square matrices of the same order.

Working the conditions

  1. ABAB is defined. Columns of AA (which is 44) must equal rows of BB (which is nn):

4=nΒ β‡’Β n=4.4=n\ \Rightarrow\ n=4.

  1. ABAB is square. With AmΓ—4A_{m\times 4} and B4Γ—3B_{4\times 3}, the product ABAB has order mΓ—3m\times 3. For a square matrix the two must be equal:

m=3.m=3.

So ABAB is 3Γ—33\times 3.

  1. ACAC is defined. Columns of AA (44) must equal rows of CC (pp): 4=pΒ β‡’Β p=4.4=p\ \Rightarrow\ p=4. …

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