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Question 180 of 182

Q.Assume XX, YY, ZZ, WW and PP are matrices of order 2×n2\times n, 3×k3\times k, 2×p2\times p, n×3n\times 3 and p×kp\times k, respectively. Then the restriction on nn, kk and pp so that PY+WYPY+WY will be defined are:
(A) k=3, p=nk=3,\ p=n
(B) kk is arbitrary, p=2p=2
(C) pp is arbitrary, k=3k=3
(D) k=2, p=3k=2,\ p=3

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For PY+WYPY+WY to be defined, both products must be compatible and the resulting matrices must have the same dimensions. This forces p=np = n and k=3k = 3, making option (A) correct.

The core idea here is matrix multiplication compatibility — the number of columns in the first matrix must equal the number of rows in the second. When adding two matrices, they must also have identical dimensions. So PY+WYPY+WY being defined means both PYPY and WYWY are individually defined, and their results are the same size.

Let’s unpack the given orders:

  • XX: 2×n2 \times n
  • YY: 3×k3 \times k
  • ZZ: 2×p2 \times p
  • WW: n×3n \times 3
  • PP: p×kp \times k

We only care about PP, WW, and YY.

  1. Check PYPY: PP is p×kp \times k, YY is 3×k3 \times k. For multiplication P⋅YP \cdot Y, the number of columns of PP must equal the number of rows of YY. That means k=3k = 3. If k≠3k \neq 3, PYPY is not defined at all, so the sum cannot exist. So k=3k = 3 is necessary.

  2. Check WYWY: WW is n×3n \times 3, YY is 3×k3 \times k. With k=3k = 3 from step 1, YY becomes 3×33 \times 3. The columns of WW (3) match the rows of YY (3), so WYWY is defined for any nn. Its resulting order is n×3n \times 3.

  3. Now PYPY with k=3k=3: PP is p×3p \times 3, YY is 3×33 \times 3, so PYPY is defined and its order is p×3p \times 3. …

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