Exercise 3.2 · Q21 (MCQ): When is $PY+WY$ defined?

Basics you need

Solution

1. When is $PY$ defined, and what is its order?

   $P$ is $p\times k$⁠, so it has $k$ columns. $Y$ is $3\times k$⁠, so it has 3 rows. The product $PY$ needs

   $$\text{columns of } P = \text{rows of } Y,$$

   that is, $k = 3$⁠.

   When $k = 3$⁠, $PY$ has the rows of $P$ and the columns of $Y$⁠: it is of order $p\times k$⁠, that is, $p\times3$⁠.

   Each entry of $PY$ pairs a row of $P$ (which has $k$ entries) with a column of $Y$ (which has 3 entries). The pairing only works if the two lengths are equal, so $k = 3$ is forced.
   $(m\times n)(n\times q) = m\times q$⁠; the inner numbers must match. Lesson: matrix multiplication

2. When is $WY$ defined, and what is its order?

   $W$ is $n\times3$⁠, so it has 3 columns; $Y$ has 3 rows. These always match, so $WY$ is defined for every $n$ and $k$⁠, and its order is $n\times k$⁠.

   Here the inner numbers are both 3 by the way the orders were given, so $WY$ imposes no condition at all.

3. When is the sum $PY + WY$ defined?

   From Steps 1 and 2, $PY$ is $p\times k$ and $WY$ is $n\times k$⁠. A sum needs equal orders:

   $$p\times k = n\times k \iff p = n.$$

   (The column counts are both $k$⁠, so they already agree; only the row counts can differ.)

   Putting the three conditions together: $PY + WY$ is defined exactly when $k = 3$ and $p = n$⁠. Both are needed, and together they are enough, since then $PY$ and $WY$ are both $n\times3$⁠.
   Matrices of different orders cannot be added. Lesson: addition of matrices

4. Examine every option

   Option	Claim	Verdict
   (A)	$k = 3$⁠, $p = n$	Correct. Exactly the two conditions from Steps 1 and 3; with them, both products are $n\times3$ and the sum is defined.
   (B)	$k$ arbitrary, $p = 2$	Wrong. If $k \ne 3$⁠, say $k = 4$⁠, then $P$ has 4 columns but $Y$ has 3 rows, so $PY$ is not defined. Also $p = 2$ does not give $p = n$ unless $n$ happens to be 2.
   (C)	$p$ arbitrary, $k = 3$	Wrong. $k = 3$ makes both products defined, but $PY$ is $p\times3$ and $WY$ is $n\times3$⁠. For example $p = 1$⁠, $n = 2$ gives orders $1\times3$ and $2\times3$⁠, which cannot be added.
   (D)	$k = 2$⁠, $p = 3$	Wrong. With $k = 2$⁠, $P$ has 2 columns but $Y$ has 3 rows, so $PY$ is not defined.

   To reject an option, one choice of values allowed by that option for which $PY + WY$ fails is enough; the table gives one for each wrong option.

Every fact used, and where it comes from