Question
Matrix \(X\) has \((a+b)\) rows and \((a+2)\) columns. Matrix \(Y\) has \((b+1)\) rows and \((a+3)\) columns. Both \(XY\) and \(YX\) exist. Find \(a\) and \(b\). Also determine whether \(XY\) and \(YX\) are of same type and equal.
Solution
Step 1: Condition for \(XY\) to exist
\[ \text{Columns of } X = \text{Rows of } Y \] \[ a + 2 = b + 1 \Rightarrow a – b = -1 \quad …(1) \]Step 2: Condition for \(YX\) to exist
\[ \text{Columns of } Y = \text{Rows of } X \] \[ a + 3 = a + b \Rightarrow b = 3 \quad …(2) \]Step 3: Solve
Substitute \(b = 3\) in (1): \[ a – 3 = -1 \Rightarrow a = 2 \]Step 4: Dimensions
\[ X = (a+b) \times (a+2) = (5 \times 4) \] \[ Y = (b+1) \times (a+3) = (4 \times 5) \]Step 5: Types of Products
\[ XY = (5 \times 4)(4 \times 5) = (5 \times 5) \] \[ YX = (4 \times 5)(5 \times 4) = (4 \times 4) \]Step 6: Compare
– \(XY\) is \(5 \times 5\) – \(YX\) is \(4 \times 4\) \[ \Rightarrow \text{Not of same type} \] \[ \Rightarrow XY \ne YX \]Final Answer
\[
a = 2,\quad b = 3
\]
\[
XY \text{ is } 5 \times 5,\quad YX \text{ is } 4 \times 4
\]
\[
\text{Hence, } XY \ne YX
\]