Question:
If
\[
A=\begin{bmatrix}3 & 4 5 & 6\end{bmatrix}
\quad \text{and} \quad
B=\begin{bmatrix}x & 0 0 & y\end{bmatrix},
\]
where \(x,y \in \mathbb{N}\), then:
If
\[
A=\begin{bmatrix}3 & 4 5 & 6\end{bmatrix}
\quad \text{and} \quad
B=\begin{bmatrix}x & 0 0 & y\end{bmatrix},
\]
where \(x,y \in \mathbb{N}\), then:
Show Hint
For a diagonal matrix
\[
\begin{bmatrix}
x & 0
0 & y
\end{bmatrix},
\]
matrix multiplication scales rows or columns separately.
To verify \(AB=BA\), always compare corresponding entries after multiplication.
Updated On: Jun 3, 2026
- \(x = 3,\ y = 6\)
- \(x = 4,\ y = 5\)
- \(x = 5,\ y = 4\)
- \(x = 6,\ y = 3\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
Two matrices \(A\) and \(B\) are said to commute if
\[
AB = BA.
\]
To check this condition, we calculate both products separately and compare the corresponding entries.
For diagonal matrices, multiplication becomes simpler because only the diagonal entries are non-zero.
Step 1: Compute the product \(AB\). Given: \[ A=\begin{bmatrix}3 & 4 5 & 6\end{bmatrix}, \qquad B=\begin{bmatrix}x & 0 0 & y\end{bmatrix} \] Now, \[ AB= \begin{bmatrix} 3 & 4 5 & 6 \end{bmatrix} \begin{bmatrix} x & 0 0 & y \end{bmatrix} \] Using matrix multiplication: \[ AB= \begin{bmatrix} 3x+4(0) & 3(0)+4y 5x+6(0) & 5(0)+6y \end{bmatrix} \] Hence, \[ AB= \begin{bmatrix} 3x & 4y 5x & 6y \end{bmatrix} \]
Step 2: Compute the product \(BA\). \[ BA= \begin{bmatrix} x & 0 0 & y \end{bmatrix} \begin{bmatrix} 3 & 4 5 & 6 \end{bmatrix} \] Multiplying: \[ BA= \begin{bmatrix} 3x & 4x 5y & 6y \end{bmatrix} \]
Step 3: Apply the condition \(AB=BA\). For two matrices to be equal, corresponding elements must be equal. Thus, \[ \begin{bmatrix} 3x & 4y 5x & 6y \end{bmatrix} = \begin{bmatrix} 3x & 4x 5y & 6y \end{bmatrix} \] Comparing corresponding entries: From the \((1,2)\)-entry: \[ 4y = 4x \] \[ x=y \] From the \((2,1)\)-entry: \[ 5x = 5y \] \[ x=y \] Thus, the condition for commutativity is: \[ x=y \] But since \(x,y \in \mathbb{N}\), infinitely many choices are possible: \[ (1,1), (2,2), (3,3), \ldots \] Hence, infinitely many matrices \(B\) satisfy \[ AB=BA. \] Therefore, option (d) is correct.
Step 4: Check the validity of the given answer choices carefully. Since infinitely many matrices satisfy \(AB=BA\), \[ \boxed{\text{Option (d) is correct}} \] and therefore options (a), (b), and (c) are incorrect.
Step 1: Compute the product \(AB\). Given: \[ A=\begin{bmatrix}3 & 4 5 & 6\end{bmatrix}, \qquad B=\begin{bmatrix}x & 0 0 & y\end{bmatrix} \] Now, \[ AB= \begin{bmatrix} 3 & 4 5 & 6 \end{bmatrix} \begin{bmatrix} x & 0 0 & y \end{bmatrix} \] Using matrix multiplication: \[ AB= \begin{bmatrix} 3x+4(0) & 3(0)+4y 5x+6(0) & 5(0)+6y \end{bmatrix} \] Hence, \[ AB= \begin{bmatrix} 3x & 4y 5x & 6y \end{bmatrix} \]
Step 2: Compute the product \(BA\). \[ BA= \begin{bmatrix} x & 0 0 & y \end{bmatrix} \begin{bmatrix} 3 & 4 5 & 6 \end{bmatrix} \] Multiplying: \[ BA= \begin{bmatrix} 3x & 4x 5y & 6y \end{bmatrix} \]
Step 3: Apply the condition \(AB=BA\). For two matrices to be equal, corresponding elements must be equal. Thus, \[ \begin{bmatrix} 3x & 4y 5x & 6y \end{bmatrix} = \begin{bmatrix} 3x & 4x 5y & 6y \end{bmatrix} \] Comparing corresponding entries: From the \((1,2)\)-entry: \[ 4y = 4x \] \[ x=y \] From the \((2,1)\)-entry: \[ 5x = 5y \] \[ x=y \] Thus, the condition for commutativity is: \[ x=y \] But since \(x,y \in \mathbb{N}\), infinitely many choices are possible: \[ (1,1), (2,2), (3,3), \ldots \] Hence, infinitely many matrices \(B\) satisfy \[ AB=BA. \] Therefore, option (d) is correct.
Step 4: Check the validity of the given answer choices carefully. Since infinitely many matrices satisfy \(AB=BA\), \[ \boxed{\text{Option (d) is correct}} \] and therefore options (a), (b), and (c) are incorrect.
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