Let and are two matrices such that AB = BA and c 0. Then the value of then K

Question

Let $A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}]$ and $B = [\begin{matrix} a & b \\ c & d \end{matrix}]$ are two matrices such that $A B = B A$ and $c \neq 0$ . Then the value of $\frac{a - d}{3 b - c} = \frac{- 14}{K}$ then $K =$

Smart Achievers · Accepted Answer

(14)

$A B = [\begin{matrix} a + 2 c & b + 2 d \\ 3 a + 4 c & 3 b + 4 d \end{matrix}]$

$B A = [\begin{matrix} a + 3 b & 2 a + 4 b \\ c + 3 d & 2 c + 4 d \end{matrix}]$

$A B = B A \Rightarrow 2 a - 2 d = - 3 b,$ $\frac{a - d}{3 b - c} = - 1$

Solution

Q.
Let $A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}]$ and $B = [\begin{matrix} a & b \\ c & d \end{matrix}]$ are two matrices such that $A B = B A$ and $c \neq 0$ . Then the value of $\frac{a - d}{3 b - c} = \frac{- 14}{K}$ then $K =$

Ans.
(14)

$A B = [\begin{matrix} a + 2 c & b + 2 d \\ 3 a + 4 c & 3 b + 4 d \end{matrix}]$

$B A = [\begin{matrix} a + 3 b & 2 a + 4 b \\ c + 3 d & 2 c + 4 d \end{matrix}]$

$A B = B A \Rightarrow 2 a - 2 d = - 3 b,$ $\frac{a - d}{3 b - c} = - 1$

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Solution

Q. Let A=[1234] and B=[abcd] are two matrices such that AB=BA and c≠0. Then the value of a-d3b-c=-14K then K=

Ans. (14) AB=[a+2cb+2d3a+4c3b+4d] BA=[a+3b2a+4bc+3d2c+4d] AB=BA⇒2a-2d=-3b, a-d3b-c=-1

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Q.
Let $A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}]$ and $B = [\begin{matrix} a & b \\ c & d \end{matrix}]$ are two matrices such that $A B = B A$ and $c \neq 0$ . Then the value of $\frac{a - d}{3 b - c} = \frac{- 14}{K}$ then $K =$

Ans.
(14)

$A B = [\begin{matrix} a + 2 c & b + 2 d \\ 3 a + 4 c & 3 b + 4 d \end{matrix}]$

$B A = [\begin{matrix} a + 3 b & 2 a + 4 b \\ c + 3 d & 2 c + 4 d \end{matrix}]$

$A B = B A \Rightarrow 2 a - 2 d = - 3 b,$ $\frac{a - d}{3 b - c} = - 1$