#### Please solve RD Sharma class 12 chapter 4 Algebra of Matrices exercise Fill in the blank question 5 maths textbook solution

x = 1

Hint:

Use the basic concept of identity matrix.

Given:

$A=\left[\begin{array}{cc} \frac{1}{3} & 2 \\ 0 & 2 x-3 \end{array}\right] \text { and } B=\left[\begin{array}{cc} 3 & 6 \\ 0 & -1 \end{array}\right],[A B]=[I]$

Find x

Solution:

\begin{aligned} &\text { Here, }[A B]=[I] \\ &\text { So, }\left[\begin{array}{cc} \frac{1}{3} & 2 \\ 0 & 2 x-3 \end{array}\right]\left[\begin{array}{cc} 3 & 6 \\ 0 & -1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &=\left[\begin{array}{ll} \frac{1}{3} \times 3+2 \times 0 & \frac{2}{3} \times 6+2 \times(-1) \\ 0 \times 3+0 \times 0 & 0+(2 x-3)(-1) \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}

$=\left[\begin{array}{cc} 1 & 0 \\ 0 & 2 x+3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$

So, here both sides have 2 x 2 matrices

Hence, by comparing respective elements,

\begin{aligned} &-2 s+3=1 \\ &3-1=2 x \\ &2=2 x \\ &x=1 \end{aligned}