Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 19 Subquestion (i) Maths Textbook Solution.

Answer: $x = 2, y = 4, z = 1, t =3$
Hint: Solve LHS part and equate with RHS part.

Given: $3 \begin{bmatrix} x &y \\ z&t \end{bmatrix} = \begin{bmatrix} x &6 \\ -1 &2t \end{bmatrix} + \begin{bmatrix} 4 &x+y \\ z+t & 2t+3 \end{bmatrix}$
Here, we have to compute x, y, z and t.
Solution:

$3 \begin{bmatrix} x &y \\ z&t \end{bmatrix} = \begin{bmatrix} x &6 \\ -1 &2t \end{bmatrix} + \begin{bmatrix} 4 &x+y \\ z+t & 2t+3 \end{bmatrix}$

$\begin{bmatrix} 3x &3y \\ 3z&3t \end{bmatrix} = \begin{bmatrix} x &6 \\ -1 &2t \end{bmatrix} + \begin{bmatrix} 4 &x+y \\ z+t & 2t+3 \end{bmatrix}$

$\begin{bmatrix} 3x &3y \\ 3z&3t \end{bmatrix} = \begin{bmatrix} x+4 &6+x+y \\ -1+z+t &2t+3 \end{bmatrix}$

Equating this, we get:

$\! \! \! \! \! \! \! \! 3 x = x + 4\\ 2x = 4 \\ \Rightarrow x = 2$

Also,

$\! \! \! \! \! \! \! \! 3y = 6 + x + y \\ 2y = 6 + x$

Putting the value of ‘x ’, we get:

$\! \! \! \! \! \! \! \! 2y = 6 + 2\\ 2y = 8\\ \Rightarrow y = 4$
Also,

$\! \! \! \! \! \! \! 3t = 2t + 3\\ \Rightarrow t = 3$

Now,

$\! \! \! \! \! \! \! \! \! 3z = -1 + z + t\\ 2z = - 1 + t$

Putting the value of ‘t’, we get:

$\! \! \! \! \! \! \! 2z = -1 + 3\\ 2z = 2\\ \Rightarrow z = 1$