Use the product $\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ to solve the system of equations $$x+3z=9,\quad -x+2y-2z=4,\quad 2x-3y+4z=-3.$$

CBSE Class 12 · 2017 · MatricesLong Answer

Use the product

103 - 1 2 - 2 2 - 3 4 - 2 96 021 1 - 3 - 2

to solve the system of equations

x + 3 z = 9, - x + 2 y - 2 z = 4, 2 x - 3 y + 4 z = - 3.

Answer: $x = 0, y = 5, z = 3$

The product equals

I

, so the second matrix is the inverse of the first; equivalently the inverse of

A = 1 - 1 2 02 - 3 3 - 2 4

(the transpose relation) is known.
Writing the system as

A^{⊤} X = C

with

C = 94 - 3

and using

X = (A^{⊤})^{- 1} C = (A^{- 1})^{⊤} C

gives

x y z = - 2 01 92 - 3 61 - 2 94 - 3 = 053 .

x = 0, y = 5, z = 3

CBSE Class 12 Board · 2017 · Subjective