If $A=\left[\begin{array}{ccc}1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1\end{array}\right]$, then find $A^{-1}$ and use it to solve the following system of equations: $$x+2y-3z=6,\quad 3x+2y-2z=3,\quad 2x-y+z=2.$$ OR Using properties of determinants, prove that $$\left|\begin{array}{lll}(b+c)^{2} & a^{2} & bc \\ (c+a)^{2} & b^{2} & ca \\ (a+b)^{2} & c^{2} & ab\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right).$$

CBSE Class 12 · 2020 · MatricesLong Answer

A = 132 22 - 1 - 3 - 2 1

, then find

A^{- 1}

and use it to solve the following system of equations:

x + 2 y - 3 z = 6, 3 x + 2 y - 2 z = 3, 2 x - y + z = 2.

OR
Using properties of determinants, prove that

(b + c)^{2} (c + a)^{2} (a + b)^{2} a^{2} b^{2} c^{2} b c c a ab = (a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2}) .

Answer: $x = 1, y = - 5, z = - 5$ (OR: identity proved)

Main part: here

∣ A ∣ = 7 \neq = 0

and

A^{- 1} = \frac{1}{7} 0 - 7 - 7 175 2 - 7 - 4

. With

B = 632

X = A^{- 1} B = \frac{1}{7} 7 - 35 - 35 = 1 - 5 - 5 .

x = 1, y = - 5, z = - 5

.
OR part: applying

C_{1} \to C_{1} - 2 C_{3}

then

C_{1} \to C_{1} + (a^{2} + b^{2} + c^{2})

-type grouping, taking

(a^{2} + b^{2} + c^{2})

common from

C_{1}

, and reducing with

R_{2} \to R_{2} - R_{1}

R_{3} \to R_{3} - R_{1}

gives the stated factorisation

(a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2})

x = 1, y = - 5, z = - 5

CBSE Class 12 Board · 2020 · Subjective