CBSE Class 12 Mathematics 2023: question 33(b)

Hints

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Step-by-step solution

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1. Write the system as $AX=B$, where $A=\left[\begin{array}{ccc}1& 2& 3\\ 2& -1& 1\\ 3& 2& -2\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$, $B=\left[\begin{array}{c}6\\ 2\\ 3\end{array}\right]$. Then $X=A^{-1}B$.

   $|A|=1(2-2)-2(-4-3)+3(4+3)=0+14+21=35\ne 0$, so $A^{-1}$ exists.

   ½ mark

2. Cofactors (first two rows):

   • $A_{11}=0,A_{12}=7,A_{13}=7$
   • $A_{21}=10,A_{22}=-11,A_{23}=4$
   1 mark

3. Cofactors (third row): $A_{31}=5,A_{32}=5,A_{33}=-5$.

   1½ marks

4. $A^{-1}=\frac{1}{35}\left[\begin{array}{ccc}0& 10& 5\\ 7& -11& 5\\ 7& 4& -5\end{array}\right]$

   1 mark

5. $X=\frac{1}{35}\left[\begin{array}{ccc}0& 10& 5\\ 7& -11& 5\\ 7& 4& -5\end{array}\right]\left[\begin{array}{c}6\\ 2\\ 3\end{array}\right]=\frac{1}{35}\left[\begin{array}{c}35\\ 35\\ 35\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

   Hence $x=1,y=1,z=1$.

   1 mark

Answer: $x=1,y=1,z=1$

Revise the chapter

• Chapter: Matrices (notes, formulas, where marks are lost and every past question)