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Provide Solutio for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise Fill in the blanks Question 19

Answers (1)

Answer  \rightarrow \lambda=-1/6

Given  \rightarrow A=\left[\begin{array}{ll} 0 & 3 \\ 2 & 0 \end{array}\right], A^{-1}=\lambda(\operatorname{adj} A)

Explanation   \rightarrow \operatorname{adj} A=\left[\begin{array}{cc} 0 & -3 \\ -2 & 0 \end{array}\right]

                          \lambda(\operatorname{adj} A)=\left[\begin{array}{llll} 0 & -3 \lambda & -2 \lambda & 0 \end{array}\right]             …(i)

Now,

              A^{-1}=\frac{1}{-6}\begin{vmatrix} 0 &-3 & -20 \end{vmatrix}

               \Rightarrow \begin{bmatrix} 0 &\frac{1}{2} &\frac{1}{3} & 0 \end{bmatrix}= \lambda\left ( adj A \right )= \begin{vmatrix} 0 & -3\lambda & -2\lambda &0 \end{vmatrix}                  … [from (i)]

               \Rightarrow-3 \lambda=\frac{1}{2} \quad \Rightarrow \lambda=-1 / 6

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