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Question
Find the adjoint of the following matrix:
\[\begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
Solution 1
Given below is the square matrix. Here, we will interchange the diagonal elements and change the signs of the off-diagonal elements.
\[\ A = \begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
\[adjA = \begin{bmatrix}4 & - 5 \\ - 2 & - 3\end{bmatrix}\]
\[(adjA)A = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[\left| A \right| = - 22\]
\[\left| A \right|I = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[A(adjA) = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[ \therefore (adjA)A = \left| A \right|I = A(adjA)\]
Hence verified.
Solution 2
Given below is the square matrix. Here, we will interchange the diagonal elements and change the signs of the off-diagonal elements.
\[\ A = \begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
\[adjA = \begin{bmatrix}4 & - 5 \\ - 2 & - 3\end{bmatrix}\]
\[(adjA)A = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[\left| A \right| = - 22\]
\[\left| A \right|I = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[A(adjA) = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[ \therefore (adjA)A = \left| A \right|I = A(adjA)\]
Hence verified.
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| To raise money for an orphanage, students of three schools A, B and C organised an exhibition in their residential colony, where they sold paper bags, scrap books and pastel sheets made by using recycled paper. Student of school A sold 30 paper bags, 20 scrap books and 10 pastel sheets and raised ₹ 410. Student of school B sold 20 paper bags, 10 scrap books and 20 pastel sheets and raised ₹ 290. Student of school C sold 20 paper bags, 20 scrap books and 20 pastel sheets and raised ₹ 440. |
Answer the following question:
- Translate the problem into a system of equations.
- Solve the system of equation by using matrix method.
- Hence, find the cost of one paper bag, one scrap book and one pastel sheet.
