Without expanding determinants show that |1366143712|+4|233212176|=10|121317326| - Mathematics and Statistics

प्रश्न

Without expanding determinants show that

$| \begin{matrix} 1 & 3 & 6 \\ 6 & 1 & 4 \\ 3 & 7 & 12 \end{matrix} | + 4 | \begin{matrix} 2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6 \end{matrix} | = 10 | \begin{matrix} 1 & 2 & 1 \\ 3 & 1 & 7 \\ 3 & 2 & 6 \end{matrix} |$

योग

उत्तर

L.H.S. = $| \begin{matrix} 1 & 3 & 6 \\ 6 & 1 & 4 \\ 3 & 7 & 12 \end{matrix} | + 4 | \begin{matrix} 2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6 \end{matrix} |$

By taking 2 common from C₃ in the first determinant, we get,

L.H.S. = $2 | \begin{matrix} 1 & 3 & 3 \\ 6 & 1 & 2 \\ 3 & 7 & 6 \end{matrix} | + 4 | \begin{matrix} 2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6 \end{matrix} |$

= $| \begin{matrix} 2 & 3 & 3 \\ 12 & 1 & 2 \\ 6 & 7 & 6 \end{matrix} | + | \begin{matrix} 8 & 3 & 3 \\ 8 & 1 & 2 \\ 4 & 7 & 6 \end{matrix} |$

= $| \begin{matrix} 2 + 8 & 3 & 3 \\ 12 + 8 & 1 & 2 \\ 6 + 4 & 7 & 6 \end{matrix} |$

as C₂ and C₃ are identical in the two determinants

= $| \begin{matrix} 10 & 3 & 3 \\ 20 & 1 & 2 \\ 10 & 7 & 6 \end{matrix} |$

By interchanging rows and columns, we get,

L.H.S. = $| \begin{matrix} 10 & 20 & 10 \\ 3 & 1 & 7 \\ 3 & 2 & 6 \end{matrix} |$

By taking 10 common from R₁, we get,

L.H.S. = $10 | \begin{matrix} 1 & 2 & 1 \\ 3 & 1 & 7 \\ 3 & 2 & 6 \end{matrix} |$

= R.H.S.

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Properties of Determinants

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 6 Q 5 Q 1. (i)

अध्याय 4: Determinants and Matrices - Exercise 4.2 [पृष्ठ ६८]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 11 Standard Maharashtra State Board

अध्याय 4 Determinants and Matrices
Exercise 4.2 | Q 6 | पृष्ठ ६८

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Without expanding determinants show that |1366143712|+4|233212176|=10|121317326| - Mathematics and Statistics

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