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Determinant
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Determinant
All Questions (Page: 13)
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All Questions (PDF)
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Study Material (PDF)
All Questions (PDF)
English Version
Self Assesment
Sort As
Serial Number
Newly Added
Language
English
Click on each question to see answer & solution.
Question No: #121
If `x^3=1`, then prove that $ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = (a+bx+cx^2) \begin{vmatrix} 1 & b & c \\ x^2 & c & a \\ x & a & b \end{vmatrix} $
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Question No: #122
Prove that $ \begin{vmatrix} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} = 3abc - (a^3+b^3+c^3)$
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Question No: #123
Prove that: $ \begin{vmatrix} b+c & a+b & a \\ c+a & b+c & b \\ a+b & c+a & c \end{vmatrix} = a^3+b^3+c^3-3abc $
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Question No: #124
If `a, b, c` are real numbers, then prove that $ \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 2 \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = -2(a^3+b^3+c^3-3abc) $ and hence show that if the value of this determinant is zero, then either `a+b+c=0` or `a=b=c`.
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Question No: #125
Prove that $ \begin{vmatrix} 2bc-a^2 & c^2 & b^2 \\ c^2 & 2ca-b^2 & a^2 \\ b^2 & a^2 & 2ab-c^2 \end{vmatrix} = (a^3+b^3+c^3-3abc)^2$
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Question No: #126
Using properties of determinant, prove that $ \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} = 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} $
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Question No: #127
Justify the truth. $ \begin{vmatrix} b+c & a-c & a-b \\ b-c & c+a & b-a \\ c-b & c-a & a+b \end{vmatrix} = 8abc $
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Question No: #128
Prove that $ \begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix} = -2 $
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Question No: #129
If `S_n = \alpha^n+\beta^n+\gamma^n` then show that $ \begin{vmatrix} S_0 & S_1 & S_2 \\ S_1 & S_2 & S_3 \\ S_2 & S_3 & S_4 \end{vmatrix} = (\alpha-\beta)^2 (\beta-\gamma)^2 (\gamma-\alpha)^2 $
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Question No: #130
Show that the condition for which equations `x+ay+az=0`, `bx+y+bz=0`, `cx+cy+z=0` have non-zero solution is `\frac{a}{1-a]+\frac{b}{1-b}+\frac{c}{1-c}=-1`
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