Determinants
13 Board Maths previous year questions on Determinants — options free on every question; 1 include the answer & explanation free, the rest unlock with PYQ Pass.
\(\left|\begin{array}{cc}\mathrm{x}+1 & \mathrm{x}-1 \\ \mathrm{x}^2+\mathrm{x}+1 & \mathrm{x}^2-\mathrm{x}+1\end{array}\right|\) is equal to :
2
Given: \(\left|\begin{matrix}x+1 & x-1 \\ {x}^{2}+x+1 & {x}^{2}-x+1\end{matrix}\right|\)
\(=\left(x+1\right)\left({x}^{2}-x+1\right)-\left(x-1\right)\left({x}^{2}+x+1\right)\)
\(=x^3-x^2+x+x^2-x+1-(x^3+x^2+x-x^2-x-1)\)
\(=\left(x^3+1\right)-\left(x^3-1\right)=2\)
If inverse of matrix \(\left[\begin{matrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{matrix}\right]\) is the matrix \(\left[\begin{matrix}1 & 3 & 3 \\ 1 & \lambda & 3 \\ 1 & 3 & 4\end{matrix}\right]\), then value of \(\lambda\) is:
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If \(\left|\begin{array}{rrr}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right|=k a b c\), then the value of \(k\) is :
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If inverse of matrix \(\left[\begin{matrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{matrix}\right]\) is the matrix \(\left[\begin{matrix}1 & 3 & 3 \\ 1 & \lambda & 3 \\ 1 & 3 & 4\end{matrix}\right]\), then value of \(\lambda\) is:
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If A is a square matrix of order 3 such that the value of \(|adj\cdot \mathrm{A}|=8\), then the value of \(\left|{\mathrm{A}}^{\mathrm{T}}\right|\) is :
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If \(A\) is a square matrix of order \(3\) such that the value of \(|adj\cdot \mathrm{A}|=8\), then the value of \(\left|{\mathrm{A}}^{\mathrm{T}}\right|\) is :
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If \(\left|\begin{array}{rrr}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right|=k a b c\), then the value of \(k\) is :
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\(\left|\begin{array}{cc}\mathrm{x}+1 & \mathrm{x}-1 \\ \mathrm{x}^2+\mathrm{x}+1 & \mathrm{x}^2-\mathrm{x}+1\end{array}\right|\) is equal to :
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If A is an invertible matrix of order 2 , then \(\det \left({A}^{-1}\right)\) is equal to :
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If A is a nonsingular square matrix of order \(3\times 3\) and \(|A|=2\), then \(|adjA|\) is equal to -
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If A is a square matrix of order \(2\times 2\), then \(|5A|\) is equal to
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Assertion (A) : For matrix \(\mathrm{A}=\left[\begin{array}{ccc}1 & \cos \theta & 1 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1\end{array}\right]\), where \(\theta \in[0,2 \pi]\), \(|\mathrm{A}| \in[2,4]\).
Reason \((R): \quad \cos \theta \in[-1,1], \forall \theta \in[0,2 \pi]\).
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If \(\left|\begin{matrix}x & 2 \\ 18 & x\end{matrix}\right|=0\), then x is equal to
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