79157
If \(\alpha\) is a non-real cube root of -2 , then the value of \(\left|\begin{array}{ccc}1 & 2 \alpha & 1 \\ \alpha^{2} & 1 & 3 \alpha^{2} \\ 2 & 2 \alpha & 1\end{array}\right|\), is
79158
If \(f(x)=\left|\begin{array}{ccc}x+\lambda & x & x \\ x & x+\lambda & \mathbf{x} \\ \mathbf{x} & \mathbf{x} & \mathbf{x}+\lambda\end{array}\right|\), then \(f(3 x)-f(x)=\)
1 \(3 x \lambda^{2}\)
2 \(4 x \lambda^{2}\)
3 \(6 x \lambda^{2}\)
4 \(x \lambda^{2}\)
Explanation:
(C) : It is given that, \(f(x)=\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|\) Operating : \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\) \(f(x)=\left|\begin{array}{ccc}3 x+\lambda & x & x \\ 3 x+\lambda & x+\lambda & x \\ 3 x+\lambda & x & x+\lambda\end{array}\right|=(3 x+\lambda)\left|\begin{array}{ccc}1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda\end{array}\right|\) Now, Operating: \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) we get, \(=(3 \mathrm{x}+\lambda)\left|\begin{array}{lll}1 & \mathrm{x} & \mathrm{x} \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda\end{array}\right|=(3 \mathrm{x}+\lambda) \lambda^{2}\) \(\therefore \mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})=(9 \mathrm{x}+\lambda) \lambda^{2}-(3 \mathrm{x}+\lambda) \lambda^{2}=6 \mathrm{x} \lambda^{2}\) \(\left|\begin{array}{lll} x^{2}+x & 3 x-1 & -x+3 \end{array}\right|\)
BCECE-2017
Matrix and Determinant
79159
If \(2 x+1 \quad 2+x^{2} \quad x^{3}-3\) \(\begin{array}{lll}x-3 & x^{2}+4 \quad 3 x\end{array}\) \(=\mathbf{a}_{0}+\mathbf{a}_{1} \mathbf{x}+\mathbf{a}_{2} \mathbf{x}^{2}+\ldots .+\mathbf{a}_{7} \mathbf{x}^{7}\), Then the value of \(a_{0}\) is
79157
If \(\alpha\) is a non-real cube root of -2 , then the value of \(\left|\begin{array}{ccc}1 & 2 \alpha & 1 \\ \alpha^{2} & 1 & 3 \alpha^{2} \\ 2 & 2 \alpha & 1\end{array}\right|\), is
79158
If \(f(x)=\left|\begin{array}{ccc}x+\lambda & x & x \\ x & x+\lambda & \mathbf{x} \\ \mathbf{x} & \mathbf{x} & \mathbf{x}+\lambda\end{array}\right|\), then \(f(3 x)-f(x)=\)
1 \(3 x \lambda^{2}\)
2 \(4 x \lambda^{2}\)
3 \(6 x \lambda^{2}\)
4 \(x \lambda^{2}\)
Explanation:
(C) : It is given that, \(f(x)=\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|\) Operating : \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\) \(f(x)=\left|\begin{array}{ccc}3 x+\lambda & x & x \\ 3 x+\lambda & x+\lambda & x \\ 3 x+\lambda & x & x+\lambda\end{array}\right|=(3 x+\lambda)\left|\begin{array}{ccc}1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda\end{array}\right|\) Now, Operating: \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) we get, \(=(3 \mathrm{x}+\lambda)\left|\begin{array}{lll}1 & \mathrm{x} & \mathrm{x} \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda\end{array}\right|=(3 \mathrm{x}+\lambda) \lambda^{2}\) \(\therefore \mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})=(9 \mathrm{x}+\lambda) \lambda^{2}-(3 \mathrm{x}+\lambda) \lambda^{2}=6 \mathrm{x} \lambda^{2}\) \(\left|\begin{array}{lll} x^{2}+x & 3 x-1 & -x+3 \end{array}\right|\)
BCECE-2017
Matrix and Determinant
79159
If \(2 x+1 \quad 2+x^{2} \quad x^{3}-3\) \(\begin{array}{lll}x-3 & x^{2}+4 \quad 3 x\end{array}\) \(=\mathbf{a}_{0}+\mathbf{a}_{1} \mathbf{x}+\mathbf{a}_{2} \mathbf{x}^{2}+\ldots .+\mathbf{a}_{7} \mathbf{x}^{7}\), Then the value of \(a_{0}\) is
79157
If \(\alpha\) is a non-real cube root of -2 , then the value of \(\left|\begin{array}{ccc}1 & 2 \alpha & 1 \\ \alpha^{2} & 1 & 3 \alpha^{2} \\ 2 & 2 \alpha & 1\end{array}\right|\), is
79158
If \(f(x)=\left|\begin{array}{ccc}x+\lambda & x & x \\ x & x+\lambda & \mathbf{x} \\ \mathbf{x} & \mathbf{x} & \mathbf{x}+\lambda\end{array}\right|\), then \(f(3 x)-f(x)=\)
1 \(3 x \lambda^{2}\)
2 \(4 x \lambda^{2}\)
3 \(6 x \lambda^{2}\)
4 \(x \lambda^{2}\)
Explanation:
(C) : It is given that, \(f(x)=\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|\) Operating : \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\) \(f(x)=\left|\begin{array}{ccc}3 x+\lambda & x & x \\ 3 x+\lambda & x+\lambda & x \\ 3 x+\lambda & x & x+\lambda\end{array}\right|=(3 x+\lambda)\left|\begin{array}{ccc}1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda\end{array}\right|\) Now, Operating: \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) we get, \(=(3 \mathrm{x}+\lambda)\left|\begin{array}{lll}1 & \mathrm{x} & \mathrm{x} \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda\end{array}\right|=(3 \mathrm{x}+\lambda) \lambda^{2}\) \(\therefore \mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})=(9 \mathrm{x}+\lambda) \lambda^{2}-(3 \mathrm{x}+\lambda) \lambda^{2}=6 \mathrm{x} \lambda^{2}\) \(\left|\begin{array}{lll} x^{2}+x & 3 x-1 & -x+3 \end{array}\right|\)
BCECE-2017
Matrix and Determinant
79159
If \(2 x+1 \quad 2+x^{2} \quad x^{3}-3\) \(\begin{array}{lll}x-3 & x^{2}+4 \quad 3 x\end{array}\) \(=\mathbf{a}_{0}+\mathbf{a}_{1} \mathbf{x}+\mathbf{a}_{2} \mathbf{x}^{2}+\ldots .+\mathbf{a}_{7} \mathbf{x}^{7}\), Then the value of \(a_{0}\) is
79157
If \(\alpha\) is a non-real cube root of -2 , then the value of \(\left|\begin{array}{ccc}1 & 2 \alpha & 1 \\ \alpha^{2} & 1 & 3 \alpha^{2} \\ 2 & 2 \alpha & 1\end{array}\right|\), is
79158
If \(f(x)=\left|\begin{array}{ccc}x+\lambda & x & x \\ x & x+\lambda & \mathbf{x} \\ \mathbf{x} & \mathbf{x} & \mathbf{x}+\lambda\end{array}\right|\), then \(f(3 x)-f(x)=\)
1 \(3 x \lambda^{2}\)
2 \(4 x \lambda^{2}\)
3 \(6 x \lambda^{2}\)
4 \(x \lambda^{2}\)
Explanation:
(C) : It is given that, \(f(x)=\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|\) Operating : \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\) \(f(x)=\left|\begin{array}{ccc}3 x+\lambda & x & x \\ 3 x+\lambda & x+\lambda & x \\ 3 x+\lambda & x & x+\lambda\end{array}\right|=(3 x+\lambda)\left|\begin{array}{ccc}1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda\end{array}\right|\) Now, Operating: \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) we get, \(=(3 \mathrm{x}+\lambda)\left|\begin{array}{lll}1 & \mathrm{x} & \mathrm{x} \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda\end{array}\right|=(3 \mathrm{x}+\lambda) \lambda^{2}\) \(\therefore \mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})=(9 \mathrm{x}+\lambda) \lambda^{2}-(3 \mathrm{x}+\lambda) \lambda^{2}=6 \mathrm{x} \lambda^{2}\) \(\left|\begin{array}{lll} x^{2}+x & 3 x-1 & -x+3 \end{array}\right|\)
BCECE-2017
Matrix and Determinant
79159
If \(2 x+1 \quad 2+x^{2} \quad x^{3}-3\) \(\begin{array}{lll}x-3 & x^{2}+4 \quad 3 x\end{array}\) \(=\mathbf{a}_{0}+\mathbf{a}_{1} \mathbf{x}+\mathbf{a}_{2} \mathbf{x}^{2}+\ldots .+\mathbf{a}_{7} \mathbf{x}^{7}\), Then the value of \(a_{0}\) is