79058
If \(\left|\begin{array}{ccc}p & q-y & r-z \\ p-x & q & r-z\end{array}\right|=0\), then the value of \(\mathbf{p}-\mathbf{x} \quad \mathbf{q}-\mathbf{y} \quad \mathbf{r}\) \(\frac{\mathbf{p}}{\mathbf{x}}+\frac{\mathbf{q}}{\mathbf{y}}+\frac{\mathbf{r}}{\mathbf{z}}\) is
1 0
2 1
3 2
4 \(4 \mathrm{pqr}\)
Explanation:
(C) : \(\left|\begin{array}{ccc}p & q-y & r-z \\ p-x & q & r-z \\ p-x & q-y & r\end{array}\right|=0\) Apply \(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\) and \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\), we get \(\left|\begin{array}{ccc} x & 0 & -z \\ 0 & y & -z \\ p-x & q-y & r \end{array}\right|=0\) \(\Rightarrow x[y r+z(q-y)]-z[0-y(p-x)]=0\) \(\Rightarrow x y r+x z q-x z y+y z p-z y x=0\) \(\Rightarrow x y r+z x q+y z p=2 x y z \Rightarrow \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2\)
BITSAT-2017
Matrix and Determinant
79059
Let \(M=\left(\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right)\). Then \(\frac{1}{3} \operatorname{det}\left(3\left(M+M^{T}\right)\right)\) is
79058
If \(\left|\begin{array}{ccc}p & q-y & r-z \\ p-x & q & r-z\end{array}\right|=0\), then the value of \(\mathbf{p}-\mathbf{x} \quad \mathbf{q}-\mathbf{y} \quad \mathbf{r}\) \(\frac{\mathbf{p}}{\mathbf{x}}+\frac{\mathbf{q}}{\mathbf{y}}+\frac{\mathbf{r}}{\mathbf{z}}\) is
1 0
2 1
3 2
4 \(4 \mathrm{pqr}\)
Explanation:
(C) : \(\left|\begin{array}{ccc}p & q-y & r-z \\ p-x & q & r-z \\ p-x & q-y & r\end{array}\right|=0\) Apply \(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\) and \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\), we get \(\left|\begin{array}{ccc} x & 0 & -z \\ 0 & y & -z \\ p-x & q-y & r \end{array}\right|=0\) \(\Rightarrow x[y r+z(q-y)]-z[0-y(p-x)]=0\) \(\Rightarrow x y r+x z q-x z y+y z p-z y x=0\) \(\Rightarrow x y r+z x q+y z p=2 x y z \Rightarrow \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2\)
BITSAT-2017
Matrix and Determinant
79059
Let \(M=\left(\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right)\). Then \(\frac{1}{3} \operatorname{det}\left(3\left(M+M^{T}\right)\right)\) is
79058
If \(\left|\begin{array}{ccc}p & q-y & r-z \\ p-x & q & r-z\end{array}\right|=0\), then the value of \(\mathbf{p}-\mathbf{x} \quad \mathbf{q}-\mathbf{y} \quad \mathbf{r}\) \(\frac{\mathbf{p}}{\mathbf{x}}+\frac{\mathbf{q}}{\mathbf{y}}+\frac{\mathbf{r}}{\mathbf{z}}\) is
1 0
2 1
3 2
4 \(4 \mathrm{pqr}\)
Explanation:
(C) : \(\left|\begin{array}{ccc}p & q-y & r-z \\ p-x & q & r-z \\ p-x & q-y & r\end{array}\right|=0\) Apply \(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\) and \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\), we get \(\left|\begin{array}{ccc} x & 0 & -z \\ 0 & y & -z \\ p-x & q-y & r \end{array}\right|=0\) \(\Rightarrow x[y r+z(q-y)]-z[0-y(p-x)]=0\) \(\Rightarrow x y r+x z q-x z y+y z p-z y x=0\) \(\Rightarrow x y r+z x q+y z p=2 x y z \Rightarrow \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2\)
BITSAT-2017
Matrix and Determinant
79059
Let \(M=\left(\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right)\). Then \(\frac{1}{3} \operatorname{det}\left(3\left(M+M^{T}\right)\right)\) is
79058
If \(\left|\begin{array}{ccc}p & q-y & r-z \\ p-x & q & r-z\end{array}\right|=0\), then the value of \(\mathbf{p}-\mathbf{x} \quad \mathbf{q}-\mathbf{y} \quad \mathbf{r}\) \(\frac{\mathbf{p}}{\mathbf{x}}+\frac{\mathbf{q}}{\mathbf{y}}+\frac{\mathbf{r}}{\mathbf{z}}\) is
1 0
2 1
3 2
4 \(4 \mathrm{pqr}\)
Explanation:
(C) : \(\left|\begin{array}{ccc}p & q-y & r-z \\ p-x & q & r-z \\ p-x & q-y & r\end{array}\right|=0\) Apply \(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\) and \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\), we get \(\left|\begin{array}{ccc} x & 0 & -z \\ 0 & y & -z \\ p-x & q-y & r \end{array}\right|=0\) \(\Rightarrow x[y r+z(q-y)]-z[0-y(p-x)]=0\) \(\Rightarrow x y r+x z q-x z y+y z p-z y x=0\) \(\Rightarrow x y r+z x q+y z p=2 x y z \Rightarrow \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2\)
BITSAT-2017
Matrix and Determinant
79059
Let \(M=\left(\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right)\). Then \(\frac{1}{3} \operatorname{det}\left(3\left(M+M^{T}\right)\right)\) is