QBManager

Matrix and Determinant

79173 What is the area of the triangle with vertices at \((\mathbf{0}, \mathbf{0}, \mathbf{0}),(\mathbf{2}, \mathbf{0}, \mathbf{0})\) and \((\mathbf{0},-\mathbf{2}, \mathbf{0})\) ?

1 \(\frac{1}{2}\) square unit

2 1 square unit

3 2 square units

4 4 square units

SCRA-2014

Matrix and Determinant

79174 The set of all values of \(t \in R\), for which the matrix
\(\left[\begin{array}{ccc}e^{t} & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^{t} & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^{t} & e^{-t} \cos t & e^{-1} \sin t\end{array}\right]\) is
invertible, is

1 \(\mathrm{R}\)

2 \(\left\{(2 \mathrm{k}+1) \frac{\pi}{2}, \mathrm{k} \in \mathrm{Z}\right\}\)

3 \(\{\mathrm{k} \pi, \mathrm{k} \in \mathrm{Z}\}\)

4 \(\left\{\mathrm{k} \pi+\frac{\pi}{4}, \mathrm{k} \in \mathrm{Z}\right\}\)

Explanation:

(A) : Given,
\(\left[\begin{array}{ccc} e^{t} & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^{t} & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^{t} & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]\)
Now, taking common \(\mathrm{e}^{\mathrm{t}}, \mathrm{e}^{-\mathrm{t}}, \mathrm{e}^{-\mathrm{t}}\) from I, II and III column respetively -
\(=\mathrm{e}^{\mathrm{t}} \cdot \mathrm{e}^{-\mathrm{t}} \cdot \mathrm{e}^{-\mathrm{t}}\left|\begin{array}{ccc} 1 & \sin \mathrm{t}-2 \cos \mathrm{t} & -2 \sin \mathrm{t}-\cos \mathrm{t} \\ 1 & 2 \sin \mathrm{t}+\cos \mathrm{t} & \sin \mathrm{t}-2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\)
\(=\mathrm{e}^{\mathrm{t}}\left[\begin{array}{ccc} 1 & (\sin \mathrm{t}-2 \cos \mathrm{t}) & (-2 \sin \mathrm{t}-\cos \mathrm{t}) \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right]\)
\(=\mathrm{e}^{-\mathrm{t}}\left|\begin{array}{ccc} 0 & -\sin \mathrm{t}-3 \cos \mathrm{t} & -3 \sin \mathrm{t}-\cos \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
\(=\mathrm{e}^{\mathrm{t}}\left|\begin{array}{ccc} 1 & -2 \cos \mathrm{t} & -2 \sin \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
\(\mathrm{e}^{\mathrm{t}}\left[\begin{array}{ccc} 1 & \sin \mathrm{t}-2 \cos \mathrm{t} & -2 \sin \mathrm{t}-\cos \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos t \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right]\)
Operating \(\mathrm{R}_{1} \rightarrow 2 \mathrm{R}_{1}-\mathrm{R}_{2}\)
\(=e^{t}\left|\begin{array}{ccc} 2 & -2 \cos t & -2 \sin t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t \end{array}\right|\)
\(4 e^{t}\left[\begin{array}{rrr}
1 & -\cos t & -\sin t \\ 0 & \sin t & -\cos t \\ 1 & \cos t & \sin t \end{array}\right]\)
Expanding along \(\mathrm{C}_{1}\),
\(4 \mathrm{e}^{\mathrm{t}}\left(1\left[\sin ^{2}+\cos ^{2} \mathrm{t}\right]+\left[\cos ^{2}+\sin ^{2} \mathrm{t}\right]\right)\)
\(4 \mathrm{e}^{\mathrm{t}}(1+1)=8 \mathrm{e}^{\mathrm{t}}\)
Now, for A being invertible then
\(8 e^{t}\) must be equal to zero which is not possible for any real numbers-
Hence, the set of all values of \(t \in R\) for which the matrix \(\Delta\) is invertible is \(\mathrm{R}\).

JEE Main-2023-29.01.2023

Matrix and Determinant

79175 If \(y=\sin m x\), then the value of the determinant
\(\left|\begin{array}{lll}y^{\prime} & \mathbf{y}_{1} & \mathbf{y}_{2} \\ \mathbf{y}_{3} & \mathbf{y}_{4} & \mathbf{y}_{5} \\ \mathbf{y}_{6} & \mathbf{y}_{7} & \mathbf{y}_{8}\end{array}\right|\) where \(\mathbf{y}_{\mathbf{n}}=\frac{\mathbf{d}^{\mathrm{n}} \mathbf{y}}{\mathbf{d x}^{\mathbf{n}}}\) is

1 \(\mathrm{m}^{9}\)

2 \(\mathrm{m}^{2}\)

3 \(\mathrm{m}^{3}\)

4 none of these

AMU-2017

Matrix and Determinant

79176 If \(f(x)=\left|\begin{array}{ccc}x-3 & 2 x^{2}-18 & 3 x^{3}-81 \\ x-5 & 2 x^{2}-50 & 4 x^{3}-500 \\ 1 & 2 & 3\end{array}\right|\), then
\(\boldsymbol{F}(\mathbf{1}) \boldsymbol{f ( 3 )}+\boldsymbol{f ( 5 )}+\boldsymbol{f ( 1 )}\) is equal to

1 \(\mathrm{f}(1)\)

2 \(\mathrm{f}(3)\)

3 \(f(1)+f(3)\)

4 \(\mathrm{f}(1)+\mathrm{f}(5)\)

AP EAMCET-2020-17.09.2020

Matrix and Determinant

79173 What is the area of the triangle with vertices at \((\mathbf{0}, \mathbf{0}, \mathbf{0}),(\mathbf{2}, \mathbf{0}, \mathbf{0})\) and \((\mathbf{0},-\mathbf{2}, \mathbf{0})\) ?

1 \(\frac{1}{2}\) square unit

2 1 square unit

3 2 square units

4 4 square units

SCRA-2014

Matrix and Determinant

79174 The set of all values of \(t \in R\), for which the matrix
\(\left[\begin{array}{ccc}e^{t} & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^{t} & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^{t} & e^{-t} \cos t & e^{-1} \sin t\end{array}\right]\) is
invertible, is

1 \(\mathrm{R}\)

2 \(\left\{(2 \mathrm{k}+1) \frac{\pi}{2}, \mathrm{k} \in \mathrm{Z}\right\}\)

3 \(\{\mathrm{k} \pi, \mathrm{k} \in \mathrm{Z}\}\)

4 \(\left\{\mathrm{k} \pi+\frac{\pi}{4}, \mathrm{k} \in \mathrm{Z}\right\}\)

Explanation:

(A) : Given,
\(\left[\begin{array}{ccc} e^{t} & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^{t} & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^{t} & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]\)
Now, taking common \(\mathrm{e}^{\mathrm{t}}, \mathrm{e}^{-\mathrm{t}}, \mathrm{e}^{-\mathrm{t}}\) from I, II and III column respetively -
\(=\mathrm{e}^{\mathrm{t}} \cdot \mathrm{e}^{-\mathrm{t}} \cdot \mathrm{e}^{-\mathrm{t}}\left|\begin{array}{ccc} 1 & \sin \mathrm{t}-2 \cos \mathrm{t} & -2 \sin \mathrm{t}-\cos \mathrm{t} \\ 1 & 2 \sin \mathrm{t}+\cos \mathrm{t} & \sin \mathrm{t}-2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\)
\(=\mathrm{e}^{\mathrm{t}}\left[\begin{array}{ccc} 1 & (\sin \mathrm{t}-2 \cos \mathrm{t}) & (-2 \sin \mathrm{t}-\cos \mathrm{t}) \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right]\)
\(=\mathrm{e}^{-\mathrm{t}}\left|\begin{array}{ccc} 0 & -\sin \mathrm{t}-3 \cos \mathrm{t} & -3 \sin \mathrm{t}-\cos \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
\(=\mathrm{e}^{\mathrm{t}}\left|\begin{array}{ccc} 1 & -2 \cos \mathrm{t} & -2 \sin \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
\(\mathrm{e}^{\mathrm{t}}\left[\begin{array}{ccc} 1 & \sin \mathrm{t}-2 \cos \mathrm{t} & -2 \sin \mathrm{t}-\cos \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos t \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right]\)
Operating \(\mathrm{R}_{1} \rightarrow 2 \mathrm{R}_{1}-\mathrm{R}_{2}\)
\(=e^{t}\left|\begin{array}{ccc} 2 & -2 \cos t & -2 \sin t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t \end{array}\right|\)
\(4 e^{t}\left[\begin{array}{rrr}
1 & -\cos t & -\sin t \\ 0 & \sin t & -\cos t \\ 1 & \cos t & \sin t \end{array}\right]\)
Expanding along \(\mathrm{C}_{1}\),
\(4 \mathrm{e}^{\mathrm{t}}\left(1\left[\sin ^{2}+\cos ^{2} \mathrm{t}\right]+\left[\cos ^{2}+\sin ^{2} \mathrm{t}\right]\right)\)
\(4 \mathrm{e}^{\mathrm{t}}(1+1)=8 \mathrm{e}^{\mathrm{t}}\)
Now, for A being invertible then
\(8 e^{t}\) must be equal to zero which is not possible for any real numbers-
Hence, the set of all values of \(t \in R\) for which the matrix \(\Delta\) is invertible is \(\mathrm{R}\).

JEE Main-2023-29.01.2023

Matrix and Determinant

79175 If \(y=\sin m x\), then the value of the determinant
\(\left|\begin{array}{lll}y^{\prime} & \mathbf{y}_{1} & \mathbf{y}_{2} \\ \mathbf{y}_{3} & \mathbf{y}_{4} & \mathbf{y}_{5} \\ \mathbf{y}_{6} & \mathbf{y}_{7} & \mathbf{y}_{8}\end{array}\right|\) where \(\mathbf{y}_{\mathbf{n}}=\frac{\mathbf{d}^{\mathrm{n}} \mathbf{y}}{\mathbf{d x}^{\mathbf{n}}}\) is

1 \(\mathrm{m}^{9}\)

2 \(\mathrm{m}^{2}\)

3 \(\mathrm{m}^{3}\)

4 none of these

AMU-2017

Matrix and Determinant

79176 If \(f(x)=\left|\begin{array}{ccc}x-3 & 2 x^{2}-18 & 3 x^{3}-81 \\ x-5 & 2 x^{2}-50 & 4 x^{3}-500 \\ 1 & 2 & 3\end{array}\right|\), then
\(\boldsymbol{F}(\mathbf{1}) \boldsymbol{f ( 3 )}+\boldsymbol{f ( 5 )}+\boldsymbol{f ( 1 )}\) is equal to

1 \(\mathrm{f}(1)\)

2 \(\mathrm{f}(3)\)

3 \(f(1)+f(3)\)

4 \(\mathrm{f}(1)+\mathrm{f}(5)\)

AP EAMCET-2020-17.09.2020

Matrix and Determinant

79173 What is the area of the triangle with vertices at \((\mathbf{0}, \mathbf{0}, \mathbf{0}),(\mathbf{2}, \mathbf{0}, \mathbf{0})\) and \((\mathbf{0},-\mathbf{2}, \mathbf{0})\) ?

1 \(\frac{1}{2}\) square unit

2 1 square unit

3 2 square units

4 4 square units

SCRA-2014

Matrix and Determinant

79174 The set of all values of \(t \in R\), for which the matrix
\(\left[\begin{array}{ccc}e^{t} & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^{t} & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^{t} & e^{-t} \cos t & e^{-1} \sin t\end{array}\right]\) is
invertible, is

1 \(\mathrm{R}\)

2 \(\left\{(2 \mathrm{k}+1) \frac{\pi}{2}, \mathrm{k} \in \mathrm{Z}\right\}\)

3 \(\{\mathrm{k} \pi, \mathrm{k} \in \mathrm{Z}\}\)

4 \(\left\{\mathrm{k} \pi+\frac{\pi}{4}, \mathrm{k} \in \mathrm{Z}\right\}\)

Explanation:

(A) : Given,
\(\left[\begin{array}{ccc} e^{t} & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^{t} & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^{t} & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]\)
Now, taking common \(\mathrm{e}^{\mathrm{t}}, \mathrm{e}^{-\mathrm{t}}, \mathrm{e}^{-\mathrm{t}}\) from I, II and III column respetively -
\(=\mathrm{e}^{\mathrm{t}} \cdot \mathrm{e}^{-\mathrm{t}} \cdot \mathrm{e}^{-\mathrm{t}}\left|\begin{array}{ccc} 1 & \sin \mathrm{t}-2 \cos \mathrm{t} & -2 \sin \mathrm{t}-\cos \mathrm{t} \\ 1 & 2 \sin \mathrm{t}+\cos \mathrm{t} & \sin \mathrm{t}-2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\)
\(=\mathrm{e}^{\mathrm{t}}\left[\begin{array}{ccc} 1 & (\sin \mathrm{t}-2 \cos \mathrm{t}) & (-2 \sin \mathrm{t}-\cos \mathrm{t}) \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right]\)
\(=\mathrm{e}^{-\mathrm{t}}\left|\begin{array}{ccc} 0 & -\sin \mathrm{t}-3 \cos \mathrm{t} & -3 \sin \mathrm{t}-\cos \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
\(=\mathrm{e}^{\mathrm{t}}\left|\begin{array}{ccc} 1 & -2 \cos \mathrm{t} & -2 \sin \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
\(\mathrm{e}^{\mathrm{t}}\left[\begin{array}{ccc} 1 & \sin \mathrm{t}-2 \cos \mathrm{t} & -2 \sin \mathrm{t}-\cos \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos t \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right]\)
Operating \(\mathrm{R}_{1} \rightarrow 2 \mathrm{R}_{1}-\mathrm{R}_{2}\)
\(=e^{t}\left|\begin{array}{ccc} 2 & -2 \cos t & -2 \sin t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t \end{array}\right|\)
\(4 e^{t}\left[\begin{array}{rrr}
1 & -\cos t & -\sin t \\ 0 & \sin t & -\cos t \\ 1 & \cos t & \sin t \end{array}\right]\)
Expanding along \(\mathrm{C}_{1}\),
\(4 \mathrm{e}^{\mathrm{t}}\left(1\left[\sin ^{2}+\cos ^{2} \mathrm{t}\right]+\left[\cos ^{2}+\sin ^{2} \mathrm{t}\right]\right)\)
\(4 \mathrm{e}^{\mathrm{t}}(1+1)=8 \mathrm{e}^{\mathrm{t}}\)
Now, for A being invertible then
\(8 e^{t}\) must be equal to zero which is not possible for any real numbers-
Hence, the set of all values of \(t \in R\) for which the matrix \(\Delta\) is invertible is \(\mathrm{R}\).

JEE Main-2023-29.01.2023

Matrix and Determinant

79175 If \(y=\sin m x\), then the value of the determinant
\(\left|\begin{array}{lll}y^{\prime} & \mathbf{y}_{1} & \mathbf{y}_{2} \\ \mathbf{y}_{3} & \mathbf{y}_{4} & \mathbf{y}_{5} \\ \mathbf{y}_{6} & \mathbf{y}_{7} & \mathbf{y}_{8}\end{array}\right|\) where \(\mathbf{y}_{\mathbf{n}}=\frac{\mathbf{d}^{\mathrm{n}} \mathbf{y}}{\mathbf{d x}^{\mathbf{n}}}\) is

1 \(\mathrm{m}^{9}\)

2 \(\mathrm{m}^{2}\)

3 \(\mathrm{m}^{3}\)

4 none of these

AMU-2017

Matrix and Determinant

79176 If \(f(x)=\left|\begin{array}{ccc}x-3 & 2 x^{2}-18 & 3 x^{3}-81 \\ x-5 & 2 x^{2}-50 & 4 x^{3}-500 \\ 1 & 2 & 3\end{array}\right|\), then
\(\boldsymbol{F}(\mathbf{1}) \boldsymbol{f ( 3 )}+\boldsymbol{f ( 5 )}+\boldsymbol{f ( 1 )}\) is equal to

1 \(\mathrm{f}(1)\)

2 \(\mathrm{f}(3)\)

3 \(f(1)+f(3)\)

4 \(\mathrm{f}(1)+\mathrm{f}(5)\)

AP EAMCET-2020-17.09.2020

Matrix and Determinant

79173 What is the area of the triangle with vertices at \((\mathbf{0}, \mathbf{0}, \mathbf{0}),(\mathbf{2}, \mathbf{0}, \mathbf{0})\) and \((\mathbf{0},-\mathbf{2}, \mathbf{0})\) ?

1 \(\frac{1}{2}\) square unit

2 1 square unit

3 2 square units

4 4 square units

SCRA-2014

Matrix and Determinant

79174 The set of all values of \(t \in R\), for which the matrix
\(\left[\begin{array}{ccc}e^{t} & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^{t} & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^{t} & e^{-t} \cos t & e^{-1} \sin t\end{array}\right]\) is
invertible, is

1 \(\mathrm{R}\)

2 \(\left\{(2 \mathrm{k}+1) \frac{\pi}{2}, \mathrm{k} \in \mathrm{Z}\right\}\)

3 \(\{\mathrm{k} \pi, \mathrm{k} \in \mathrm{Z}\}\)

4 \(\left\{\mathrm{k} \pi+\frac{\pi}{4}, \mathrm{k} \in \mathrm{Z}\right\}\)

Explanation:

(A) : Given,
\(\left[\begin{array}{ccc} e^{t} & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^{t} & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^{t} & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]\)
Now, taking common \(\mathrm{e}^{\mathrm{t}}, \mathrm{e}^{-\mathrm{t}}, \mathrm{e}^{-\mathrm{t}}\) from I, II and III column respetively -
\(=\mathrm{e}^{\mathrm{t}} \cdot \mathrm{e}^{-\mathrm{t}} \cdot \mathrm{e}^{-\mathrm{t}}\left|\begin{array}{ccc} 1 & \sin \mathrm{t}-2 \cos \mathrm{t} & -2 \sin \mathrm{t}-\cos \mathrm{t} \\ 1 & 2 \sin \mathrm{t}+\cos \mathrm{t} & \sin \mathrm{t}-2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\)
\(=\mathrm{e}^{\mathrm{t}}\left[\begin{array}{ccc} 1 & (\sin \mathrm{t}-2 \cos \mathrm{t}) & (-2 \sin \mathrm{t}-\cos \mathrm{t}) \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right]\)
\(=\mathrm{e}^{-\mathrm{t}}\left|\begin{array}{ccc} 0 & -\sin \mathrm{t}-3 \cos \mathrm{t} & -3 \sin \mathrm{t}-\cos \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
\(=\mathrm{e}^{\mathrm{t}}\left|\begin{array}{ccc} 1 & -2 \cos \mathrm{t} & -2 \sin \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos \mathrm{t} \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right|\)
\(\mathrm{e}^{\mathrm{t}}\left[\begin{array}{ccc} 1 & \sin \mathrm{t}-2 \cos \mathrm{t} & -2 \sin \mathrm{t}-\cos \mathrm{t} \\ 0 & 2 \sin \mathrm{t} & -2 \cos t \\ 1 & \cos \mathrm{t} & \sin \mathrm{t} \end{array}\right]\)
Operating \(\mathrm{R}_{1} \rightarrow 2 \mathrm{R}_{1}-\mathrm{R}_{2}\)
\(=e^{t}\left|\begin{array}{ccc} 2 & -2 \cos t & -2 \sin t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t \end{array}\right|\)
\(4 e^{t}\left[\begin{array}{rrr}
1 & -\cos t & -\sin t \\ 0 & \sin t & -\cos t \\ 1 & \cos t & \sin t \end{array}\right]\)
Expanding along \(\mathrm{C}_{1}\),
\(4 \mathrm{e}^{\mathrm{t}}\left(1\left[\sin ^{2}+\cos ^{2} \mathrm{t}\right]+\left[\cos ^{2}+\sin ^{2} \mathrm{t}\right]\right)\)
\(4 \mathrm{e}^{\mathrm{t}}(1+1)=8 \mathrm{e}^{\mathrm{t}}\)
Now, for A being invertible then
\(8 e^{t}\) must be equal to zero which is not possible for any real numbers-
Hence, the set of all values of \(t \in R\) for which the matrix \(\Delta\) is invertible is \(\mathrm{R}\).

JEE Main-2023-29.01.2023

Matrix and Determinant

79175 If \(y=\sin m x\), then the value of the determinant
\(\left|\begin{array}{lll}y^{\prime} & \mathbf{y}_{1} & \mathbf{y}_{2} \\ \mathbf{y}_{3} & \mathbf{y}_{4} & \mathbf{y}_{5} \\ \mathbf{y}_{6} & \mathbf{y}_{7} & \mathbf{y}_{8}\end{array}\right|\) where \(\mathbf{y}_{\mathbf{n}}=\frac{\mathbf{d}^{\mathrm{n}} \mathbf{y}}{\mathbf{d x}^{\mathbf{n}}}\) is

1 \(\mathrm{m}^{9}\)

2 \(\mathrm{m}^{2}\)

3 \(\mathrm{m}^{3}\)

4 none of these

AMU-2017

Matrix and Determinant

79176 If \(f(x)=\left|\begin{array}{ccc}x-3 & 2 x^{2}-18 & 3 x^{3}-81 \\ x-5 & 2 x^{2}-50 & 4 x^{3}-500 \\ 1 & 2 & 3\end{array}\right|\), then
\(\boldsymbol{F}(\mathbf{1}) \boldsymbol{f ( 3 )}+\boldsymbol{f ( 5 )}+\boldsymbol{f ( 1 )}\) is equal to

1 \(\mathrm{f}(1)\)

2 \(\mathrm{f}(3)\)

3 \(f(1)+f(3)\)

4 \(\mathrm{f}(1)+\mathrm{f}(5)\)

AP EAMCET-2020-17.09.2020

Question Bank Series

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