QBManager

Matrix and Determinant

79090 Matrix \(\mathbf{A}_{\mathrm{r}}=\left[\begin{array}{cc}\mathbf{r} & \mathbf{r}-\mathbf{1} \\ \mathbf{r}-\mathbf{1} & \mathbf{r}\end{array}\right] ; \mathbf{r}=\mathbf{1}, \mathbf{2}, \mathbf{3}, \ldots \ldots\).
If \(\sum_{\mathbf{r}=1}^{100}\left|\mathbf{A}_{\mathbf{r}}\right|=(\sqrt{\mathbf{1 0}})^{\mathrm{K}}\), then \(\mathrm{K}=\square ;\left|\mathbf{A}_{\mathbf{r}}\right|=\operatorname{det}\)

1 6

2 4

3 2

4 8

GUJCET-2019

Matrix and Determinant

79091 If \(A=\left|\begin{array}{lll}4 & 1 & 4 \\ 4 & 4 & 1\end{array}\right|\), then \(A^{2}-6 A=\)

1 \(27 \mathrm{I}_{3}\)

2 \(5 \mathrm{I}_{3}\)

3 \(20 \mathrm{I}_{3}\)

4 \(30 \mathrm{I}_{3}\)

GUJCET-2017

Matrix and Determinant

79092 \(\left|\begin{array}{cc}\sin ^2 \theta & \cos ^2 \theta \\ -\cos ^2 \theta & \sin ^2 \theta\end{array}\right|=\)______.

1 \(\frac{1}{2}\left(1+\cos ^{2} 2 \theta\right)\)

2 \(\frac{1}{2}\left(1-\sin ^{2} 2 \theta\right)\)

3 \(\cos 2 \theta\)

4 \(\frac{1}{2} \sin ^{2} 2 \theta\)

GUJCET-2019

Matrix and Determinant

79093 If \(k=p+q+r\), then the value of
\(\left|\begin{array}{ccc} \mathbf{k}+\mathbf{r} & \mathbf{p} & \mathbf{q} \\ \mathbf{r} & \mathbf{k}+\mathbf{p} & \mathbf{q} \\ \mathbf{r} & \mathbf{p} & \mathbf{k}+\mathbf{q} \end{array}\right| \text { is }\)

1 \(2 \mathrm{k}^{2}\)

2 \(2 \mathrm{k}^{3}\)

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

4 \(3 \mathrm{k}^{2}\)

Explanation:

(B) : Given,
\(\mathrm{k}=\mathrm{p}+\mathrm{q}+\mathrm{r} \text {, then }\)
\(\left|\begin{array}{clc} \mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
* Firstly we change column \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\)
Then,
\(\left|\begin{array}{clc} \mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|=\left|\begin{array}{ccc} \mathrm{k}+\mathrm{p}+\mathrm{r}+\mathrm{q} & \mathrm{p} & \mathrm{q} \\ \mathrm{k}+\mathrm{p}+\mathrm{q}+\mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{k}+\mathrm{p}+\mathrm{q}+\mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
\(=\left|\begin{array}{ccr}
2 \mathrm{k} & \mathrm{p} & \mathrm{q} \\ 2 \mathrm{k} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ 2 \mathrm{k} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
\(=2 \mathrm{k}\left|\begin{array}{rrr}
1 & \mathrm{p} & \mathrm{q} \\ 1 & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ 1 & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right| \text { (taking common) }\)
Again we change Row, \(\begin{aligned} & \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \\ & \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\end{aligned}\)
\(=2 \mathrm{k}\left|\begin{array}{ccc} 1 & \mathrm{p} & \mathrm{q} \\ 0 & \mathrm{k} & 0 \\ 0 & 0 & \mathrm{k} \end{array}\right|\)
\(=2 \mathrm{k}\left\{1\left(\mathrm{k}^{2}-0\right)-\mathrm{p}(0-0)+\mathrm{q}(0-0)\right\}\)
\(\text { * } \quad=2 \mathrm{k}\left(\mathrm{k}^{2}+0+0\right)=2 \mathrm{k}^{3} \text {. }\)
Hence,
\(\left|\begin{array}{ccc}\mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q}\end{array}\right|=2 \mathrm{k}^{3}(\) when \(\mathrm{k}=\mathrm{p}+\mathrm{q}+\mathrm{r})\)

GUJCET-2017

Matrix and Determinant

79090 Matrix \(\mathbf{A}_{\mathrm{r}}=\left[\begin{array}{cc}\mathbf{r} & \mathbf{r}-\mathbf{1} \\ \mathbf{r}-\mathbf{1} & \mathbf{r}\end{array}\right] ; \mathbf{r}=\mathbf{1}, \mathbf{2}, \mathbf{3}, \ldots \ldots\).
If \(\sum_{\mathbf{r}=1}^{100}\left|\mathbf{A}_{\mathbf{r}}\right|=(\sqrt{\mathbf{1 0}})^{\mathrm{K}}\), then \(\mathrm{K}=\square ;\left|\mathbf{A}_{\mathbf{r}}\right|=\operatorname{det}\)

1 6

2 4

3 2

4 8

GUJCET-2019

Matrix and Determinant

79091 If \(A=\left|\begin{array}{lll}4 & 1 & 4 \\ 4 & 4 & 1\end{array}\right|\), then \(A^{2}-6 A=\)

1 \(27 \mathrm{I}_{3}\)

2 \(5 \mathrm{I}_{3}\)

3 \(20 \mathrm{I}_{3}\)

4 \(30 \mathrm{I}_{3}\)

GUJCET-2017

Matrix and Determinant

79092 \(\left|\begin{array}{cc}\sin ^2 \theta & \cos ^2 \theta \\ -\cos ^2 \theta & \sin ^2 \theta\end{array}\right|=\)______.

1 \(\frac{1}{2}\left(1+\cos ^{2} 2 \theta\right)\)

2 \(\frac{1}{2}\left(1-\sin ^{2} 2 \theta\right)\)

3 \(\cos 2 \theta\)

4 \(\frac{1}{2} \sin ^{2} 2 \theta\)

GUJCET-2019

Matrix and Determinant

79093 If \(k=p+q+r\), then the value of
\(\left|\begin{array}{ccc} \mathbf{k}+\mathbf{r} & \mathbf{p} & \mathbf{q} \\ \mathbf{r} & \mathbf{k}+\mathbf{p} & \mathbf{q} \\ \mathbf{r} & \mathbf{p} & \mathbf{k}+\mathbf{q} \end{array}\right| \text { is }\)

1 \(2 \mathrm{k}^{2}\)

2 \(2 \mathrm{k}^{3}\)

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

4 \(3 \mathrm{k}^{2}\)

Explanation:

(B) : Given,
\(\mathrm{k}=\mathrm{p}+\mathrm{q}+\mathrm{r} \text {, then }\)
\(\left|\begin{array}{clc} \mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
* Firstly we change column \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\)
Then,
\(\left|\begin{array}{clc} \mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|=\left|\begin{array}{ccc} \mathrm{k}+\mathrm{p}+\mathrm{r}+\mathrm{q} & \mathrm{p} & \mathrm{q} \\ \mathrm{k}+\mathrm{p}+\mathrm{q}+\mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{k}+\mathrm{p}+\mathrm{q}+\mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
\(=\left|\begin{array}{ccr}
2 \mathrm{k} & \mathrm{p} & \mathrm{q} \\ 2 \mathrm{k} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ 2 \mathrm{k} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
\(=2 \mathrm{k}\left|\begin{array}{rrr}
1 & \mathrm{p} & \mathrm{q} \\ 1 & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ 1 & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right| \text { (taking common) }\)
Again we change Row, \(\begin{aligned} & \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \\ & \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\end{aligned}\)
\(=2 \mathrm{k}\left|\begin{array}{ccc} 1 & \mathrm{p} & \mathrm{q} \\ 0 & \mathrm{k} & 0 \\ 0 & 0 & \mathrm{k} \end{array}\right|\)
\(=2 \mathrm{k}\left\{1\left(\mathrm{k}^{2}-0\right)-\mathrm{p}(0-0)+\mathrm{q}(0-0)\right\}\)
\(\text { * } \quad=2 \mathrm{k}\left(\mathrm{k}^{2}+0+0\right)=2 \mathrm{k}^{3} \text {. }\)
Hence,
\(\left|\begin{array}{ccc}\mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q}\end{array}\right|=2 \mathrm{k}^{3}(\) when \(\mathrm{k}=\mathrm{p}+\mathrm{q}+\mathrm{r})\)

GUJCET-2017

Matrix and Determinant

79090 Matrix \(\mathbf{A}_{\mathrm{r}}=\left[\begin{array}{cc}\mathbf{r} & \mathbf{r}-\mathbf{1} \\ \mathbf{r}-\mathbf{1} & \mathbf{r}\end{array}\right] ; \mathbf{r}=\mathbf{1}, \mathbf{2}, \mathbf{3}, \ldots \ldots\).
If \(\sum_{\mathbf{r}=1}^{100}\left|\mathbf{A}_{\mathbf{r}}\right|=(\sqrt{\mathbf{1 0}})^{\mathrm{K}}\), then \(\mathrm{K}=\square ;\left|\mathbf{A}_{\mathbf{r}}\right|=\operatorname{det}\)

1 6

2 4

3 2

4 8

GUJCET-2019

Matrix and Determinant

79091 If \(A=\left|\begin{array}{lll}4 & 1 & 4 \\ 4 & 4 & 1\end{array}\right|\), then \(A^{2}-6 A=\)

1 \(27 \mathrm{I}_{3}\)

2 \(5 \mathrm{I}_{3}\)

3 \(20 \mathrm{I}_{3}\)

4 \(30 \mathrm{I}_{3}\)

GUJCET-2017

Matrix and Determinant

79092 \(\left|\begin{array}{cc}\sin ^2 \theta & \cos ^2 \theta \\ -\cos ^2 \theta & \sin ^2 \theta\end{array}\right|=\)______.

1 \(\frac{1}{2}\left(1+\cos ^{2} 2 \theta\right)\)

2 \(\frac{1}{2}\left(1-\sin ^{2} 2 \theta\right)\)

3 \(\cos 2 \theta\)

4 \(\frac{1}{2} \sin ^{2} 2 \theta\)

GUJCET-2019

Matrix and Determinant

79093 If \(k=p+q+r\), then the value of
\(\left|\begin{array}{ccc} \mathbf{k}+\mathbf{r} & \mathbf{p} & \mathbf{q} \\ \mathbf{r} & \mathbf{k}+\mathbf{p} & \mathbf{q} \\ \mathbf{r} & \mathbf{p} & \mathbf{k}+\mathbf{q} \end{array}\right| \text { is }\)

1 \(2 \mathrm{k}^{2}\)

2 \(2 \mathrm{k}^{3}\)

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

4 \(3 \mathrm{k}^{2}\)

Explanation:

(B) : Given,
\(\mathrm{k}=\mathrm{p}+\mathrm{q}+\mathrm{r} \text {, then }\)
\(\left|\begin{array}{clc} \mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
* Firstly we change column \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\)
Then,
\(\left|\begin{array}{clc} \mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|=\left|\begin{array}{ccc} \mathrm{k}+\mathrm{p}+\mathrm{r}+\mathrm{q} & \mathrm{p} & \mathrm{q} \\ \mathrm{k}+\mathrm{p}+\mathrm{q}+\mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{k}+\mathrm{p}+\mathrm{q}+\mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
\(=\left|\begin{array}{ccr}
2 \mathrm{k} & \mathrm{p} & \mathrm{q} \\ 2 \mathrm{k} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ 2 \mathrm{k} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
\(=2 \mathrm{k}\left|\begin{array}{rrr}
1 & \mathrm{p} & \mathrm{q} \\ 1 & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ 1 & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right| \text { (taking common) }\)
Again we change Row, \(\begin{aligned} & \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \\ & \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\end{aligned}\)
\(=2 \mathrm{k}\left|\begin{array}{ccc} 1 & \mathrm{p} & \mathrm{q} \\ 0 & \mathrm{k} & 0 \\ 0 & 0 & \mathrm{k} \end{array}\right|\)
\(=2 \mathrm{k}\left\{1\left(\mathrm{k}^{2}-0\right)-\mathrm{p}(0-0)+\mathrm{q}(0-0)\right\}\)
\(\text { * } \quad=2 \mathrm{k}\left(\mathrm{k}^{2}+0+0\right)=2 \mathrm{k}^{3} \text {. }\)
Hence,
\(\left|\begin{array}{ccc}\mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q}\end{array}\right|=2 \mathrm{k}^{3}(\) when \(\mathrm{k}=\mathrm{p}+\mathrm{q}+\mathrm{r})\)

GUJCET-2017

Matrix and Determinant

79090 Matrix \(\mathbf{A}_{\mathrm{r}}=\left[\begin{array}{cc}\mathbf{r} & \mathbf{r}-\mathbf{1} \\ \mathbf{r}-\mathbf{1} & \mathbf{r}\end{array}\right] ; \mathbf{r}=\mathbf{1}, \mathbf{2}, \mathbf{3}, \ldots \ldots\).
If \(\sum_{\mathbf{r}=1}^{100}\left|\mathbf{A}_{\mathbf{r}}\right|=(\sqrt{\mathbf{1 0}})^{\mathrm{K}}\), then \(\mathrm{K}=\square ;\left|\mathbf{A}_{\mathbf{r}}\right|=\operatorname{det}\)

1 6

2 4

3 2

4 8

GUJCET-2019

Matrix and Determinant

79091 If \(A=\left|\begin{array}{lll}4 & 1 & 4 \\ 4 & 4 & 1\end{array}\right|\), then \(A^{2}-6 A=\)

1 \(27 \mathrm{I}_{3}\)

2 \(5 \mathrm{I}_{3}\)

3 \(20 \mathrm{I}_{3}\)

4 \(30 \mathrm{I}_{3}\)

GUJCET-2017

Matrix and Determinant

79092 \(\left|\begin{array}{cc}\sin ^2 \theta & \cos ^2 \theta \\ -\cos ^2 \theta & \sin ^2 \theta\end{array}\right|=\)______.

1 \(\frac{1}{2}\left(1+\cos ^{2} 2 \theta\right)\)

2 \(\frac{1}{2}\left(1-\sin ^{2} 2 \theta\right)\)

3 \(\cos 2 \theta\)

4 \(\frac{1}{2} \sin ^{2} 2 \theta\)

GUJCET-2019

Matrix and Determinant

79093 If \(k=p+q+r\), then the value of
\(\left|\begin{array}{ccc} \mathbf{k}+\mathbf{r} & \mathbf{p} & \mathbf{q} \\ \mathbf{r} & \mathbf{k}+\mathbf{p} & \mathbf{q} \\ \mathbf{r} & \mathbf{p} & \mathbf{k}+\mathbf{q} \end{array}\right| \text { is }\)

1 \(2 \mathrm{k}^{2}\)

2 \(2 \mathrm{k}^{3}\)

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

4 \(3 \mathrm{k}^{2}\)

Explanation:

(B) : Given,
\(\mathrm{k}=\mathrm{p}+\mathrm{q}+\mathrm{r} \text {, then }\)
\(\left|\begin{array}{clc} \mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
* Firstly we change column \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\)
Then,
\(\left|\begin{array}{clc} \mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|=\left|\begin{array}{ccc} \mathrm{k}+\mathrm{p}+\mathrm{r}+\mathrm{q} & \mathrm{p} & \mathrm{q} \\ \mathrm{k}+\mathrm{p}+\mathrm{q}+\mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{k}+\mathrm{p}+\mathrm{q}+\mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
\(=\left|\begin{array}{ccr}
2 \mathrm{k} & \mathrm{p} & \mathrm{q} \\ 2 \mathrm{k} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ 2 \mathrm{k} & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right|\)
\(=2 \mathrm{k}\left|\begin{array}{rrr}
1 & \mathrm{p} & \mathrm{q} \\ 1 & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ 1 & \mathrm{p} & \mathrm{k}+\mathrm{q} \end{array}\right| \text { (taking common) }\)
Again we change Row, \(\begin{aligned} & \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \\ & \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\end{aligned}\)
\(=2 \mathrm{k}\left|\begin{array}{ccc} 1 & \mathrm{p} & \mathrm{q} \\ 0 & \mathrm{k} & 0 \\ 0 & 0 & \mathrm{k} \end{array}\right|\)
\(=2 \mathrm{k}\left\{1\left(\mathrm{k}^{2}-0\right)-\mathrm{p}(0-0)+\mathrm{q}(0-0)\right\}\)
\(\text { * } \quad=2 \mathrm{k}\left(\mathrm{k}^{2}+0+0\right)=2 \mathrm{k}^{3} \text {. }\)
Hence,
\(\left|\begin{array}{ccc}\mathrm{k}+\mathrm{r} & \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{k}+\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{p} & \mathrm{k}+\mathrm{q}\end{array}\right|=2 \mathrm{k}^{3}(\) when \(\mathrm{k}=\mathrm{p}+\mathrm{q}+\mathrm{r})\)

GUJCET-2017

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: