QBManager

Matrix and Determinant

79094 If
\(\begin{array}{lll}
1! & 2! & 3! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 ! \end{array}=2016K \text {, then } K=\)

1 84

2 \(\frac{1}{24}\)

3 24

4 \(\frac{1}{84}\)

GUJCET-2019

Matrix and Determinant

79095 If \(A\) is a \(3 \times 3\) matrix and the matrix obtained by replacing the elements of \(A\) with their
corresponding cofactors is \(\left[\begin{array}{lll}1 & -2 & 1 \\ 4 & -5 & -2 \\ -2 & 4 & 1\end{array}\right]\)
then a possible value of the determinant of \(A\) is

1 4

2 3

3 2

4 1

TS EAMCET-2020-09.09.2020

Matrix and Determinant

79096 If \(\Delta_{1}=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|\) and \(\Delta_{2}=\left|\begin{array}{lll}b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1\end{array}\right|\) then \(\frac{\Delta_{1}}{\Delta_{2}}=\)

1 \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\)

2 abc

3 \(2(a b+b c+c a)\)

4 \((a+b+c)^{2}\)

Explanation:

(A) : \(\Delta_{1}=\left|\begin{array}{lll}1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 1 & \mathrm{~b}^{2} & \mathrm{~b}^{3} \\ 1 & \mathrm{c}^{2} & \mathrm{c}^{3}\end{array}\right|\)
Applying, \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\)
\(\Delta_{1}=\left|\begin{array}{ccc} 1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 0 & \mathrm{~b}^{2}-\mathrm{a}^{2} & \mathrm{~b}^{3}-\mathrm{a}^{3} \\ 0 & \mathrm{c}^{2}-\mathrm{a}^{2} & \mathrm{c}^{3}-\mathrm{a}^{3} \end{array}\right|\)
\(=(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{a})\left|\begin{array}{ccc} 1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 0 & \mathrm{~b}+\mathrm{a} & \mathrm{b}^{2}+\mathrm{a}^{2}+\mathrm{ab} \\ 0 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2}+\mathrm{ac} \end{array}\right|\)
\(=(b-a)(c-a)\left[(b+a)\left(c^{2}+a^{2}+a c\right)-(c+a)\left(b^{2}+a^{2}+a b\right)\right]\)
\(=(b-a)(c-a)\left[b c^{2}+a^{2} b+a b c+a c^{2}+a^{3}+a^{2} c-b^{2} c-\right.\) \(\left.a^{2} c-a b c-a b^{2}-a^{3}-a^{2} b\right]\)
\(\Delta_{1}=(b-a)(c-a)[(b-c)(a b+b c+c a)]\)
\(\Delta_{1}=-(b-a)(b-c)(c-a)(a b+b c+c a)\)
Now,
\(\Delta_{2}=\left|\begin{array}{lll}
\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{ab} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right|\)
Applying, \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}\)
\(\Delta_{2}=\left|\begin{array}{ccc}\mathrm{bc} & (\mathrm{b}+\mathrm{c}) & 1 \\ \mathrm{c}(\mathrm{a}-\mathrm{b}) & (\mathrm{a}-\mathrm{b}) & 0 \\ \mathrm{~b}(\mathrm{c}-\mathrm{a}) & (\mathrm{a}-\mathrm{c}) & 0\end{array}\right|\)
\(\Delta_{2}=(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})\left|\begin{array}{ccc}\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{c} & 1 & 0 \\ \mathrm{~b} & 1 & 0\end{array}\right|\)
\(\Delta_{2}=(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})(\mathrm{c}-\mathrm{b})=-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})\)
Now,
\(\frac{\Delta_{1}}{\Delta_{2}}=\frac{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})}\)
\(\frac{\Delta_{1}}{\Delta_{2}}=(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\)

TS EAMCET-2018-05.05.2018

Matrix and Determinant

79097 If the determinant of the matrix
\(A=\left[\begin{array}{ccc}0 & a & \mathbf{b} \\ -\mathbf{a} & \mathbf{0} & \boldsymbol{\beta} \\ -\mathbf{b} & \boldsymbol{\alpha} & \mathbf{0}\end{array}\right]\) is zero for all \(a, b\) then \(\alpha+\boldsymbol{\beta}=\)

1 0

2 1

3 -1

4 2

TS EAMCET-2019-03.05.2019

Matrix and Determinant

79094 If
\(\begin{array}{lll}
1! & 2! & 3! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 ! \end{array}=2016K \text {, then } K=\)

1 84

2 \(\frac{1}{24}\)

3 24

4 \(\frac{1}{84}\)

GUJCET-2019

Matrix and Determinant

79095 If \(A\) is a \(3 \times 3\) matrix and the matrix obtained by replacing the elements of \(A\) with their
corresponding cofactors is \(\left[\begin{array}{lll}1 & -2 & 1 \\ 4 & -5 & -2 \\ -2 & 4 & 1\end{array}\right]\)
then a possible value of the determinant of \(A\) is

1 4

2 3

3 2

4 1

TS EAMCET-2020-09.09.2020

Matrix and Determinant

79096 If \(\Delta_{1}=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|\) and \(\Delta_{2}=\left|\begin{array}{lll}b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1\end{array}\right|\) then \(\frac{\Delta_{1}}{\Delta_{2}}=\)

1 \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\)

2 abc

3 \(2(a b+b c+c a)\)

4 \((a+b+c)^{2}\)

Explanation:

(A) : \(\Delta_{1}=\left|\begin{array}{lll}1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 1 & \mathrm{~b}^{2} & \mathrm{~b}^{3} \\ 1 & \mathrm{c}^{2} & \mathrm{c}^{3}\end{array}\right|\)
Applying, \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\)
\(\Delta_{1}=\left|\begin{array}{ccc} 1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 0 & \mathrm{~b}^{2}-\mathrm{a}^{2} & \mathrm{~b}^{3}-\mathrm{a}^{3} \\ 0 & \mathrm{c}^{2}-\mathrm{a}^{2} & \mathrm{c}^{3}-\mathrm{a}^{3} \end{array}\right|\)
\(=(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{a})\left|\begin{array}{ccc} 1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 0 & \mathrm{~b}+\mathrm{a} & \mathrm{b}^{2}+\mathrm{a}^{2}+\mathrm{ab} \\ 0 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2}+\mathrm{ac} \end{array}\right|\)
\(=(b-a)(c-a)\left[(b+a)\left(c^{2}+a^{2}+a c\right)-(c+a)\left(b^{2}+a^{2}+a b\right)\right]\)
\(=(b-a)(c-a)\left[b c^{2}+a^{2} b+a b c+a c^{2}+a^{3}+a^{2} c-b^{2} c-\right.\) \(\left.a^{2} c-a b c-a b^{2}-a^{3}-a^{2} b\right]\)
\(\Delta_{1}=(b-a)(c-a)[(b-c)(a b+b c+c a)]\)
\(\Delta_{1}=-(b-a)(b-c)(c-a)(a b+b c+c a)\)
Now,
\(\Delta_{2}=\left|\begin{array}{lll}
\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{ab} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right|\)
Applying, \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}\)
\(\Delta_{2}=\left|\begin{array}{ccc}\mathrm{bc} & (\mathrm{b}+\mathrm{c}) & 1 \\ \mathrm{c}(\mathrm{a}-\mathrm{b}) & (\mathrm{a}-\mathrm{b}) & 0 \\ \mathrm{~b}(\mathrm{c}-\mathrm{a}) & (\mathrm{a}-\mathrm{c}) & 0\end{array}\right|\)
\(\Delta_{2}=(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})\left|\begin{array}{ccc}\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{c} & 1 & 0 \\ \mathrm{~b} & 1 & 0\end{array}\right|\)
\(\Delta_{2}=(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})(\mathrm{c}-\mathrm{b})=-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})\)
Now,
\(\frac{\Delta_{1}}{\Delta_{2}}=\frac{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})}\)
\(\frac{\Delta_{1}}{\Delta_{2}}=(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\)

TS EAMCET-2018-05.05.2018

Matrix and Determinant

79097 If the determinant of the matrix
\(A=\left[\begin{array}{ccc}0 & a & \mathbf{b} \\ -\mathbf{a} & \mathbf{0} & \boldsymbol{\beta} \\ -\mathbf{b} & \boldsymbol{\alpha} & \mathbf{0}\end{array}\right]\) is zero for all \(a, b\) then \(\alpha+\boldsymbol{\beta}=\)

1 0

2 1

3 -1

4 2

TS EAMCET-2019-03.05.2019

Matrix and Determinant

79094 If
\(\begin{array}{lll}
1! & 2! & 3! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 ! \end{array}=2016K \text {, then } K=\)

1 84

2 \(\frac{1}{24}\)

3 24

4 \(\frac{1}{84}\)

GUJCET-2019

Matrix and Determinant

79095 If \(A\) is a \(3 \times 3\) matrix and the matrix obtained by replacing the elements of \(A\) with their
corresponding cofactors is \(\left[\begin{array}{lll}1 & -2 & 1 \\ 4 & -5 & -2 \\ -2 & 4 & 1\end{array}\right]\)
then a possible value of the determinant of \(A\) is

1 4

2 3

3 2

4 1

TS EAMCET-2020-09.09.2020

Matrix and Determinant

79096 If \(\Delta_{1}=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|\) and \(\Delta_{2}=\left|\begin{array}{lll}b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1\end{array}\right|\) then \(\frac{\Delta_{1}}{\Delta_{2}}=\)

1 \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\)

2 abc

3 \(2(a b+b c+c a)\)

4 \((a+b+c)^{2}\)

Explanation:

(A) : \(\Delta_{1}=\left|\begin{array}{lll}1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 1 & \mathrm{~b}^{2} & \mathrm{~b}^{3} \\ 1 & \mathrm{c}^{2} & \mathrm{c}^{3}\end{array}\right|\)
Applying, \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\)
\(\Delta_{1}=\left|\begin{array}{ccc} 1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 0 & \mathrm{~b}^{2}-\mathrm{a}^{2} & \mathrm{~b}^{3}-\mathrm{a}^{3} \\ 0 & \mathrm{c}^{2}-\mathrm{a}^{2} & \mathrm{c}^{3}-\mathrm{a}^{3} \end{array}\right|\)
\(=(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{a})\left|\begin{array}{ccc} 1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 0 & \mathrm{~b}+\mathrm{a} & \mathrm{b}^{2}+\mathrm{a}^{2}+\mathrm{ab} \\ 0 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2}+\mathrm{ac} \end{array}\right|\)
\(=(b-a)(c-a)\left[(b+a)\left(c^{2}+a^{2}+a c\right)-(c+a)\left(b^{2}+a^{2}+a b\right)\right]\)
\(=(b-a)(c-a)\left[b c^{2}+a^{2} b+a b c+a c^{2}+a^{3}+a^{2} c-b^{2} c-\right.\) \(\left.a^{2} c-a b c-a b^{2}-a^{3}-a^{2} b\right]\)
\(\Delta_{1}=(b-a)(c-a)[(b-c)(a b+b c+c a)]\)
\(\Delta_{1}=-(b-a)(b-c)(c-a)(a b+b c+c a)\)
Now,
\(\Delta_{2}=\left|\begin{array}{lll}
\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{ab} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right|\)
Applying, \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}\)
\(\Delta_{2}=\left|\begin{array}{ccc}\mathrm{bc} & (\mathrm{b}+\mathrm{c}) & 1 \\ \mathrm{c}(\mathrm{a}-\mathrm{b}) & (\mathrm{a}-\mathrm{b}) & 0 \\ \mathrm{~b}(\mathrm{c}-\mathrm{a}) & (\mathrm{a}-\mathrm{c}) & 0\end{array}\right|\)
\(\Delta_{2}=(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})\left|\begin{array}{ccc}\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{c} & 1 & 0 \\ \mathrm{~b} & 1 & 0\end{array}\right|\)
\(\Delta_{2}=(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})(\mathrm{c}-\mathrm{b})=-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})\)
Now,
\(\frac{\Delta_{1}}{\Delta_{2}}=\frac{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})}\)
\(\frac{\Delta_{1}}{\Delta_{2}}=(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\)

TS EAMCET-2018-05.05.2018

Matrix and Determinant

79097 If the determinant of the matrix
\(A=\left[\begin{array}{ccc}0 & a & \mathbf{b} \\ -\mathbf{a} & \mathbf{0} & \boldsymbol{\beta} \\ -\mathbf{b} & \boldsymbol{\alpha} & \mathbf{0}\end{array}\right]\) is zero for all \(a, b\) then \(\alpha+\boldsymbol{\beta}=\)

1 0

2 1

3 -1

4 2

TS EAMCET-2019-03.05.2019

Matrix and Determinant

79094 If
\(\begin{array}{lll}
1! & 2! & 3! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 ! \end{array}=2016K \text {, then } K=\)

1 84

2 \(\frac{1}{24}\)

3 24

4 \(\frac{1}{84}\)

GUJCET-2019

Matrix and Determinant

79095 If \(A\) is a \(3 \times 3\) matrix and the matrix obtained by replacing the elements of \(A\) with their
corresponding cofactors is \(\left[\begin{array}{lll}1 & -2 & 1 \\ 4 & -5 & -2 \\ -2 & 4 & 1\end{array}\right]\)
then a possible value of the determinant of \(A\) is

1 4

2 3

3 2

4 1

TS EAMCET-2020-09.09.2020

Matrix and Determinant

79096 If \(\Delta_{1}=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|\) and \(\Delta_{2}=\left|\begin{array}{lll}b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1\end{array}\right|\) then \(\frac{\Delta_{1}}{\Delta_{2}}=\)

1 \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\)

2 abc

3 \(2(a b+b c+c a)\)

4 \((a+b+c)^{2}\)

Explanation:

(A) : \(\Delta_{1}=\left|\begin{array}{lll}1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 1 & \mathrm{~b}^{2} & \mathrm{~b}^{3} \\ 1 & \mathrm{c}^{2} & \mathrm{c}^{3}\end{array}\right|\)
Applying, \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\)
\(\Delta_{1}=\left|\begin{array}{ccc} 1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 0 & \mathrm{~b}^{2}-\mathrm{a}^{2} & \mathrm{~b}^{3}-\mathrm{a}^{3} \\ 0 & \mathrm{c}^{2}-\mathrm{a}^{2} & \mathrm{c}^{3}-\mathrm{a}^{3} \end{array}\right|\)
\(=(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{a})\left|\begin{array}{ccc} 1 & \mathrm{a}^{2} & \mathrm{a}^{3} \\ 0 & \mathrm{~b}+\mathrm{a} & \mathrm{b}^{2}+\mathrm{a}^{2}+\mathrm{ab} \\ 0 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2}+\mathrm{ac} \end{array}\right|\)
\(=(b-a)(c-a)\left[(b+a)\left(c^{2}+a^{2}+a c\right)-(c+a)\left(b^{2}+a^{2}+a b\right)\right]\)
\(=(b-a)(c-a)\left[b c^{2}+a^{2} b+a b c+a c^{2}+a^{3}+a^{2} c-b^{2} c-\right.\) \(\left.a^{2} c-a b c-a b^{2}-a^{3}-a^{2} b\right]\)
\(\Delta_{1}=(b-a)(c-a)[(b-c)(a b+b c+c a)]\)
\(\Delta_{1}=-(b-a)(b-c)(c-a)(a b+b c+c a)\)
Now,
\(\Delta_{2}=\left|\begin{array}{lll}
\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{ab} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right|\)
Applying, \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}\)
\(\Delta_{2}=\left|\begin{array}{ccc}\mathrm{bc} & (\mathrm{b}+\mathrm{c}) & 1 \\ \mathrm{c}(\mathrm{a}-\mathrm{b}) & (\mathrm{a}-\mathrm{b}) & 0 \\ \mathrm{~b}(\mathrm{c}-\mathrm{a}) & (\mathrm{a}-\mathrm{c}) & 0\end{array}\right|\)
\(\Delta_{2}=(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})\left|\begin{array}{ccc}\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{c} & 1 & 0 \\ \mathrm{~b} & 1 & 0\end{array}\right|\)
\(\Delta_{2}=(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})(\mathrm{c}-\mathrm{b})=-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})\)
Now,
\(\frac{\Delta_{1}}{\Delta_{2}}=\frac{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})}\)
\(\frac{\Delta_{1}}{\Delta_{2}}=(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\)

TS EAMCET-2018-05.05.2018

Matrix and Determinant

79097 If the determinant of the matrix
\(A=\left[\begin{array}{ccc}0 & a & \mathbf{b} \\ -\mathbf{a} & \mathbf{0} & \boldsymbol{\beta} \\ -\mathbf{b} & \boldsymbol{\alpha} & \mathbf{0}\end{array}\right]\) is zero for all \(a, b\) then \(\alpha+\boldsymbol{\beta}=\)

1 0

2 1

3 -1

4 2

TS EAMCET-2019-03.05.2019