Suppose vectors x_{1}, x_{2} and x_{3} are solutions of system of linear equations, Ax = b when vect

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3 and 4 .Determinants and Matrices

hard

Suppose the vectors $x_{1}, x_{2}$ and $x_{3}$ are the solutions of the system of linear equations, $Ax = b$ when the vector $b$ on the right side is equal to $b _{1}, b _{2}$ and $b _{3}$ respectively. If $x =\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right], x _{2}=\left[\begin{array}{l}0 \\ 2 \\ 1\end{array}\right], x _{3}=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], b _{1}=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$ $b _{2}=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]$ and $b _{3}=\left[\begin{array}{l}0 \\ 0 \\ 2\end{array}\right],$ then the determinant of $A$ is equal to

$\frac{1}{2}$

$4$

$\frac{3}{2}$

$2 $

(JEE MAIN-2020)

Solution

$A x_{1}=b_{1}$

$A x_{2}=b_{2}$

$A x_{3}=b_{3}$

$\Rightarrow\left|\begin{array}{lll}1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 1\end{array}\right|=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right|$

$\Rightarrow|A|=\frac{4}{2}=2$

Standard 12

Mathematics

If $A = \left[ {\begin{array}{*{20}{c}}\lambda &1\\{ – 1}&{ – \lambda }\end{array}} \right]$, then for what value of $\lambda ,\,{A^2} = O$

easy

$AB = 0$, if and only if

normal

Let $A = \left[ {\begin{array}{{20}{c}}1&2\\3&4\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}a&0\\0&b\end{array}} \right]\;,a,b \in N$ then

easy

(AIEEE-2006)

If $A = \left[ {\begin{array}{*{20}{c}}4&2\\{ – 1}&1\end{array}} \right]$and $I$ is the identity matrix of order $2$, then $(A – 2I)(A – 3I) = $

easy

If $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],$ find $k$ so that $A^{2}=k A-2 I$

medium

Solution

Similar Questions

If $A = \left[ {\begin{array}{*{20}{c}}\lambda &1\\{ – 1}&{ – \lambda }\end{array}} \right]$, then for what value of $\lambda ,\,{A^2} = O$

$AB = 0$, if and only if

Let $A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}a&0\\0&b\end{array}} \right]\;,a,b \in N$ then

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If $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],$ find $k$ so that $A^{2}=k A-2 I$

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Let $A = \left[ {\begin{array}{{20}{c}}1&2\\3&4\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}a&0\\0&b\end{array}} \right]\;,a,b \in N$ then