Let S₁ and S₂ be respectively sets of all a ∈ R -{0} for which system of linear equations a x+

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3 and 4 .Determinants and Matrices

hard

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R -\{0\}$ for which the system of linear equations

$a x+2 a y-3 a z=1$

$(2 a+1) x+(2 a+3) y+(a+1) z=2$

$(3 a+5) x+(a+5) y+(a+2) z=3$

has unique solution and infinitely many solutions. Then

$n \left( S _1\right)=2$ and $S _2$ is an infinite set

$S_1$ is an infinite set an $n\left(S_2\right)=2$

$S _1=\Phi$ and $S _2= R -\{0\}$

$S _1= R -\{0\}$ and $S _2=\Phi$

(JEE MAIN-2023)

Solution

$\begin{array}{l}\Delta=\left|\begin{array}{lll}a & 2 a & -3 a \\ 2 a+1 & 2 a+3 & a+1 \\ 3 a+5 & a+5 & a+2\end{array}\right| \\ =a\left(15 a^2+31 a+36\right)=0 \Rightarrow a=0 \\ \Delta \neq 0 \text { for all } a \in R-\{0\} \\ \text { Hence } S_1=R-\{0\} \quad S_2=\Phi\end{array}$

Standard 12

Mathematics

Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N=N M$. Further, if $M \neq N^2$ and $M^2=N^4$, then

$(A)$ determinant of $\left( M ^2+ MN ^2\right)$ is $0$

$(B)$ there is a $3 \times 3$ non-zero matrix $U$ such that $\left( M ^2+ MN ^2\right) U$ is the zero matrix

$(C)$ determinant of $\left( M ^2+ MN ^2\right) \geq 1$

$(D)$ for a $3 \times 3$ matrix $U$, if $\left( M ^2+ MN ^2\right) U$ equals the zero matrix then $U$ is the zero matrix

normal

(IIT-2014)

If $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha – b}\\b&c&{b\alpha – c}\\2&1&0\end{array}\,} \right| = 0$ and $\alpha \ne \frac{1}{2},$ then

medium

Find values of $x$, if $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$

medium

If $A$, $B$ and $C$ are square matrices of order $3$ such that $A = \left[ {\begin{array}{*{20}{c}} x&0&1 \\ 0&y&0 \\ 0&0&z \end{array}} \right]$ and $\left| B \right| = 36$, $\left| C \right| = 4$, $\left( {x,y,z \in N} \right)$ and $\left| {ABC} \right| = 1152$ then the minimum value of $x + y + z$ is

normal

$\left| {\,\begin{array}{*{20}{c}}{11}&{12}&{13}\\{12}&{13}&{14}\\{13}&{14}&{15}\end{array}\,} \right| = $

easy

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R -\{0\}$ for which the system of linear equations $a x+2 a y-3 a z=1$ $(2 a+1) x+(2 a+3) y+(a+1) z=2$ $(3 a+5) x+(a+5) y+(a+2) z=3$ has unique solution and infinitely many solutions. Then

Solution

Similar Questions

If $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha – b}\\b&c&{b\alpha – c}\\2&1&0\end{array}\,} \right| = 0$ and $\alpha \ne \frac{1}{2},$ then

Find values of $x$, if $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$

$\left| {\,\begin{array}{*{20}{c}}{11}&{12}&{13}\\{12}&{13}&{14}\\{13}&{14}&{15}\end{array}\,} \right| = $

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Let $S_1$ and $S_2$ be respectively the sets of all $a \in R -\{0\}$ for which the system of linear equations

$a x+2 a y-3 a z=1$

$(2 a+1) x+(2 a+3) y+(a+1) z=2$

$(3 a+5) x+(a+5) y+(a+2) z=3$

has unique solution and infinitely many solutions. Then