Roots of equation left| {,begin{array}{*{20}{c}}{1 + x}&1&11&{1 + x}&11&1&{1

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3 and 4 .Determinants and Matrices

easy

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + x}&1\\1&1&{1 + x}\end{array}\,} \right| = 0$ are

$0, -3$

$0, 0, -3$

$0, 0, 0, -3$

None of these

Solution

(b) $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + x}&1\\1&1&{1 + x}\end{array}\,} \right|\, = \,0$

$ \Rightarrow $$\left| {\,\begin{array}{*{20}{c}}
{3 + x}&0&1\\
{3 + x}&x&1\\
{3 + x}&{ – x}&{1 + x}
\end{array}\,} \right|\, = \,0$, $\left( \begin{array}{l}{C_1} \to {C_1} + {C_2} + {C_3}\\{C_2} \to {C_2} – {C_3}\end{array} \right)$

$ \Rightarrow $ $(x + 3)\,\left| {\,\begin{array}{*{20}{c}}1&0&1\\1&x&1\\1&{ – x}&{1 + x}\end{array}\,} \right|\, = 0$

$ \Rightarrow $ $(x + 3)\,\left| {\,\begin{array}{*{20}{c}}1&0&1\\0&x&0\\0&{ – x}&x\end{array}\,} \right|\, = 0$, $\left( \begin{array}{l}{R_2} \to {R_2} – {R_1}\\{R_3} \to {R_3} – {R_1}\end{array} \right)$

$ \Rightarrow $ $(x + 3){x^2} = 0 \Rightarrow x = 0,\,0,\, – 3$.

Trick : Obviously the equation is of degree three, therefore it must have three solutions.

So check for option $ (b).$

Standard 12

Mathematics

The value of $x,$ if $\left| {\,\begin{array}{*{20}{c}}{ – x}&1&0\\1&{ – x}&1\\0&1&{ – x}\end{array}\,} \right| = 0 $ is equal to

easy

If $n$ be the number of values of $x$ for which
matrix $\Delta (x) =\left[ {\begin{array}{*{20}{c}}
{ – x}&x&2\\
2&x&{ – x}\\
x&{ – 2}&{ – x}
\end{array}} \right]$ will be singular, then $det(\Delta\,(n))$ is

$($ where $det(B)$ denotes determinant of Matrix $B) -$

normal

If the lines $x + 2ay + a = 0, x + 3by + b = 0$ and $x + 4cy + c = 0$ are concurrent, then $a, b$ and $c$ are in :-

normal

In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$ has the value

medium

For the system of linear equations

$2 x-y+3 z=5$

$3 x+2 y-z=7$

$4 x+5 y+\alpha z=\beta$

Which of the following is NOT correct ?

hard

(JEE MAIN-2023)

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + x}&1\\1&1&{1 + x}\end{array}\,} \right| = 0$ are

Solution

Similar Questions

The value of $x,$ if $\left| {\,\begin{array}{*{20}{c}}{ – x}&1&0\\1&{ – x}&1\\0&1&{ – x}\end{array}\,} \right| = 0 $ is equal to

If $n$ be the number of values of $x$ for which matrix $\Delta (x) =\left[ {\begin{array}{*{20}{c}} { – x}&x&2\\ 2&x&{ – x}\\ x&{ – 2}&{ – x} \end{array}} \right]$ will be singular, then $det(\Delta\,(n))$ is $($ where $det(B)$ denotes determinant of Matrix $B) -$

If the lines $x + 2ay + a = 0, x + 3by + b = 0$ and $x + 4cy + c = 0$ are concurrent, then $a, b$ and $c$ are in :-

In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$ has the value

For the system of linear equations $2 x-y+3 z=5$ $3 x+2 y-z=7$ $4 x+5 y+\alpha z=\beta$ Which of the following is NOT correct ?

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If $n$ be the number of values of $x$ for which
matrix $\Delta (x) =\left[ {\begin{array}{*{20}{c}}
{ – x}&x&2\\
2&x&{ – x}\\
x&{ – 2}&{ – x}
\end{array}} \right]$ will be singular, then $det(\Delta\,(n))$ is

$($ where $det(B)$ denotes determinant of Matrix $B) -$

For the system of linear equations

$2 x-y+3 z=5$

$3 x+2 y-z=7$

$4 x+5 y+\alpha z=\beta$

Which of the following is NOT correct ?