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Maths-

General

Easy

Question

The system of linear equations x+y-z=6, x+2y-3z=14 and $2 x plus 5 y minus lambda z equals 9 left parenthesis lambda element of R right parenthesis$ has a unique solution if

$λ = 8$
$λ \neq 8$
$λ = 7$
$λ \neq 7$

The correct answer is: $lambda not equal to 8$

The system of linear equations x+y-z=6, x+2y-3z=14 and $2 x plus 5 y minus lambda z equals 9 left parenthesis lambda element of R right parenthesis$ has a unique solution if the determinant of the coefficients is not equal to zero
Hence given equations are
x+y-z=6
x+2y-3z=14
$2 x plus 5 y minus lambda z equals 9 left parenthesis lambda element of R right parenthesis$
We can draw theirr matrix as
let lamda=x
$open vertical bar table row 1 1 1 row 1 2 3 row 2 2 x end table close vertical bar not equal to 0$
$left parenthesis 2 x minus 15 right parenthesis minus left parenthesis x minus 6 right parenthesis plus 1 not equal to 0$
$x minus 8 not equal to 0$
$x not equal to 8$

Hence option B is the correct option

Related Questions to study

If $D equals$ $open vertical bar table row cell a to the power of 2 end exponent plus 1 end cell cell a b end cell cell a c end cell row cell b a end cell cell b to the power of 2 end exponent plus 1 end cell cell b c end cell row cell c a end cell cell c b end cell cell c to the power of 2 end exponent plus 1 end cell end table close vertical bar$ then $D equals$

If $D equals$ $open vertical bar table row cell a to the power of 2 end exponent plus 1 end cell cell a b end cell cell a c end cell row cell b a end cell cell b to the power of 2 end exponent plus 1 end cell cell b c end cell row cell c a end cell cell c b end cell cell c to the power of 2 end exponent plus 1 end cell end table close vertical bar$ then $D equals$

Let $A equals open curly brackets table attributes columnalign left end attributes row cell c o s to the power of negative 1 end exponent x end cell cell c o s to the power of negative 1 end exponent y end cell cell c o s to the power of negative 1 end exponent end cell z row cell c o s to the power of negative 1 end exponent y end cell cell c o s to the power of negative 1 end exponent z end cell cell c o s to the power of negative 1 end exponent x end cell blank row cell c o s to the power of negative 1 end exponent z end cell cell c o s to the power of negative 1 end exponent x end cell cell c o s to the power of negative 1 end exponent y end cell blank end table close curly brackets$ such that $vertical line A vertical line equals 0$ , then maximum value of $x plus y plus z$ is

Let $A equals open curly brackets table attributes columnalign left end attributes row cell c o s to the power of negative 1 end exponent x end cell cell c o s to the power of negative 1 end exponent y end cell cell c o s to the power of negative 1 end exponent end cell z row cell c o s to the power of negative 1 end exponent y end cell cell c o s to the power of negative 1 end exponent z end cell cell c o s to the power of negative 1 end exponent x end cell blank row cell c o s to the power of negative 1 end exponent z end cell cell c o s to the power of negative 1 end exponent x end cell cell c o s to the power of negative 1 end exponent y end cell blank end table close curly brackets$ such that $vertical line A vertical line equals 0$ , then maximum value of $x plus y plus z$ is

Statement‐I : The inverse of the matrix $A equals left square bracket a subscript i j end subscript right square bracket subscript n cross times n end subscript$ where $a subscript i j end subscript equals 0 comma$ $i greater or equal than j$ is $B equals left square bracket a subscript i j to the power of negative 1 end exponent end subscript right square bracket subscript n cross times n end subscript$
Statement‐II : The inverse of singular matrix does not exist.

Hence option D is the suitable option

Statement‐I : The inverse of the matrix $A equals left square bracket a subscript i j end subscript right square bracket subscript n cross times n end subscript$ where $a subscript i j end subscript equals 0 comma$ $i greater or equal than j$ is $B equals left square bracket a subscript i j to the power of negative 1 end exponent end subscript right square bracket subscript n cross times n end subscript$
Statement‐II : The inverse of singular matrix does not exist.

Hence option D is the suitable option

parallel

Let $P equals open square brackets a subscript i j end subscript close square brackets$ be a $3 cross times 3$ matrix and let $Q equals left square bracket b subscript i j end subscript right square bracket$ , where $b subscript i j end subscript equals 2 to the power of i plus j end exponent a subscript i j end subscript$ for $1 less than i comma$ $j less than 3$ If the determinant of $P$ is 2, then the determinant of the matrix $Q$ is ‐

Let $P equals open square brackets a subscript i j end subscript close square brackets$ be a $3 cross times 3$ matrix and let $Q equals left square bracket b subscript i j end subscript right square bracket$ , where $b subscript i j end subscript equals 2 to the power of i plus j end exponent a subscript i j end subscript$ for $1 less than i comma$ $j less than 3$ If the determinant of $P$ is 2, then the determinant of the matrix $Q$ is ‐

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