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Maths-

General

Easy

Question

If $D equals$ $open vertical bar table row cell a squared plus 1 end cell cell a b end cell cell a c end cell row cell b a end cell cell b squared plus 1 end cell cell b c end cell row cell c a end cell cell c b end cell cell c squared plus 1 end cell end table close vertical bar$ then $D equals$

$1 + a^{2} + b^{2} + c^{2}$
$a^{2} + b^{2} + c^{2}$
$(a + b + c)^{2}$
none

The correct answer is: $1 plus a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent$

In the question we were asked find the determinant of a given matrix
$open vertical bar table row cell a squared plus 1 end cell cell a b end cell cell a c end cell row cell b a end cell cell b squared plus 1 end cell cell b c end cell row cell c a end cell cell c b end cell cell c squared plus 1 end cell end table close vertical bar$
Let us solve the matrix
Given matrix can be written as
$a squared plus 1 left parenthesis left parenthesis b squared plus 1 right parenthesis left parenthesis c squared plus 1 right parenthesis minus left parenthesis b c right parenthesis left parenthesis c b right parenthesis right parenthesis minus a b left parenthesis b a left parenthesis c squared plus 1 right parenthesis minus b c left parenthesis c a right parenthesis right parenthesis plus a c left parenthesis b a left parenthesis c squared plus 1 right parenthesis minus b c left parenthesis c a right parenthesis right parenthesis$

After solving the expression we get $1 plus a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent$

Hence option A is the correct option

Related Questions to study

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

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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