If $\Delta$ = $$\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
{2x - 3} & {3x - 4} & {4x - 5} \cr
{3x - 5} & {5x - 8} & {10x - 17} \cr
} } \right|$$ =
Ax3 + Bx2 + Cx + D, then B + C is equal to :
Solution
$\Delta$ = $$\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
{2x - 3} & {3x - 4} & {4x - 5} \cr
{3x - 5} & {5x - 8} & {10x - 17} \cr
} } \right|$$
<br><br>R<sub>2</sub> $\to$ R<sub>2</sub> – R<sub>1</sub>
<br>R<sub>3</sub> $\to$ R<sub>3</sub> – R<sub>2</sub>
<br><br>= $$\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
{x - 1} & {x - 1} & {x - 1} \cr
{x - 2} & {2\left( {x - 2} \right)} & {6\left( {x - 2} \right)} \cr
} } \right|$$
<br><br>= $$\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
1 & 1 & 1 \cr
1 & 2 & 6 \cr
} } \right|$$
<br><br>C<sub>1</sub> $\to$ C<sub>1</sub> - C<sub>2</sub>
<br>C<sub>2</sub> $\to$ C<sub>2</sub> - C<sub>3</sub>
<br><br>= $$\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{
{ - x + 1} & { - x + 1} & {3x - 4} \cr
0 & 0 & 1 \cr
{ - 1} & { - 4} & 6 \cr
} } \right|$$
<br><br>= -(x - 1)(x - 2)[-4(1 - x) + 1(1 - x)]
<br><br>= -(x<sup>2</sup> - 3x + 2)[3x - 3]
<br><br>= -3x<sup>3</sup> + 9x<sup>2</sup> - 6x + 3x<sup>2</sup> - 9x + 6
<br><br>= -3x<sup>3</sup> + 12x<sup>2</sup> - 15x + 6 = Ax<sup>3</sup> + Bx<sup>2</sup> + Cx + D
<br><br>$\therefore$ A = -3, B = 12, C = -15
<br><br>$\therefore$ B + C = 12 – 15 = – 3
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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