The ordered pair (a, b), for which the system of linear equations
3x $-$ 2y + z = b
5x $-$ 8y + 9z = 3
2x + y + az = $-$1
has no solution, is :
Solution
<p>$$\left| {\matrix{
3 & { - 2} & 1 \cr
5 & { - 8} & 9 \cr
2 & 1 & a \cr
} } \right| = 0 \Rightarrow - 14a - 42 = 0 \Rightarrow a = - 3$$</p>
<p>Now 3 (equation (1)) $-$ (equation (2)) $-$ 2 (equation (3)) is</p>
<p>$3(3x - 2y + z - b) - (5x - 8y + 9z - 3) - 2(2x + y + az + 1) = 0$</p>
<p>$\Rightarrow - 3b + 3 - 2 = 0 \Rightarrow b = {1 \over 3}$</p>
<p>So for no solution $a = - 3$ and $b \ne {1 \over 3}$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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