Question

If A = $\left[\begin{array}{ccc}0&2\\3&-4\\\end{array}\right]$ , kA = $\left[\begin{array}{ccc}0&3a\\2b&24\\\end{array}\right]$ then find the values of k, a, and b

Answer 1

Given:

$\sf A = \left[\begin{array}{ccc}0&2\\3&-4\\ \end{array}\right] \: and \: kA = \left[\begin{array}{ccc} 0&3a\\2b&24\\\end{array}\right]$

$\rule{81}{2}$

Find:

⁍ values of k,a and b

$\rule{81}{2}$

Solution:

$\sf If, A = \left[\begin{array}{ccc}0&2\\3&-4\\ \end{array}\right]$

$\sf Then \: kA = k\left[\begin{array}{ccc}0&2\\3&-4\\ \end{array}\right] = \left[\begin{array}{ccc}0&2k\\3k&-4k\\ \end{array}\right]$

$\sf But \: it \: also \: given, kA = \left[\begin{array}{ccc} 0&3a\\2b&24\\\end{array}\right]$

$\sf So, \left[\begin{array}{ccc}0&2k\\3k&-4k\\ \end{array}\right] = \left[\begin{array}{ccc} 0&3a\\2b&24\\\end{array}\right]$

$\sf \implies 2k = 3a.....(i)$

$\sf \implies 3k = 2b.....(ii)$

$\sf \implies - 4k = 24.....(iii)$

$\sf From \: eq(iii) \: k = \frac{ - 24}{4} = - 6$

$\sf \implies k = - 6$

Now, putting value of k in eq(i) we, get

$\sf 2k = 3a$

$\sf \implies 2( - 6) = 3a$

$\sf \implies - 12 = 3a$

$\sf \implies a = \frac{ - 12}{3} = - 4$

$\sf \implies a = - 4$

Now, putting value of k in eq(ii) we, get

$\sf 3k = 2b$

$\sf \implies 3( - 6) = 2b$

$\sf \implies - 18 = 2b$

$\sf \implies b = \frac{ - 18}{2} = - 9$

$\sf \implies b = - 9$

$\rule{122}{2}$

Hence, the value of

• k = -6
• a = -4
• b = -9