If $A = \left[ {\begin{array}{*{20}{c}} {1\,\,\,\; - 2}\\ {4\,\,\,\;\;5} \end{array}} \right]{\rm{\;and\;}}f\left( t \right) = {t^2} - 3t + 7,{\rm{\;then\;}}f\left( A \right) + \left[ {\begin{array}{*{20}{c}} {3\;\;\;\;\;\;\;\;6}\\ { - 12\,\,\,\,\; - 9} \end{array}} \right]{\rm{\;}}$ is equal to

$\left[ {\begin{array}{*{20}{c}} 0&0\\ 0&0 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&1\\ 0&0 \end{array}} \right]$

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Application +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} {1\,\,\,\; - 2}\\ {4\,\,\,\;\;5} \end{array}} \right]\mathrm{\;and\;}f\left( t \right) = {t^2} - 3t + 7,\mathrm{\;then\;}f\left( A \right) + \left[ {\begin{array}{cccccccccccccccccccc} {3\;\;\;\;\;\;\;\;6}\\ { - 12\,\,\,\,\; - 9} \end{array}} \right]\mathrm{\;}

is equal to

View answer and explanation

Correct option: B

\left[ {\begin{array}{cccccccccccccccccccc} 0&0\\ 0&0 \end{array}} \right]

The correct answer is B. By calculating f(A) and adding the given matrix, the result is a zero matrix.