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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
The matrix $A^2 + 4A - 5I$, where $I$ is identity matrix and $A = \left[ {\begin{array}{*{20}{c}}
1&2\\
4&{ - 3}
\end{array}} \right]$ , equals
  • ✓
    $4\left[ {\begin{array}{*{20}{c}}
    2&1\\
    2&0
    \end{array}} \right]$
  • B
    $4\left[ {\begin{array}{*{20}{c}}
    0&{ - 1}\\
    2&2
    \end{array}} \right]$
  • C
    $32\left[ {\begin{array}{*{20}{c}}
    2&1\\
    2&0
    \end{array}} \right]$
  • D
    $32\left[ {\begin{array}{*{20}{c}}
    1&1\\
    1&0
    \end{array}} \right]$
✓

Answer

Correct option: A.
$4\left[ {\begin{array}{*{20}{c}}
2&1\\
2&0
\end{array}} \right]$
a
${A^2} + 4A - 51 = A \times A + 4A - 5I$

$ = \left[ {\begin{array}{*{20}{c}}
1&2\\
4&{ - 3}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
1&2\\
4&{ - 3}
\end{array}} \right] + 4\left[ {\begin{array}{*{20}{c}}
1&2\\
4&{ - 3}
\end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
9&{ - 4}\\
{ - 8}&{17}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
4&8\\
{16}&{ - 12}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
5&0\\
0&5
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
{9 + 4 - 5}&{ - 4 + 8 - 0}\\
{ - 8 + 16 - 0}&{17 - 12 - 5}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
8&4\\
8&0
\end{array}} \right]$

$ = 4\left[ {\begin{array}{*{20}{c}}
2&1\\
2&0
\end{array}} \right]$

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