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3 and 4 .Determinants and Matrices
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Construct a $2 \times 2$ matrix, $A=\left[a_{ij}\right]$, whose elements are given by : $a_{i j}=\frac{(i+2 j)^{2}}{2}$
Option A
Option B
Option C
Option D
Solution
$(iii)$ Since it is a $2 \times 2$ matrix it has $2$ rows and $2$ column. Let matrix be $A$
Where $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$
Now it is given that
$a_{i j}=\frac{(i+2 j)^2}{2}$
| $a_{i j}$ | $i=, j=$ | $a_{ ij }=\frac{( i +2 j )^2}{2}$ |
| $a_{11}$ | $i=1, j=1$ | $a_{11}=\frac{(1+2(1))^2}{2}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$ |
| $a_{12}$ | $i=1, j=2$ | $a_{12}=\frac{(1+2(2))^2}{2}=\frac{(1+4)^2}{2}=\frac{(5)^2}{2}=\frac{25}{2}$ |
| $a_{21}$ | $i=2, j=1$ | $a_{21}=\frac{(2+2(1))^2}{2}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$ |
| $a_{22}$ | $i=2, j=2$ | $a_{22}=\frac{(2+2(2))^2}{2}=\frac{(2+4)^2}{2}=\frac{(6)^2}{2}=\frac{36}{2}=18$ |
Hence, the required matrix $A$ ia
$A\left[\begin{array}{ll}a_{11} & a_{12} \\a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{cc}\frac{9}{2} & \frac{25}{2} \\8 & 18\end{array}\right]$
Standard 12
Mathematics
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