Question
Construct a 2$\times$2 matrix A = [aij] whose element aij is given by: $\frac{(i+j)^{2}}{2}$
✓
Answer
Let A = [aij]2$\times$2
So, the elements in a $2\times2$ matrix are
a11, a12, a21, a22,
A = $\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}$ ...(i)
a11 =$\frac{\left(1+1\right)^2}2$ = $\frac42$ = 2
a12 = $\frac{(1+2)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2}$
a21 = $\frac{(2+1)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2}$
a22 = $\frac{(2+2)^{2}}{2}=\frac{4^{2}}{2}=\frac{16}{2}$ = 8
So, (i) becomes
A = $\left(\begin{array}{cc} {2} & {\frac 92} \\ {\frac92} & {8} \end{array}\right)$
So, the elements in a $2\times2$ matrix are
a11, a12, a21, a22,
A = $\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}$ ...(i)
a11 =$\frac{\left(1+1\right)^2}2$ = $\frac42$ = 2
a12 = $\frac{(1+2)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2}$
a21 = $\frac{(2+1)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2}$
a22 = $\frac{(2+2)^{2}}{2}=\frac{4^{2}}{2}=\frac{16}{2}$ = 8
So, (i) becomes
A = $\left(\begin{array}{cc} {2} & {\frac 92} \\ {\frac92} & {8} \end{array}\right)$
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