91Ó°ÊÓ

a)

Construct matrix A from four blocks as \({C_{11}},{C_{12}},{C_{21}},\) and \({C_{22}}\). Consider an example with \({C_{11}}\) as a \(30 \times 30\) matrix and \({C_{22}}\) as a \(20 \times 20\) matrix.

Use the MATLAB code to compute the matrix \(A + B\), as shown below:

\[\begin{array}{l} > > {\mathop{\rm C}\nolimits} 11 = {\mathop{\rm A}\nolimits} \left( {1:30,\,\,1:30} \right) + {\mathop{\rm B}\nolimits} \left( {1:30,\,\,1:30} \right)\\ > > {\mathop{\rm C}\nolimits} 11 = {\mathop{\rm A}\nolimits} \left( {1:30,\,\,1:30} \right) + {\mathop{\rm B}\nolimits} \left( {1:30,\,\,1:30} \right)\\ > > {\mathop{\rm C}\nolimits} 21 = {\mathop{\rm A}\nolimits} \left( {31:50,\,\,1:30} \right) + {\mathop{\rm B}\nolimits} \left( {31:50,\,\,1:30} \right)\\ > > {\mathop{\rm C}\nolimits} 22 = {\mathop{\rm A}\nolimits} \left( {31:50,\,\,31:50} \right) + {\mathop{\rm B}\nolimits} \left( {31:50,\,\,31:50} \right)\\ > > {\mathop{\rm C}\nolimits} = \left[ {{\mathop{\rm C}\nolimits} 11\,\,{\mathop{\rm C}\nolimits} 12;\,\,{\mathop{\rm C}\nolimits} 21\,\,{\mathop{\rm C}\nolimits} 22} \right]\end{array}\]

b)

The required algebra comes from block matrix multiplication, as shown below:

\[\begin{array}{c}AB = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{{B_{21}}}&{{B_{22}}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{{A_{11}}{B_{11}} + {A_{12}}{B_{21}}}&{{A_{11}}{B_{12}} + {A_{12}}{B_{22}}}\\{{A_{21}}{B_{11}} + {A_{22}}{B_{21}}}&{{A_{21}}{B_{12}} + {A_{22}}{B_{22}}}\end{array}} \right]\end{array}\]

Both matrices \(A\) and \(B\) can be partitioned, for example, in \(30 \times 30\)\(\left( {1,1} \right)\) blocks and \[20 \times 20\left( {2,2} \right)\] blocks. The four necessary submatrix calculations follow the same syntax as part (a).

c)

The required algebra comes from the block matrix equation shown below:

\(\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&0\\{{A_{21}}}&{{A_{22}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\)

Here, \({x_1}\) and \({{\mathop{\rm b}\nolimits} _1}\) are in \({\mathbb{R}^{30}}\)and \({x_2}\), \({{\mathop{\rm b}\nolimits} _2}\) are in \({\mathbb{R}^{20}}\). Hence, \({A_{11}}{{\mathop{\rm x}\nolimits} _1} = {{\mathop{\rm b}\nolimits} _1}\) and \({x_1}\) can be found by solving for \({{\mathop{\rm b}\nolimits} _1}\).

After determining \({x_1}\) , rewrite the equation \({A_{21}}{{\mathop{\rm x}\nolimits} _1} + {A_{22}}{{\mathop{\rm x}\nolimits} _2} = {{\mathop{\rm b}\nolimits} _2}\) as \({A_{22}}{{\mathop{\rm x}\nolimits} _2} = {\mathop{\rm c}\nolimits} \), where \({\mathop{\rm c}\nolimits} = {{\mathop{\rm b}\nolimits} _2} - {A_{21}}{x_1}\) , and solve the equation \({A_{22}}{{\mathop{\rm x}\nolimits} _2} = {\mathop{\rm c}\nolimits} \) for \({x_2}\).