Question

The inverse of \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\C&I&{\bf{0}}\\A&B&I\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\Z&I&{\bf{0}}\\X&Y&I\end{array}} \right]\). Find X, Y, and Z.

Short answer

The values are \(X = BC - A\), \[Y = - B\], and \[Z = - C\].

Step-by-step solution

State the row-column rule

If the sum of the products of matching entries from row\(i\)of matrix A and column\(j\)of matrix B equals the item in row\(i\)and column\(j\)of AB, then it can be said that product AB is defined.

The product is\({\left( {AB} \right)_{ij}} = {a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + ... + {a_{in}}{b_{nj}}\).

Also, the product of matrix A with its inverse always gives theidentity matrix I.

Obtain the product

Compute the product of\(\left[ {\begin{array}{*{20}{c}}I&0&0\\C&I&0\\A&B&I\end{array}} \right]\)with\(\left[ {\begin{array}{*{20}{c}}I&0&0\\Z&I&0\\X&Y&I\end{array}} \right]\) to get theidentity matrix\(\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\).

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}I&0&0\\C&I&0\\A&B&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&0&0\\Z&I&0\\X&Y&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0&0\\0&I&0\\0&0&I\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{I\left( I \right) + 0\left( Z \right) + 0\left( X \right)}&{I\left( 0 \right) + 0\left( I \right) + 0\left( Y \right)}&{I\left( 0 \right) + 0\left( 0 \right) + 0\left( I \right)}\\{C\left( I \right) + I\left( Z \right) + 0\left( X \right)}&{C\left( 0 \right) + I\left( I \right) + 0\left( Y \right)}&{C\left( 0 \right) + I\left( 0 \right) + 0\left( I \right)}\\{A\left( I \right) + B\left( Z \right) + I\left( X \right)}&{A\left( 0 \right) + B\left( I \right) + I\left( Y \right)}&{A\left( 0 \right) + B\left( 0 \right) + I\left( I \right)}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0&0\\0&I&0\\0&0&I\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}I&0&0\\{C + Z}&I&0\\{A + BZ + X}&{B + Y}&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0&0\\0&I&0\\0&0&I\end{array}} \right]\end{array}\)

Thus, \(\left[ {\begin{array}{*{20}{c}}I&0&0\\{C + Z}&I&0\\{A + BZ + X}&{B + Y}&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0&0\\0&I&0\\0&0&I\end{array}} \right]\).

Equate both the sides

Equate both the matrices, as shown below:

\(\left[ {\begin{array}{*{20}{c}}I&0&0\\{C + Z}&I&0\\{A + BZ + X}&{B + Y}&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0&0\\0&I&0\\0&0&I\end{array}} \right]\)

By comparing, the formulas obtained are shown below:

\[\begin{array}{c}C + Z = 0\\Z = - C\end{array}\]

And

\[\begin{array}{c}B + Y = 0\\Y = - B\end{array}\]

And

\(\begin{array}{c}A + BZ + X = 0\\A + B\left( { - C} \right) + X = 0\\A - BC + X = 0\\X = BC - A\end{array}\)

Therefore, the values are \(X = BC - A\), \[Y = - B\], and \[Z = - C\].