Matrices

Understand the concept, operations and properties of matrices

  • Theory
    [TheoryMissing]
  • Examples
    Let \(A = \left( {\begin{array}{*{20}{c}}0&x\\y&z\end{array}} \right)\) and \(A\;\left( {\begin{array}{*{20}{c}}8\\4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}8\\{52}\end{array}} \right)\).
    (a) Find the value of x.
    (b) Prove that \(2y + z = 13\). Also, if \(y + z = 8\), find A.
  • Solutions
    (a)
    \(\begin{array}{1}A\,\left( {\begin{array}{*{20}{c}}8\\4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}8\\{52}\end{array}} \right)\\
    \left( {\begin{array}{*{20}{c}}0&x\\y&z\end{array}} \right)\;\left( {\begin{array}{*{20}{c}}8\\4\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}8\\{52}\end{array}} \right)^{}}\\\left( {\begin{array}{*{20}{c}}{4x}\\{8y + 4z}\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}8\\{52}\end{array}} \right)^{}}\end{array}\)
    Consider the definition of equality of two matrices,
    \(\begin{array}{1}4x = 8\\x = {\underline{\underline 2} ^{}}\end{array}\)

    (b)\(\left( {\begin{array}{*{20}{c}}{4x}\\{8y + 4z}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}8\\{52}\end{array}} \right)\)
    Consider the definition of equality of two matrices,
    \(8y + 4z = 52\)
    ∴ \(2y + z = 13\,......................\;(1)\)
    Also, \(\begin{array}{c}y + z = 8\\z
    = 8 - y\;.................\;(2)\end{array}\)
    Substitute (2) into (1), \(\begin{array}{c}2y + 8 - y = 13\\
    y = {5^{}}\end{array}\)
    Substitute \(y = 5\) into (2), \(\begin{array}{c}z = 8 - 5\\
    = {3^{}}\end{array}\)
    ∴ \(A = \underline{\underline {\left( {\begin{array}{*{20}{c}}0&2\\5&3\end{array}} \right)}} \)

Understand the concept, operations and properties of inverses of square matrices of order 2 and order 3

  • Theory
    [TheoryMissing]
  • Examples
    It is given that P, Q and R are square matrices of order 2, where R is invertible, \({P^2}R = P{Q^T}\), \({P^{ - 1}}Q = \sqrt 3 I\) and \(\left| {{\kern 1pt} Q{\kern 1pt} } \right| = 15\). Find the values of \(\left| {{\kern 1pt} P{\kern 1pt} } \right|\) and \(\left| {{\kern 1pt} {R^{ - 1}}{\kern 1pt} } \right|\).
  • Solutions
    \(\begin{array}{1}{P^{ - 1}}Q = \sqrt 3 I\\\left| {{\kern 1pt} {P^{ - 1}}Q{\kern 1pt} } \right| = {\left| {{\kern 1pt} \sqrt 3 I{\kern 1pt} } \right|^{}}\\\left| {{\kern 1pt} {P^{ - 1}}{\kern 1pt} } \right|\;\left| {{\kern 1pt} Q{\kern 1pt} } \right| = {(\sqrt 3 )^2}\,{\left| {{\kern 1pt} I{\kern 1pt} } \right|^{^{}}}\\{\left| {{\kern 1pt} P{\kern 1pt} } \right|^{ - 1}} \times 15 = 3 \times {1^{^{}}}\\\left| {{\kern 1pt} P{\kern 1pt} } \right| = \frac{{15}}{3}\\ = {\underline{\underline 5} ^{}}\end{array}\)
    \(\begin{array}{1}{P^2}R = P{Q^T}\\\left| {{\kern 1pt} {P^2}R{\kern 1pt} } \right| = {\left| {{\kern 1pt} P{Q^T}{\kern 1pt} } \right|^{}}\\\left| {{\kern 1pt} {P^2}{\kern 1pt} } \right|\;\left| {{\kern 1pt} R{\kern 1pt} } \right| = \left| {{\kern 1pt} P{\kern 1pt} } \right|\,{\left| {{\kern 1pt} {Q^T}{\kern 1pt} } \right|^{}}\\{\left| {{\kern 1pt} P{\kern 1pt} } \right|^2}\;\left| {{\kern 1pt} R{\kern 1pt} } \right| = \left| {{\kern 1pt} P{\kern 1pt} } \right|\;{\left| {{\kern 1pt} Q{\kern 1pt} } \right|^{^{}}}\\{5^2}\;\left| {{\kern 1pt} R{\kern 1pt} } \right| = 5 \times {15^{^{}}}\\\left| {{\kern 1pt} R{\kern 1pt} } \right| = {3^{^{}}}\end{array}\)
    ∵ R is invertible.
    ∴ \(\begin{array}{1}\left| {{\kern 1pt} {R^{ - 1}}{\kern 1pt} } \right| = {\left| {{\kern 1pt} R{\kern 1pt} } \right|^{ - 1}}\\ = {3^{ - 1}}\\ = \underline{\underline {\frac{1}{3}}} \end{array}\)