### Matrices

#### Understand the concept, operations and properties of matrices

• [TheoryMissing]
• 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.
• (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

• [TheoryMissing]
• 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 $$\left| {{\kern 1pt} {R^{ - 1}}{\kern 1pt} } \right|$$.
• $$\begin{array}{1}{P^{ - 1}}Q = \sqrt 3 I\\\left| {{\kern 1pt} {P^{ - 1}}Q{\kern 1pt} } \right| = {\left| {{\kern 1pt} \sqrt 3 I{\kern 1pt} } \right|^{}}\\\left| {{\kern 1pt} {P^{ - 1}}{\kern 1pt} } \right|\;\left| {{\kern 1pt} Q{\kern 1pt} } \right| = {(\sqrt 3 )^2}\,{\left| {{\kern 1pt} I{\kern 1pt} } \right|^{^{}}}\\{\left| {{\kern 1pt} P{\kern 1pt} } \right|^{ - 1}} \times 15 = 3 \times {1^{^{}}}\\\left| {{\kern 1pt} P{\kern 1pt} } \right| = \frac{{15}}{3}\\ = {\underline{\underline 5} ^{}}\end{array}$$
$$\begin{array}{1}{P^2}R = P{Q^T}\\\left| {{\kern 1pt} {P^2}R{\kern 1pt} } \right| = {\left| {{\kern 1pt} P{Q^T}{\kern 1pt} } \right|^{}}\\\left| {{\kern 1pt} {P^2}{\kern 1pt} } \right|\;\left| {{\kern 1pt} R{\kern 1pt} } \right| = \left| {{\kern 1pt} P{\kern 1pt} } \right|\,{\left| {{\kern 1pt} {Q^T}{\kern 1pt} } \right|^{}}\\{\left| {{\kern 1pt} P{\kern 1pt} } \right|^2}\;\left| {{\kern 1pt} R{\kern 1pt} } \right| = \left| {{\kern 1pt} P{\kern 1pt} } \right|\;{\left| {{\kern 1pt} Q{\kern 1pt} } \right|^{^{}}}\\{5^2}\;\left| {{\kern 1pt} R{\kern 1pt} } \right| = 5 \times {15^{^{}}}\\\left| {{\kern 1pt} R{\kern 1pt} } \right| = {3^{^{}}}\end{array}$$
∵ R is invertible.
∴ $$\begin{array}{1}\left| {{\kern 1pt} {R^{ - 1}}{\kern 1pt} } \right| = {\left| {{\kern 1pt} R{\kern 1pt} } \right|^{ - 1}}\\ = {3^{ - 1}}\\ = \underline{\underline {\frac{1}{3}}} \end{array}$$