 # 4.10 Matrices, SPM Paper 2 (Long Questions)

Question 1:
It is given that matrix A = $\left(\begin{array}{cc}3& -1\\ 5& -2\end{array}\right)$
(a) Find the inverse matrix of A.
(b) Write the following simultaneous linear equations as matrix equation:
3uv = 9
5u – 2v = 13
Hence, using matrix method, calculate the value of u and v.

Solution
:

(a)
$\begin{array}{l}{A}^{-1}=\frac{1}{3\left(-2\right)-\left(5\right)\left(-1\right)}\left(\begin{array}{cc}-2& 1\\ -5& 3\end{array}\right)\\ =-1\left(\begin{array}{cc}-2& 1\\ -5& 3\end{array}\right)=\left(\begin{array}{cc}2& -1\\ 5& -3\end{array}\right)\end{array}$

(b)

$\begin{array}{l}\left(\begin{array}{cc}3& -1\\ 5& -2\end{array}\right)\left(\begin{array}{l}u\\ v\end{array}\right)=\left(\begin{array}{l}9\\ 13\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=-1\left(\begin{array}{cc}-2& 1\\ -5& 3\end{array}\right)\left(\begin{array}{l}9\\ 13\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=-1\left(\begin{array}{l}\left(-2\right)\left(9\right)+\left(1\right)\left(13\right)\\ \left(-5\right)\left(9\right)+\left(3\right)\left(13\right)\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=-1\left(\begin{array}{l}-5\\ -6\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=\left(\begin{array}{l}5\\ 6\end{array}\right)\\ \therefore u=5,v=6\end{array}$

Question 2:
It is given that matrix A = $\left(\begin{array}{cc}2& -5\\ 1& 3\end{array}\right)$ and matrix B = $m\left(\begin{array}{cc}3& k\\ -1& 2\end{array}\right)$ such that AB = $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$
(a) Find the value of m and of k.
(b) Write the following simultaneous linear equations as matrix equation:
2u – 5v = –15
u + 3v = –2
Hence, using matrix method, calculate the value of u and v.

Solution:

(a) Since AB = $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ , B is the inverse of A.

$m=\frac{1}{\left(2\right)\left(3\right)-\left(-5\right)\left(1\right)}=\frac{1}{11}$
k = 5

(b)
$\begin{array}{l}\left(\begin{array}{cc}2& -5\\ 1& 3\end{array}\right)\left(\begin{array}{l}u\\ v\end{array}\right)=\left(\begin{array}{l}-15\\ -2\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=\frac{1}{11}\left(\begin{array}{cc}3& 5\\ -1& 2\end{array}\right)\left(\begin{array}{l}-15\\ -2\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=\frac{1}{11}\left(\begin{array}{l}\left(3\right)\left(-15\right)+\left(5\right)\left(-2\right)\\ \left(-1\right)\left(-15\right)+\left(2\right)\left(-2\right)\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=\frac{1}{11}\left(\begin{array}{l}-55\\ 11\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=\left(\begin{array}{l}-5\\ 1\end{array}\right)\\ \therefore u=-5,v=1\end{array}$

Question 3:
It is given that Q  $\left(\begin{array}{cc}3& 2\\ 6& 5\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ , where Q is a 2 x 2 matrix.
(a) Find Q.
(b) Write the following simultaneous linear equations as matrix equation:
3u + 2v = 5
6u + 5v = 2
Hence, using matrix method, calculate the value of u and v.

Solution
:

(a)
$\begin{array}{l}Q={\left(\begin{array}{cc}3& 2\\ 6& 5\end{array}\right)}^{-1}\\ Q=\frac{1}{3\left(5\right)-2\left(6\right)}\left(\begin{array}{cc}5& -2\\ -6& 3\end{array}\right)\\ Q=\frac{1}{3}\left(\begin{array}{cc}5& -2\\ -6& 3\end{array}\right)\\ Q=\left(\begin{array}{cc}\frac{5}{3}& -\frac{2}{3}\\ -2& 1\end{array}\right)\end{array}$

(b)

$\begin{array}{l}\left(\begin{array}{cc}3& 2\\ 6& 5\end{array}\right)\left(\begin{array}{l}u\\ v\end{array}\right)=\left(\begin{array}{l}5\\ 2\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=\frac{1}{3}\left(\begin{array}{cc}5& -2\\ -6& 3\end{array}\right)\left(\begin{array}{l}5\\ 2\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=\frac{1}{3}\left(\begin{array}{l}\left(5\right)\left(5\right)+\left(-2\right)\left(2\right)\\ \left(-6\right)\left(5\right)+\left(3\right)\left(2\right)\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=\frac{1}{3}\left(\begin{array}{l}21\\ -24\end{array}\right)\\ \text{}\left(\begin{array}{l}u\\ v\end{array}\right)=\left(\begin{array}{l}7\\ -8\end{array}\right)\\ \therefore u=7,v=-8\end{array}$

Question 4:
It is given that $Q\left(\begin{array}{cc}3& -2\\ 5& -4\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$  , where Q is a 2 × 2 matrix.
(a) Find the matrix Q.
(b) Write the following simultaneous linear equations as matrix equation:
3x – 2y = 7
5x – 4y = 9
Hence, using matrix method, calculate the value of x and y.

Solution:

(a)
$\begin{array}{l}Q=\frac{1}{3\left(-4\right)-\left(5\right)\left(-2\right)}\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\\ =-\frac{1}{2}\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\\ =\left(\begin{array}{cc}2& -1\\ \frac{5}{2}& -\frac{3}{2}\end{array}\right)\end{array}$

(b)

$\begin{array}{l}\left(\begin{array}{cc}3& -2\\ 5& -4\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}7\\ 9\end{array}\right)\\ \text{}\left(\begin{array}{l}x\\ y\end{array}\right)=-\frac{1}{2}\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\left(\begin{array}{l}7\\ 9\end{array}\right)\\ \text{}\left(\begin{array}{l}x\\ y\end{array}\right)=-\frac{1}{2}\left(\begin{array}{l}\left(-4\right)\left(7\right)+\left(2\right)\left(9\right)\\ \left(-5\right)\left(7\right)+\left(3\right)\left(9\right)\end{array}\right)\\ \text{}\left(\begin{array}{l}x\\ y\end{array}\right)=-\frac{1}{2}\left(\begin{array}{l}-10\\ -8\end{array}\right)\\ \text{}\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}5\\ 4\end{array}\right)\\ \therefore x=5,\text{}y=4\end{array}$