A matrix \(\displaystyle A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}\) lacks an inverse if and only if its determinant \(\displaystyle ad-bc\) is equal to zero. The determinant of \(\displaystyle A = \begin{bmatrix}x & 8 \\ 18 & x \end{bmatrix}\) is 

\(\displaystyle D = x \cdot x - 8 \cdot 18 = x^{2}- 144\).

Setting this equal to 0:

\(\displaystyle x^{2}- 144 = 0\)

\(\displaystyle x^{2}- 144+144 = 0 +144\)

\(\displaystyle x^{2} = 144\)

Taking the square root of both sides:

\(\displaystyle x = \pm \sqrt{144}\)

\(\displaystyle x = \pm 12\)

The matrix therefore has no inverse if either \(\displaystyle x = -12\) or \(\displaystyle x = 12\).

\(\displaystyle A = \begin{bmatrix} 4& -7\\ 8& 3 \end{bmatrix}\); the two-by-two identity matrix is \(\displaystyle I = \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}\). Add the two by adding elements in corresponding positions:

\(\displaystyle A +I = \begin{bmatrix} 4+1& -7+0\\ 8+0& 3+1 \end{bmatrix} = \begin{bmatrix} 5& -7 \\ 8 & 4 \end{bmatrix}\).

The inverse of a two-by-two matrix \(\displaystyle B = \begin{bmatrix} a& b\\ c& d \end{bmatrix}\) is \(\displaystyle B ^{-1}= \begin{bmatrix} \frac{ d}{\det B}& -\frac{b}{\det B}\\-\frac{ c}{\det B}& \frac{a }{\det B}\end{bmatrix}\), where 

\(\displaystyle \det B = ad -bc\).

We can find \(\displaystyle (A+I)^{-1}\) by setting \(\displaystyle a= 5, b = -7 , c = 8, d = 4\). The determinant of \(\displaystyle (A+I)^{-1}\) is

\(\displaystyle ad -bc = 5\cdot 4 - (-7) \cdot 8 = 20 - (-56 ) = 76\)

Replacing:

\(\displaystyle (A+I)^{-1}= \begin{bmatrix} \frac{ 4}{76}& \frac{7}{76}\\-\frac{ 8}{76}& \frac{5 }{76}\end{bmatrix}\);

simplifying the fractions, this is 

\(\displaystyle (A+I)^{-1}= \begin{bmatrix} \frac{ 1}{19}& \frac{7}{76}\\-\frac{ 2}{19}& \frac{5 }{76}\end{bmatrix}\)

The inverse \(\displaystyle A ^{-1}\) of any two-by-two matrix \(\displaystyle A\) can be found according to this pattern:

If \(\displaystyle A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}\)

then 

\(\displaystyle A ^{-1}= \frac{1}{D} \begin{bmatrix}d& - b\\ -c& a \end{bmatrix} = \begin{bmatrix}\frac{d}{D}& - \frac{b}{D}\\ -\frac{c}{D}& \frac{a }{D} \end{bmatrix}\),

where determinant \(\displaystyle D\) is equal to \(\displaystyle ad-cb\).

Therefore, if \(\displaystyle B = A ^{-1}\), then \(\displaystyle b_{11}\), the first row/first column entry in the matrix \(\displaystyle B\), can be found by setting \(\displaystyle a = 5, b= 4, c= 10, d=-8\), then evaluating:

\(\displaystyle b_{11} = \frac{d}{D} = \frac{d}{ad-bc} = \frac{-8}{5(-8)-4(10)} = \frac{-8}{-40 -40} = \frac{-8}{-80} = \frac{1}{10}\)

A matrix \(\displaystyle A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}\) lacks an inverse if and only if its determinant \(\displaystyle ad-bc\) is equal to zero. The determinant of \(\displaystyle A = \begin{bmatrix} 8& x\\ x& 4 \end{bmatrix}\) is

\(\displaystyle D = 8 \cdot 4 - x \cdot x = 32 - x^{2}\)

We seek the value of \(\displaystyle x\) that sets this quantity equal to 16. Setting it as such then solving for \(\displaystyle x\):

\(\displaystyle 32 - x^{2} = 16\)

\(\displaystyle 32 - x^{2} - 32 = 16 - 32\)

\(\displaystyle - x^{2} = -16\)

\(\displaystyle x^{2} = 16\)

\(\displaystyle x = \pm 4\)

Therefore, either \(\displaystyle x= -4\) or \(\displaystyle x = 4\).

A matrix \(\displaystyle A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}\) lacks an inverse if and only if its determinant \(\displaystyle ad-bc\) is equal to zero. The determinant of \(\displaystyle A = \begin{bmatrix} x & 8 \\ x&32 \end{bmatrix}\) is

\(\displaystyle D = x \cdot 32 - 8 \cdot x = 32x - 8x = 24x\)

Set this equal to 0 and solve for \(\displaystyle x\):

\(\displaystyle 24x = 0\)

\(\displaystyle 24x\div 24 = 0 \div 24\)

\(\displaystyle x = 0\),

the only such value.

An infinite series \(\displaystyle \sum_{i = 1}^{\infty}a^{i}\) converges to a sum if and only if \(\displaystyle |a| < 1\). However, in the series \(\displaystyle \sum_{i = 1}^{\infty} \left (\frac{4}{3} \right )^{i}\), this is not the case, as \(\displaystyle \left | \frac{4}{3} \right |= \frac{4}{3} > 1\). This series diverges.

The key to finding the next term lies in the denominators of the third term onwards. They are terms of the Fibonacci sequence, which begin with the terms 1 and 1 and whose subsequent terms are each formed by adding the previous two.

The \(\displaystyle n\)th term of the sequence is the number \(\displaystyle \frac{F_{n}-1}{F_{n}}\), where \(\displaystyle F_{n}\) is the \(\displaystyle n\)th number in the Fibonacci sequence (since the first two Fibonacci numbers are both 1, the first two terms being 0 fits this pattern). The Fibonacci number following 13 and 21 is their sum, 34, so the next number in the sequence is

\(\displaystyle \frac{34-1}{34}= \frac{33}{34}\).

\(\displaystyle 12\textbf{ + 1} = 13\)

\(\displaystyle 13\textbf{ + 4}= 17\)

\(\displaystyle 17\textbf{ + 9}= 26\)

\(\displaystyle 26\textbf{ + 16} = 42\)

\(\displaystyle 42\textbf{ + 25}=67\)

Each term is derived from the next by adding a perfect square integer; the increment increases from one square to the next higher one each time. To maintain the pattern, add the next perfect square, 36:

\(\displaystyle 67\textbf{ + 36 } =103\)

Each term is derived from the previous term by doubling the latter and alternately adding and subtracting 1, as follows:

\(\displaystyle 1 \textbf{ x 2 - 1 } = 1\)

\(\displaystyle 1 \textbf{ x 2 + 1 } = 3\)

\(\displaystyle 3 \textbf{ x 2 - 1 } = 5\)

\(\displaystyle 5 \textbf{ x 2 + 1 } = 11\)

\(\displaystyle 11 \textbf{ x 2 - 1 } =21\)

\(\displaystyle 21 \textbf{ x 2 + 1 } = 43\)

\(\displaystyle 43\textbf{ x 2 - 1 } = 85\)

\(\displaystyle 85 \textbf{ x 2 + 1 } = 171\)

The next term is derived as follows:

\(\displaystyle 171 \textbf{ x 2 - 1 } = 341\)

The pattern becomes more clear if each term is rewritten as a single radical expression:

\(\displaystyle 1 = \sqrt {1} = \sqrt{1!}\)

\(\displaystyle \sqrt{2} = \sqrt{2!}\)

\(\displaystyle \sqrt{6} = \sqrt{3!}\)

\(\displaystyle 2\sqrt{6} =\sqrt{4} \cdot \sqrt{6} = \sqrt {24}=\sqrt{4!}\)

\(\displaystyle 2 \sqrt{30}= \sqrt{4} \cdot \sqrt{30}= \sqrt{120} = \sqrt {5!}\)

\(\displaystyle 12\sqrt{5} = \sqrt{144} \cdot \sqrt{5} = \sqrt{720} = \sqrt{6!}\)

The \(\displaystyle n\)th term is \(\displaystyle \sqrt{n!}\) . The next (seventh) term is therefore

\(\displaystyle \sqrt{7!}= \sqrt{5,040}= \sqrt{144} \times \sqrt{35} = 12 \sqrt{35}\)