《Only Holy Story》的歌詞如下：
Verse 1:
Only holy story
Tells the truth about love
It's a story of hope
It's a story of pain
Chorus:
But it's a story worth telling
It's a story worth singing
Only holy story
It's the truth about our lives
Verse 2:
Only holy story
Shows us what it means
To love and to be loved
To give and to receive
Chorus:
But it's a story worth telling
It's a story worth singing
Only holy story
It's the truth about our lives
Bridge:
So let's open our hearts
And let the words sink in
We'll be singing this song
Till we've made it a cinch
Chorus:
Only holy story
The one that will never end
The one that will forever
In our hearts remain
Outro:
Only holy story
Is the truth about love
The one that will forever be求解下列行列式，可以使用展開方法求解： 23x + 46, -23x - 38, -14x + 39, -47x - 64; {49x + 76, 37x - 57, 67x + 89, 98x - 124; {-35x - 67, 79x + 56, -29x + 98, -78x - 56.
把給出的四階行列式，按照第一列展開，即可得到它的值。根據行列式的展開規則：$(a_{i,j}b_{j,k} = a_{i,k}b_{j,k} + \lbrack i \times (i - 1)\rbrack a_{j,k}b_{i + 1,k} + \lbrack i \times (i - 1)\rbrack a_{k,j}b_{i + 1,k}$。對給定的行列式按第一列展開後得到的三階行列式如下：$- a_{j,1}b_{j,3}$ 和 $- \lbrack j \times (j - 1)\rbrack a_{i + 1,2}b_{i + 2,3}$ $+ \lbrack j \times (j - 1)\rbrack a_{k,2}b_{k,3}$ ，按照第二列展開後得到的三階行列式如下：$- a_{j,2}b_{j,2}$ 和 $- \lbrack j \times (j - 1)\rbrack a_{i + 1,3}b_{i + 2,2}$ $+ \lbrack j \times (j - 1)\rbrack a_{k,3}b_{k,2}$，可以發現無論哪種展開方法都會導致中間出現 $3$ 個相同的 $0$。因為矩陣只有 $3$ 個 $1$，其它均為 $0$，故可以將上述三階行列式求和後再取模 $7$ 後作為該行列式的值，由於第三列是 $- \lbrack j \times (j - 1)\rbrack b$ 而與列下標的關係有 $- (n - i) \cdot i$ 項之差，導致無法找到除 $(i, j)$ 的數字對應的組合結果與餘下的相同，故該行列式的值為 $0$。所以給出的四階行列式等於 $0$。