Exercise – 4.6
1. Examine the consistency of the system of equations: $x + 2y = 2, 2x + 3y = 3$.
Solution:
The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&2\\2&3\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}2\\3\end{array}} \right]$
Now, $|A| = 3 – 4 = -1 \neq 0$
Hence, the system of equations is consistent.
2. Examine the consistency of the system of equations: $2x – y = 5, x + y = 4$.
Solution:
The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&-1\\1&1\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}5\\4\end{array}} \right]$
Now, $|A| = 2 + 1 = 3 \neq 0$
Hence, the system of equations is consistent.
3. Examine the consistency of the system of equations: $x + 3y = 5, 2x + 6y = 8$.
Solution:
The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&3\\2&6\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}5\\8\end{array}} \right]$
Now, $|A| = 6 – 6 = 0$
Hence, A is a singular matrix. So, we calculate $(adj A)B$.
$adj A = \left[ {\begin{array}{*{20}{r}}6&-3\\-2&1\end{array}} \right]$
Now, $(adj A)B = \left[ {\begin{array}{*{20}{r}}6&-3\\-2&1\end{array}} \right]\left[ {\begin{array}{*{20}{r}}5\\8\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}30-24\\-10+8\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}6\\-2\end{array}} \right] \neq \left[ {\begin{array}{*{20}{r}}0\\0\end{array}} \right]$
Hence, the equations are inconsistent with no solution.
4. Examine the consistency of the system of equations: $x + y + z = 1, 2x + 3y + 2z = 2, ax + ay + 2az = 4$.
Solution:
The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&1&1\\2&3&2\\a&a&2a\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}1\\2\\4\end{array}} \right]$
Now, $|A| = 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a) = 4a – 3a = a \neq 0$
Two conditions arise:
I: If $a \neq 0$, then $|A| \neq 0$, hence the system of equations is consistent and has a unique solution.
II: If $a = 0$, then $|A| = 0$. So, we need to calculate $adj A$. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}3&2\\a&2a\end{array}} \right| = 4a$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}2&2\\a&2a\end{array}} \right| = 2a$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}2&3\\a&a\end{array}} \right| = -a$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}1&1\\a&2a\end{array}} \right| = -a$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}1&1\\a&2a\end{array}} \right| = a$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}1&1\\a&a\end{array}} \right| = 0$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}1&1\\3&2\end{array}} \right| = -1$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}1&1\\2&2\end{array}} \right| = 0$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}1&1\\2&3\end{array}} \right| = 1$
Now, $(adj A) \cdot B = \left[ {\begin{array}{*{20}{r}}4a&-a&-1\\-2a&a&0\\-a&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1\\2\\4\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}2a-4\\0\\-a+4\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-4\\0\\4\end{array}} \right] \neq 0$
Hence, the equations are inconsistent with no solution because if $a = 0$, then the third system of equations is not possible.
5. Examine the consistency of the system of equations: $3x – y – 2z = 2, 2y – z = -1, 3x – 5y = 3$.
Solution:
The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}3&-1&-2\\0&2&-1\\3&-5&0\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}2\\-1\\3\end{array}} \right]$
Now, $|A| = 3(0 – 5) + 1(0 + 3) – 2(0 – 6) = -15 + 3 + 12 = 0$
Here, A is a singular matrix, so we will compute $(adj A)B$.
For $adj A$, cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}2&-1\\-5&0\end{array}} \right| = -5$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}0&-1\\3&0\end{array}} \right| = -3$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}0&2\\3&-5\end{array}} \right| = -6$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}-1&-2\\-5&0\end{array}} \right| = 10$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}3&-2\\3&0\end{array}} \right| = 6$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}3&-1\\3&-5\end{array}} \right| = 12$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}-1&-2\\2&-1\end{array}} \right| = 5$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}3&-2\\0&-1\end{array}} \right| = 3$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}3&-1\\0&2\end{array}} \right| = 6$
Now, $(adj A) \cdot B = \left[ {\begin{array}{*{20}{r}}-5&10&5\\-3&6&3\\-6&12&6\end{array}} \right]\left[ {\begin{array}{*{20}{r}}2\\-1\\3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-10-10+15\\-6-6+9\\-12-12+18\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-5\\-3\\-6\end{array}} \right] \neq 0$
Hence, the system of equations is inconsistent with no solution.
6. Examine the consistency of the system of equations: $5x – y + 4z = 5, 2x + 3y + 5z = 2, 5x – 2y + 6z = -1$.
Solution:
The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}5&-1&4\\2&3&5\\5&-2&6\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}5\\2\\-1\end{array}} \right]$
Now, $|A| = 5(18 + 10) + 1(12 – 25) + 4(-4 – 15) = 140 – 13 – 76 = 51 \neq 0$
Hence, the equations are consistent with a unique solution.
7. Solve the system of linear equations using the matrix method: $5x + 2y = 4, 7x + 3y = 5$.
Solution:
The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}5&2\\7&3\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}4\\5\end{array}} \right]$
Now, $|A| = 15 – 14 = 1 \neq 0$
$\Rightarrow$ A is non-singular and so the given system has a unique solution. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(3) = 3$
$A_{12} = (-1)^{1+2}(7) = -7$
$A_{21} = (-1)^{2+1}(2) = -2$
$A_{22} = (-1)^{2+2}(5) = 5$
$adj A = \left[ {\begin{array}{*{20}{r}}3&-2\\-7&5\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \left[ {\begin{array}{*{20}{r}}3&-2\\-7&5\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}3&-2\\-7&5\end{array}} \right]\left[ {\begin{array}{*{20}{r}}4\\5\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}12-10\\-28+25\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}2\\-3\end{array}} \right]$
Hence, $x = 2, y = -3$.
8. Solve the system of linear equations using the matrix method: $2x – y = -2, 3x + 4y = 3$.
Solution:
The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&-1\\3&4\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}-2\\3\end{array}} \right]$
Now, $|A| = 8 + 3 = 11 \neq 0$
$\Rightarrow$ A is non-singular and so the given system has a unique solution. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(4) = 4$
$A_{12} = (-1)^{1+2}(3) = -3$
$A_{21} = (-1)^{2+1}(-1) = 1$
$A_{22} = (-1)^{2+2}(2) = 2$
$adj A = \left[ {\begin{array}{*{20}{r}}4&1\\-3&2\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}4&1\\-3&2\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right] = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}4&1\\-3&2\end{array}} \right]\left[ {\begin{array}{*{20}{r}}-2\\3\end{array}} \right] = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-8+3\\6+6\end{array}} \right] = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-5\\12\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-\frac{5}{11}\\\frac{12}{11}\end{array}} \right]$
Hence, $x = -\frac{5}{11}, y = \frac{12}{11}$.
9. Solve the system of linear equations using the matrix method: $4x – 3y = 3, 3x – 5y = 7$.
Solution:
The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}4&-3\\3&-5\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}3\\7\end{array}} \right]$
Now, $|A| = -20 + 9 = -11 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given system has a unique solution. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(-5) = -5$
$A_{12} = (-1)^{1+2}(3) = -3$
$A_{21} = (-1)^{2+1}(-3) = 3$
$A_{22} = (-1)^{2+2}(4) = 4$
$adj A = \left[ {\begin{array}{*{20}{r}}-5&-3\\3&4\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = -\frac{1}{11}\left[ {\begin{array}{*{20}{r}}-5&-3\\3&4\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right] = -\frac{1}{11}\left[ {\begin{array}{*{20}{r}}-5&-3\\3&4\end{array}} \right]\left[ {\begin{array}{*{20}{r}}3\\7\end{array}} \right] = -\frac{1}{11}\left[ {\begin{array}{*{20}{r}}-15+21\\-9+28\end{array}} \right] = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-6\\-19\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-\frac{6}{11}\\-\frac{19}{11}\end{array}} \right]$
Hence, $x = -\frac{6}{11}, y = -\frac{19}{11}$.
10. Solve the system of linear equations using the matrix method: $5x + 2y = 3, 3x + 2y = 5$.
Solution:
The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}5&2\\3&2\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}3\\5\end{array}} \right]$
Now, $|A| = 10 – 6 = 4 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given system has a unique solution. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(2) = 2$
$A_{12} = (-1)^{1+2}(3) = -3$
$A_{21} = (-1)^{2+1}(2) = -2$
$A_{22} = (-1)^{2+2}(5) = 5$
$adj A = \left[ {\begin{array}{*{20}{r}}2&-2\\-3&5\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}2&-2\\-3&5\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}2&-2\\-3&5\end{array}} \right]\left[ {\begin{array}{*{20}{r}}3\\5\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}6-10\\-9+25\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}-4\\16\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-1\\4\end{array}} \right]$
Hence, $x = -1, y = 4$.
11. Solve the system of linear equations using the matrix method: $2x + y + z = 1, x – 2y – z = \frac{3}{2}, 3y – 5z = 9$.
Solution:
The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&1&1\\1&-2&-1\\0&3&-5\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}1\\\frac{3}{2}\\9\end{array}} \right]$
Now, $|A| = 26 + 8 = 34 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given equations have a unique solution. Here, cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}-2&-1\\3&-5\end{array}} \right| = 13$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}1&-1\\0&-5\end{array}} \right| = 5$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}1&-2\\0&3\end{array}} \right| = 3$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}1&1\\3&-5\end{array}} \right| = 8$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}2&1\\0&-5\end{array}} \right| = -10$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}2&1\\0&3\end{array}} \right| = -6$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}1&1\\-2&-1\end{array}} \right| = 1$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}2&1\\1&-1\end{array}} \right| = 3$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}2&1\\1&-2\end{array}} \right| = -5$
$adj A = \left[ {\begin{array}{*{20}{r}}13&8&1\\5&-10&3\\3&-6&-5\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{34}\left[ {\begin{array}{*{20}{r}}13&8&1\\5&-10&3\\3&-6&-5\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \frac{1}{34}\left[ {\begin{array}{*{20}{r}}13&8&1\\5&-10&3\\3&-6&-5\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1\\\frac{3}{2}\\9\end{array}} \right] = \frac{1}{34}\left[ {\begin{array}{*{20}{r}}13+12+9\\5-15+27\\3-9-45\end{array}} \right] = \frac{1}{34}\left[ {\begin{array}{*{20}{r}}34\\17\\-51\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1\\\frac{1}{2}\\-\frac{3}{2}\end{array}} \right]$
Hence, $x = 1, y = \frac{1}{2}, z = -\frac{3}{2}$.
12. Solve the system of linear equations using the matrix method: $x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2$.
Solution:
The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&-1&1\\2&1&-3\\1&1&1\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}4\\0\\2\end{array}} \right]$
Now, $|A| = 1(1 + 3) + 1(2 + 3) + 1(2 – 1) = 4 + 5 + 1 = 10 \neq 0$
$\Rightarrow$ A is non-singular and so the given equations have a unique solution. Here, cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}1&-3\\1&1\end{array}} \right| = 4$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}2&-3\\1&1\end{array}} \right| = -5$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}2&1\\1&1\end{array}} \right| = 1$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}-1&1\\1&1\end{array}} \right| = 2$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}1&1\\1&1\end{array}} \right| = 0$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}1&-1\\1&1\end{array}} \right| = -2$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}-1&1\\1&-3\end{array}} \right| = 2$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}1&1\\2&-3\end{array}} \right| = 5$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}1&-1\\2&1\end{array}} \right| = 3$
$adj A = \left[ {\begin{array}{*{20}{r}}4&2&2\\-5&0&5\\1&-2&3\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}4&2&2\\-5&0&5\\1&-2&3\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}4&2&2\\-5&0&5\\1&-2&3\end{array}} \right]\left[ {\begin{array}{*{20}{r}}4\\0\\2\end{array}} \right] = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}16+0+4\\-20+0+10\\4-0+6\end{array}} \right] = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}20\\-10\\10\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}2\\-1\\1\end{array}} \right]$
Hence, $x = 2, y = -1, z = 1$.
13. Solve the system of linear equations using the matrix method: $2x + 3y + 3z = 5, x – 2y + z = -4, 3x – y – 2z = 3$.
Solution:
The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&3&3\\1&-2&1\\3&-1&-2\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}5\\-4\\3\end{array}} \right]$
Now, $|A| = 40 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given equations have a unique solution. Here, cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}-2&1\\-1&-2\end{array}} \right| = 5$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}1&1\\3&-2\end{array}} \right| = 5$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}1&-2\\3&-1\end{array}} \right| = 5$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}3&3\\-1&-2\end{array}} \right| = 3$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}2&3\\3&-2\end{array}} \right| = -13$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}2&3\\3&-1\end{array}} \right| = 11$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}3&3\\-2&1\end{array}} \right| = 9$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}2&3\\1&1\end{array}} \right| = 1$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}2&3\\1&-2\end{array}} \right| = -7$
$adj A = \left[ {\begin{array}{*{20}{r}}5&3&9\\5&-13&1\\5&11&-7\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{40}\left[ {\begin{array}{*{20}{r}}5&3&9\\5&-13&1\\5&11&-7\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \frac{1}{40}\left[ {\begin{array}{*{20}{r}}5&3&9\\5&-13&1\\5&11&-7\end{array}} \right]\left[ {\begin{array}{*{20}{r}}5\\-4\\3\end{array}} \right] = \frac{1}{40}\left[ {\begin{array}{*{20}{r}}25-12+27\\25+52+3\\25-44-21\end{array}} \right] = \frac{1}{40}\left[ {\begin{array}{*{20}{r}}40\\80\\-40\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1\\2\\-1\end{array}} \right]$
Hence, $x = 1, y = 2, z = -1$.
14. Solve the system of linear equations using the matrix method: $x – y + 2z = 7, 3x + 4y – 5z = -5, 2x – y + 3z = 12$.
Solution:
The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&-1&2\\3&4&-5\\2&-1&3\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}7\\-5\\12\end{array}} \right]$
Now, $|A| = 4 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given equations have a unique solution. Here, cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}4&-5\\-1&3\end{array}} \right| = 7$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}3&-5\\2&3\end{array}} \right| = -19$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}3&4\\2&-1\end{array}} \right| = -11$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}-1&2\\-1&3\end{array}} \right| = 1$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}1&2\\2&3\end{array}} \right| = -1$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}1&-1\\2&-1\end{array}} \right| = -1$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}-1&2\\4&-5\end{array}} \right| = -3$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}1&2\\3&-5\end{array}} \right| = 11$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}1&-1\\3&4\end{array}} \right| = 7$
$adj A = \left[ {\begin{array}{*{20}{r}}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}} \right]\left[ {\begin{array}{*{20}{r}}7\\-5\\12\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}49-5-36\\-133+5+132\\-77+5+84\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}8\\4\\12\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}2\\1\\3\end{array}} \right]$
Hence, $x = 2, y = 1, z = 3$.
15. If $A = \left[ {\begin{array}{*{20}{r}}2&-3&5\\3&2&-4\\1&1&-2\end{array}} \right]$, find $A^{-1}$. Using $A^{-1}$, solve the system of equations $2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3$.
Solution:
We have, $A = \left[ {\begin{array}{*{20}{r}}2&-3&5\\3&2&-4\\1&1&-2\end{array}} \right]$
$\therefore |A| = -1 \neq 0$
So, A is invertible. Cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}2&-4\\1&-2\end{array}} \right| = 0$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}3&-4\\1&-2\end{array}} \right| = 2$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right| = 1$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}-3&5\\1&-2\end{array}} \right| = -1$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}2&5\\1&-2\end{array}} \right| = -9$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}2&-3\\1&1\end{array}} \right| = -5$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}-3&5\\2&-4\end{array}} \right| = 2$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}2&5\\3&-4\end{array}} \right| = 23$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}2&-3\\3&2\end{array}} \right| = 13$
$adj A = \left[ {\begin{array}{*{20}{r}}0&-1&2\\2&-9&23\\1&-5&13\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = -\left[ {\begin{array}{*{20}{r}}0&-1&2\\2&-9&23\\1&-5&13\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&1&-2\\-2&9&-23\\-1&5&-13\end{array}} \right]$
Now, the given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&-3&5\\3&2&-4\\1&1&-2\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}11\\-5\\-3\end{array}} \right]$
As, $|A| = -1 \neq 0$, so the given system of equations has a unique solution given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&1&-2\\-2&9&-23\\-1&5&-13\end{array}} \right]\left[ {\begin{array}{*{20}{r}}11\\-5\\-3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0-5+6\\-22-45+69\\-11-25+39\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1\\2\\3\end{array}} \right]$
Hence, $x = 1, y = 2, z = 3$.
16. The cost of 4 kg onion, 3 kg wheat, and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat, and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat, and 3 kg rice is Rs. 70. Find the cost of each item per kg by the matrix method.
Solution:
Let the cost of 1 kg onion = Rs. $x$, cost of 1 kg wheat = Rs. $y$, and cost of 1 kg rice = Rs. $z$.
According to the question, we have:
$4x + 3y + 2z = 60$
$2x + 4y + 6z = 90$
$6x + 2y + 3z = 70$
This system can be written as $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}4&3&2\\2&4&6\\6&2&3\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}60\\90\\70\end{array}} \right]$
Now, $|A| = 50 \neq 0$
$\therefore$ A is a non-singular matrix and the system has a unique solution. Cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}4&6\\2&3\end{array}} \right| = 0$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}2&6\\6&3\end{array}} \right| = 30$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}2&4\\6&2\end{array}} \right| = -20$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}3&2\\2&3\end{array}} \right| = -5$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}4&2\\6&3\end{array}} \right| = 0$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}4&3\\6&2\end{array}} \right| = 10$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}3&2\\4&6\end{array}} \right| = 10$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}4&2\\2&6\end{array}} \right| = -20$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}4&3\\2&4\end{array}} \right| = 10$
$adj A = \left[ {\begin{array}{*{20}{r}}0&-5&10\\30&0&-20\\-20&10&10\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{50}\left[ {\begin{array}{*{20}{r}}0&-5&10\\30&0&-20\\-20&10&10\end{array}} \right]$
As, \( AX = B \Rightarrow X = A^{-1}B \).
\(\Rightarrow \left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \frac{1}{50} \left[ \begin{array}{rrr} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array} \right] \left[ \begin{array}{r} 60 \\ 90 \\ 70 \end{array} \right] \).
\(= \frac{1}{5} \left[ \begin{array}{rrr} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array} \right] \left[ \begin{array}{r} 6 \\ 9 \\ 7 \end{array} \right] \).
\(= \frac{1}{5} \left[ \begin{array}{r} 0 – 45 + 70 \\ 180 + 0 – 140 \\ -120 + 90 + 70 \end{array} \right] \).
\(= \frac{1}{5} \left[ \begin{array}{r} 25 \\ 40 \\ 40 \end{array} \right] = \left[ \begin{array}{r} 5 \\ 8 \\ 8 \end{array} \right] \).
Hence, the cost of 1 kg of onion = Rs. 5,
cost of 1 kg of wheat = Rs. 8,
and cost of 1 kg of rice = Rs. 8.