24.5 Solving Systems Of Equations

We can solve system of equations using matrices by

(a).

Using the inverse of the matrix \begin {align*} ax +by + cz & = m\\ dx + ey + fz & = n\\ gx + hy + jz & =p\\ \end {align*}

We set the system as one equation \(AX=B\) where \[A= \begin {pmatrix} a & b & c\\ d & e & f\\ g & h & j\\ \end {pmatrix} ,\quad X= \begin {pmatrix} x\\ y\\ z\\ \end {pmatrix} ,\quad B= \begin {pmatrix} m\\n\\p\\ \end {pmatrix} \]

\begin {align*} AX &=B\\ A^{-1}AX &=A^{-1}B\\ IX & = A^{-1}B\\\\ \implies \quad \begin {pmatrix} x\\ y\\ z\\ \end {pmatrix} & = A^{-1}B.\\ \end {align*}

Example 24.8.

Solve the system of equations: \begin {align*} 2x+y-z & =5\\ x + y -z & =4\\ -x-2y+3z & = -8\\ \end {align*}

Solution.

\[AX=B\cdots \cdots \cdots (1)\]

\[A= \begin {pmatrix} 2 & 1 & -1\\ 1 & 1 & -1\\ -1 & -2 & 3\\ \end {pmatrix} ,\quad X= \begin {pmatrix} x\\y\\z\\ \end {pmatrix} ,\quad B= \begin {pmatrix} 5\\ 4 \\ -8\\ \end {pmatrix} \]

We find \(A^{-1}\),

\(A^{-1}= \begin {pmatrix} 1 & -1 & 0\\ -2 & 5 & 1\\ -1 & 3 & 1\\ \end {pmatrix} \)

Substitute in equation (1), we find that \begin {align*} AX & =B\\ A^{-1}AX & =A^{-1}B\\ \begin {pmatrix} 1 & -1 & 0\\ -2 & 5 & 1\\ -1 & 3 & 1\\ \end {pmatrix} \begin {pmatrix} 2 & 1 & -1\\ 1 & 1 & -1\\ -1 & -2 & 3\\ \end {pmatrix} \begin {pmatrix} x\\y\\z \end {pmatrix} & = \begin {pmatrix} 1 & -1 & 0\\ -2 & 5 & 1\\ -1 & 3 & 1\\ \end {pmatrix} \begin {pmatrix} 5\\4\\-8\\ \end {pmatrix}\\\\ \begin {pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end {pmatrix} \begin {pmatrix} x\\y\\z\\ \end {pmatrix} &= \begin {pmatrix} 1\\ 2\\ -1\\ \end {pmatrix}\\\\ \begin {pmatrix} x\\y\\z\\ \end {pmatrix} &= \begin {pmatrix} 1\\2\\-1\\ \end {pmatrix} \end {align*}

\[\therefore \quad x=1,\quad y=2,\quad z=-1\]

Note 24.9. Always substitute the answer back into the original equations. Here \[2(1)+2-(-1)=5,\quad 1+2-(-1)=4,\quad -1-2(2)+3(-1)=-8,\] all three checking out. The check costs a few seconds and catches both arithmetic slips in the inverse and misreadings of the question.

Note also why the method works at all: \(A^{-1}\) exists precisely when \(\left |A\right |\neq 0\), and that is exactly the condition for the system to have a unique solution. If \(\left |A\right |=0\) the system has either no solution or infinitely many, and no inverse can distinguish the two — a different method is needed.

(b). Cramer’s Rule
Let the system of equation be \begin {align*} ax+by+cz &=m\\ dx+ey+fz &=n\\ gx+hy+jz &=p\\ \end {align*}

Cramer’s rule state that you set \(A= \begin {pmatrix} a & b & c\\ d & e & f\\ g & h & j\\ \end {pmatrix} \) and find \(|A|= \begin {vmatrix} a & b & c\\ d & e & f\\ g & h & j\\ \end {vmatrix} \)

\[A= \begin {pmatrix} a & b & c\\ d & e & f\\ g & h & j\\ \end {pmatrix} , X= \begin {pmatrix} x\\y\\z\\ \end {pmatrix} , B= \begin {pmatrix} m\\ n\\ p\\ \end {pmatrix} \]

Then \[|A_1|= \begin {vmatrix} m & b & c\\ n & e & f\\ p & h & j\\ \end {vmatrix} ,\quad |A_2|= \begin {vmatrix} a & m & c\\ d & n & f\\ g & p & j\\ \end {vmatrix} ,\quad |A_3|= \begin {vmatrix} a & b & m\\ d & e & n\\ g & h & p\\ \end {vmatrix} \]

Then \[x=\frac {|A_1|}{|A|},\quad y=\frac {|A_2|}{|A|},\quad z=\frac {|A_3|}{|A|}\]

1.

Find the determinants of the following matrices \[B= \begin {pmatrix} 1 & -1 & 1\\ 0 & 2 & -1\\ -1 & 3 & 0\\ \end {pmatrix} ,\qquad C= \begin {pmatrix} 3 & 2 & -1\\ 4 & 5 & 5\\ 1 & -3 & 2\\ \end {pmatrix} \]

2.

Solve the following system of equations:

(a)

\( \begin {matrix} x & + & 3y & + & 5z & =4\\ -2x & + & 2y & + &4z & =3\\ 5x & + & y & + & 3z & =2\\ \end {matrix} \)

(b)

\( \begin {matrix} x & + & 2y & + & 3z & =2\\ & - & 4z & + & 2y & = -6\\ y & - & x & + & 2z & = 2\\ \end {matrix} \)

3.

Use Cramer’s Rule to solve the following:

(a)

\( \begin {matrix} x & + & 2y & + & 3z & =11\\ 3x & + & 8y & + & 5z & =27\\ -x & + & y & + & 2z & = 2\\ \end {matrix} \)

(b)

\( \begin {matrix} x & - & 2y & + & 3z & = 9\\ 2x & + & 3y & - & z & =4\\ x & + & 5y & - & 4z & =2\\ \end {matrix} \)

4.

(a)

Find the value of \(P\) such that the matrix \(B= \begin {pmatrix} 1 & -1 & 0\\ 0 & P(1 +P^2) & 4\\ 5 & 2 & 1\\ \end {pmatrix} \) is a singular matrix.

(b)

Given that \( A= \begin {pmatrix} 3 & 1\\ 0 & 7\\ \end {pmatrix} , \quad B= \begin {pmatrix} 2 & -1 & 0\\ 4 & 6 & -1\\ \end {pmatrix} \), find \(AB\).

5.

(a)

Given that the matrix \(A= \begin {pmatrix} 1 & -1 & 1\\ 3 & -9 & 5\\ 1 & -3 & 3\\ \end {pmatrix} \) evaluate \(A^{-1}\).

(b)

Use Cramer’s rule to solve the system of equations

i.

\( \begin {matrix} x & - & 2y & + & z & =-1\\ 2x & + & y & - & 3z & = 8\\ && 2y & + & z & =1\\ \end {matrix} \)

ii.

\( \begin {matrix} x & - & y & + & 2z & =-5\\ &&-x & + & 3z & =0\\ && 2x & + & y & = 1\\ \end {matrix} \)

6.

(a)

Let \(P= \begin {pmatrix} 6 & a & 1\\ -1 & 4 & b \\ 0 & 1 & c\\ \end {pmatrix} \) and \( Q= \begin {pmatrix} 4\\ 2\\3\\ \end {pmatrix} \).

i.

If \(\det P =2\), find the determinant of the \[ \begin {pmatrix} 6 & a & 1\\ 6 & 1+a & 1+c\\ -1 & 4 & b\\ \end {pmatrix} \]

ii.

Find \(PQ\).

(b)

Find the adjoint of the matrix \[B= \begin {pmatrix} 3 & 1 & 2\\ -1 & 1 & -1\\ -4 & 3 & 8\\ \end {pmatrix} \]

7.

Given the following matrices \(A= \begin {pmatrix} -1 & 2 & 4\\ 3 & 1 & 3\\ \end {pmatrix} \) and \(B= \begin {pmatrix} 1 & 2 & 3\\ 2 & 1 & 1\\ \end {pmatrix} \).
Find the following:

(a)

\(2A +3B\)

(b)

\(A^tB\)

(c)

Inverse of \(B^tA\)

8.

Given the following system of linear equations,
\( \begin {matrix} x & + & z & - & y & = -1\\ 5z & - & 9y & +& 3x & = 8\\ && z & + & 2y & =1\\ \end {matrix} \)

(a)

Write the system of linear equation in a matrix form.

(b)

Solve the system of linear equations using the inverse method.

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