Suppose the vectors x1, x2 and x3 are the
solutions of the system of linear equations,
Ax = b when the vector b on the right side is equal to b1, b2 and b3 respectively. if
${x_1} = \left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right]$, ${x_2} = \left[ {\matrix{
0 \cr
2 \cr
1 \cr
} } \right]$, ${x_3} = \left[ {\matrix{
0 \cr
0 \cr
1 \cr
} } \right]$
${b_1} = \left[ {\matrix{
1 \cr
0 \cr
0 \cr
} } \right]$, ${b_2} = \left[ {\matrix{
0 \cr
2 \cr
0 \cr
} } \right]$ and ${b_3} = \left[ {\matrix{
0 \cr
0 \cr
2 \cr
} } \right]$,
then the determinant of A is equal to :
Solution
Let A = $$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right]$$
<br><br>For Ax<sub>1</sub> = b<sub>1</sub> :
<br><br>$\Rightarrow$ $$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right]\left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right] = \left[ {\matrix{
1 \cr
0 \cr
0 \cr
} } \right]$$
<br><br>$\therefore$ ${a_1} + {a_2} + {a_3} = 1$ ....(1)
<br><br>${a_4} + {a_5} + {a_6} = 0$ ......(2)
<br><br>${a_7} + {a_8} + {a_9} = 0$ .....(3)
<br><br>For Ax<sub>2</sub> = b<sub>2</sub> :
<br><br>$$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right]\left[ {\matrix{
0 \cr
2 \cr
1 \cr
} } \right] = \left[ {\matrix{
0 \cr
2 \cr
0 \cr
} } \right]$$
<br><br>$\therefore$ $2{a_2} + {a_3} = 0$ .....(4)
<br><br>$2{a_5} + {a_6} = 2$ ....(5)
<br><br>$2{a_8} + {a_9} = 0$ ....(6)
<br><br>For Ax<sub>3</sub> = b<sub>3</sub> :
<br><br>$$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right]\left[ {\matrix{
0 \cr
0 \cr
1 \cr
} } \right] = \left[ {\matrix{
0 \cr
0 \cr
2 \cr
} } \right]$$
<br><br>$\therefore$ ${{a_3} = 0}$
<br><br>${{a_6} = 0}$
<br><br>${{a_9} = 2}$
<br><br>Putting value of ${{a_3}}$ in equation (4), we get
<br><br>${{a_2}}$ = 0
<br><br>Putting value of ${{a_6}}$ in equation (5), we get
<br><br>${{a_5}}$ = 1
<br><br>Putting value of ${{a_9}}$ in equation (6), we get
<br><br>${{a_8}}$ = -1
<br><br>Putting value of ${{a_2}}$ and ${{a_3}}$ in equation (1), we get
<br><br>${{a_1}}$ = 1
<br><br>Putting value of ${{a_5}}$ and ${{a_6}}$ in equation (6), we get
<br><br>${{a_4}}$ = -1
<br><br>Putting value of ${{a_5}}$ and ${{a_6}}$ in equation (6), we get
<br><br>${{a_4}}$ = -1
<br><br>Putting value of ${{a_8}}$ and ${{a_9}}$ in equation (6), we get
<br><br>${{a_7}}$ = -1
<br><br>$\therefore$ A = $$\left[ {\matrix{
1 & 0 & 0 \cr
{ - 1} & 1 & 0 \cr
{ - 1} & { - 1} & 2 \cr
} } \right]$$
<br><br>So, |A| = 2(1) = 2
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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