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This is the second of our many quizzes on 3 x 3 systems. Click on this to access more such competitive quizzes on Algebra.

3 x 3 systems Quiz 2

mathematics

 Find the value of $\left| A \right|$ for the matrix $\left( {\begin{array}{*{20}{c}}  2&3&0 \\   4&2&0 \\   5&1&0 \end{array}} \right)$.

 When the given matrix is  such$\left( {\begin{array}{*{20}{c}}  {{a_{11}}}&{{a_{12}}}&0 \\   {{a_{21}}}&{{a_{22}}}&0 \\   {{a_{31}}}&{{a_{32}}}&0 \end{array}} \right)$ ,then its determinant is $\left| A \right| = 0$ 

  $ {\text{Given }} \\  {\text{A}} = \left[ {\begin{array}{*{20}{c}}  2&3&0 \\   4&2&0 \\   5&1&0 \end{array}} \right] \\  \left| A \right| = 2\left( {0 - 0} \right) - 3\left( {0 - 0} \right) - 0\left( {4 - 10} \right) \  \left| A \right| = 0 \\  $

 Solve the system $:\left\{ {\begin{array}{*{20}{c}}  {2x + y - z = 0} \\   {4x - y + 3z = 12} \\   {3x - 2y + z = 5} \end{array}} \right.$ for $x$, $y$ and $z$.

 Apply the property ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)$ and solve.

  $ \text{ Let, }{\text{A =  }}\left[ {\begin{array}{*{20}{c}}  2&1&{ - 1} \\   4&{ - 1}&3 \\   3&{ - 2}&1 \end{array}} \right],{\text{ X = }}\left[ {\begin{array}{*{20}{c}}  x \\   y \\   z \end{array}} \right],{\text{ B = }}\left[ {\begin{array}{*{20}{c}}  0 \\   {12} \\   5 \end{array}} \right] \\  \left| A \right|{\text{  =  2}}\left( { - 1 + 6} \right) - 1\left( {4 - 9} \right) + \left( { - 1} \right)\left( { - 8 + 3} \right) \\   = 10{\text{  + 5 + 5 = 20}} \\  {\text{adj}}\left| A \right|{\text{  =  }}\left[ {\begin{array}{*{20}{c}}  5&1&2 \\   5&5&{ - 10} \\   { - 5}&7&{ - 6} \end{array}} \right] \\  {A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right){\text{              [}}\because {\text{By using co - facter]}} \\  {\text{ = }}\dfrac{1}{{20}}\left| {\begin{array}{*{20}{c}}  5&1&2 \\   5&5&{ - 10} \\   { - 5}&{7 - }&{ - 6} \end{array}} \right| \\  \left[ {\begin{array}{*{20}{c}}  x \\   y \\   z \end{array}} \right] = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}}  5&1&2 \\   5&5&{ - 10} \\   { - 5}&7&{ - 6} \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  0 \\   {12} \\   5 \end{array}} \right] \\   = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}}  {0 + 12 + 10} \\   {0 + 60 - 50} \\   {0 + 84 - 30} \end{array}} \right] = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}}  {22} \\   {10} \\   {54} \end{array}} \right] \\  X = {\text{ 1}}{\text{.1, Y = 0}}{\text{.5, Z = 2}}{\text{.7}} \\  $

 Write the augmented form of system $\left\{ {\begin{array}{*{20}{c}}  {2x + 2y - z = 2} \\   {x - z = 5} \\   {2y + 3z = 4} \end{array}} \right.$

$  \text{Let, }  A = \left[ {\begin{array}{*{20}{c}}  2&2&{ - 1} \\   1&0&{ - 1} \\   0&2&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  2 \\   5 \\   4 \end{array}} \right] \\  {\text{Apply }}{{\text{R}}_1} \leftrightarrow {R_2} \\  \therefore {\text{ A = }}\left[ {\begin{array}{*{20}{c}}  1&0&{ - 1} \\   2&2&{ - 1} \\   0&2&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  5 \\   2 \\   4 \end{array}} \right] \\  {\text{Apply }}{{\text{R}}_2} \to {{\text{R}}_2}{\text{ - 2}}{{\text{R}}_1} \\  \therefore {\text{A = }}\left[ {\begin{array}{*{20}{c}}  1&0&{ - 1} \\   0&2&1 \\   0&2&3 \end{array}} \right]{\text{ }}\left[ {\begin{array}{*{20}{c}}  5 \\   { - 8} \\   4 \end{array}} \right] \\  {\text{Apply }}{{\text{R}}_2} \to \dfrac{{{R_2}}}{2} \\  A = \left[ {\begin{array}{*{20}{c}}  1&0&{ - 1} \\   0&1&{\dfrac{1}{2}} \\   0&2&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  5 \\   { - 4} \\   4 \end{array}} \right] \\ {\text{ Apply }}{{\text{R}}_3} \to {R_3} - 2{R_2} \\  A = \left[ {\begin{array}{*{20}{c}}  1&0&{ - 1} \\   0&1&{\dfrac{1}{2}} \\   0&0&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  5 \\   { - 4} \\   {12} \end{array}} \right] \\  {\text{Apply }}{{\text{R}}_3} = \dfrac{{{R_3}}}{2} \\  \left[ {\begin{array}{*{20}{c}}  1&0&{ - 1} \\   0&1&{\dfrac{1}{2}} \\   0&0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  5 \\   { - 4} \\   6 \end{array}} \right] \\  \therefore x - z = 5 \\  y + \dfrac{z}{2} =  - 4 \\  z = 6 \\  \therefore x = 11,{\text{ y  =  - 7, z  =  6}} \\  $

 What is the value ${a_{31}}{A_{11}} + {a_{32}}{A_{12}} + {a_{33}}{A_{13}}$ if the system $\left\{ {\begin{array}{*{20}{c}}  {x + y - 3z = 9} \\   { - x + 2y = 6} \\   {x - y + z =  - 5} \end{array}} \right.$.

 Find the values of the cofactors and substitute. 

 $  {\text{Let }}A = \left[ {\begin{array}{*{20}{c}}  1&1&{ - 3} \\   { - 1}&2&0 \\   1&{ - 1}&1 \end{array}} \right],x = \left[ {\begin{array}{*{20}{c}}  x \\   y \\   z \end{array}} \right]{\text{and }}B = \left[ {\begin{array}{*{20}{c}}  9 \\   6 \\   { - 5} \end{array}} \right] \\   \Rightarrow {A_{11}} = 2,{A_{12}} = 1{\text{ and }}{A_{13}} =  - 1 \\  {a_{31}} = 1,{a_{32}} =  - 1{\text{ and }}{a_{33}} = 1 \\  {a_{31}}{A_{11}} + {a_{32}}{A_{12}} + {a_{33}}{A_{13}} = 2 \times 1 + 1 + \left( { - 1} \right) + \left( { - 1} \right) \times 1 \\   \Rightarrow 2 - 1 - 1 \Rightarrow 0 \\  $

 Write ${A^{ - 1}}$ of the system $\left\{ {\begin{array}{*{20}{c}}  {x - 3y + 4z = 5} \\   {2x + y - z = 7} \\   {3x + 5y - 2z = 9} \end{array}} \right.$.

 Find the value of $\left| A \right|$. If $\left| A \right| = 0$ this means ${A^{ - 1}}$ exists.  If $\left| A \right|$ is not equal to zero this means ${A^{ - 1}}$ exists.

 $  {\text{Let }}A = \left[ {\begin{array}{*{20}{c}}  1&{ - 3}&{ - 4} \\   2&1&{ - 1} \\   3&5&{ - 2} \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}}  x \\   y \\   z \end{array}} \right]{\text{ and }}Z = \left[ {\begin{array}{*{20}{c}}  5 \\   7 \\   9 \end{array}} \right] \\  \left| A \right| = 1\left( { - 2 + 5} \right) + 3\left( { - 4 + 3} \right) - 4\left( {10 - 3} \right) \\  \left| A \right| = 3 - 3 - 28 \\  \left| A \right| =  - 28 \\  \therefore {A^{ - 1}}{\text{ does not exist}} \\  $