Notes On Application of Determinants and Matrices - CBSE Class 12 Maths
Solution of a system of linear equations using inverse of a matrix Consider the system of equations a1x + a2y + a3z = k1 b1x + b2y + b3z = k2 c1x + c2y + c3z = k3 In matrix form, the system of equations can be expressed as $\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$  $\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$  = $\left[\begin{array}{c}{\text{k}}_{\text{1}}\\ {\text{k}}_{\text{2}}\\ {\text{k}}_{\text{3}}\end{array}\right]$        …(i) Let A = $\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$  ; X = $\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$  and B = $\left[\begin{array}{c}{\text{k}}_{\text{1}}\\ {\text{k}}_{\text{2}}\\ {\text{k}}_{\text{3}}\end{array}\right]$ From (i), we have AX = B Case (i): A is a non-singular matrix. ⇒ A-1 exists. We have AX = B ⇒ A-1 (AX) = A-1B ⇒ (A-1 A)X = A-1B (by associative property) ⇒ (I)X = A-1B ('.' A-1A = I , inverse property) ⇒ X = A-1B ⇒ X = A-1B ('.' IX = X , identity property) This method of solving a system of linear equations is known as the matrix method. Case (ii): A is a singular matrix. ⇒ |A| = 0 We have X = A-1B ⇒ X = (adj A / Det A)B Sub-case (i): (Adj A)B is a non-zero matrix. In this case, the system of linear equations is inconsistent. In other words, the system of linear equations has no solution. Sub-case (ii): (Adj A)B is a zero matrix. In this case, the system of linear equations has infinitely many solutions or no solution at all.

#### Summary

Solution of a system of linear equations using inverse of a matrix Consider the system of equations a1x + a2y + a3z = k1 b1x + b2y + b3z = k2 c1x + c2y + c3z = k3 In matrix form, the system of equations can be expressed as $\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$  $\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$  = $\left[\begin{array}{c}{\text{k}}_{\text{1}}\\ {\text{k}}_{\text{2}}\\ {\text{k}}_{\text{3}}\end{array}\right]$        …(i) Let A = $\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$  ; X = $\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$  and B = $\left[\begin{array}{c}{\text{k}}_{\text{1}}\\ {\text{k}}_{\text{2}}\\ {\text{k}}_{\text{3}}\end{array}\right]$ From (i), we have AX = B Case (i): A is a non-singular matrix. ⇒ A-1 exists. We have AX = B ⇒ A-1 (AX) = A-1B ⇒ (A-1 A)X = A-1B (by associative property) ⇒ (I)X = A-1B ('.' A-1A = I , inverse property) ⇒ X = A-1B ⇒ X = A-1B ('.' IX = X , identity property) This method of solving a system of linear equations is known as the matrix method. Case (ii): A is a singular matrix. ⇒ |A| = 0 We have X = A-1B ⇒ X = (adj A / Det A)B Sub-case (i): (Adj A)B is a non-zero matrix. In this case, the system of linear equations is inconsistent. In other words, the system of linear equations has no solution. Sub-case (ii): (Adj A)B is a zero matrix. In this case, the system of linear equations has infinitely many solutions or no solution at all.

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