We have matrix A = [(2,-3,5)(3,2,-4)(1,1,-2)] find A^(−1) and hence using A^(−1) , solve the system of equations :

Question

if A = \(\begin{bmatrix} 2& -3 & 5 \\[0.3em] 3 & 2 & -4 \\[0.3em] 1 & 1 & -2 \end{bmatrix}\) find A⁻¹ and hence using A⁻¹ , solve the system of equations : 

2x − 3y + 5z = 11 

3x + 2y − 4z = −5 

And x + y − 2z = −3.

Answer 1

We have matrix A =\(\begin{bmatrix} 2& -3 & 5 \\[0.3em] 3 & 2 & -4 \\[0.3em] 1 & 1 & -2 \end{bmatrix}\).

Now, determinant of matrix A is \(\begin{bmatrix} 2& -3 & 5 \\[0.3em] 3 & 2 & -4 \\[0.3em] 1 & 1 & -2 \end{bmatrix}\) = 2 \(\begin{vmatrix} 2& -4 \\[0.3em] 1 & -2 \end{vmatrix}\) - (-3) \(\begin{vmatrix} 3& -4 \\[0.3em] 1 & -2 \end{vmatrix}\) + 5\(\begin{vmatrix} 3& 2 \\[0.3em] 1 &1 \end{vmatrix}\)

= 2(−4 + 4) + 3(−6 + 4) + 5(3 − 2) = 2 × 0 + 3 × −2 + 5 × 1 = 0 − 6 + 5 = −1.

(By expanding determinant along row R₁ and \(\begin{vmatrix} a& b \\[0.3em] c &d \end{vmatrix}\) = ad - bc)

Hence, |A| = −1 ≠ 0. i.e., the inverse of matrix exists. 

We know that the cofactor of element a_ij of matrix A is given by A_ij= (−1)^i+j M_ij, where M_ij is minor of element a_ij of matrix A.

Now, the cofactor of element a₁₁ is A₁₁ = (−1)¹⁺¹ \(\begin{vmatrix} 2& -4 \\[0.3em] 1 &-2 \end{vmatrix}\) = (−4 + 4) = 0, 

The cofactor of element a₁₂ is A₁₂ = (−1)¹⁺² \(\begin{vmatrix} 3& -4 \\[0.3em] 1 &-2 \end{vmatrix}\) = −(−6 + 4) = −(−2) = 2, 

The cofactor of element a₁₃ is A₁₃ = (−1)¹⁺³ \(\begin{vmatrix} 3& 2 \\[0.3em] 1 &1 \end{vmatrix}\) = (3 − 2) = 1, 

The cofactor of element a₂₁ is A₂₁ = (−1) 2+1 \(\begin{vmatrix} -3& 5 \\[0.3em] 1 &-2 \end{vmatrix}\) = −(6 − 5) = −1, 

The cofactor of element a₂₂ is A₂₂ = (−1)²⁺² \(\begin{vmatrix} 2& 5 \\[0.3em] 1 &-2 \end{vmatrix}\) = (−4 − 5) = −9, 

The cofactor of element a₂₃ is A₂₃ = (−1)²⁺³ \(\begin{vmatrix} 2& -3 \\[0.3em] 1 &1 \end{vmatrix}\) = −(2 + 3) = −5, 

The cofactor of element a₃₁ is A₃₁ = (−1)³⁺¹ \(\begin{vmatrix} -3& 5 \\[0.3em] 2 &-4 \end{vmatrix}\) = (12 − 10) = 2,

The cofactor of element a₃₂ is A₃₂ = (−1)³⁺² \(\begin{vmatrix} 2& 5 \\[0.3em] 3 &-4 \end{vmatrix}\) = −(−8 − 15) = −(−23) = 23, 

The cofactor of element a₃₃ is A₃₃ = (−1)³⁺³ \(\begin{vmatrix}2& -3 \\[0.3em] 3 & 2 \end{vmatrix}\) = (4 + 9) = 13.

Therefore, the cofactor matrix of matrix A is \(\begin{bmatrix} 0& 2 & 1 \\[0.3em] -1 & -9 & -5 \\[0.3em] 2 &23 & 13 \end{bmatrix}\)

Now, the adjoint of matrix which is the transpose of cofactor matrix of matrix A

Therefore, a_dj A = \(\begin{bmatrix} 0& 2 & 1 \\[0.3em] -1 & -9 & -5 \\[0.3em] 2 &23 & 13 \end{bmatrix}'\) = \(\begin{bmatrix} 0& -1 & 2 \\[0.3em] 2 & -9 &23 \\[0.3em] 1 &-5 & 13 \end{bmatrix}\).

Therefore, the inverse matrix of matrix A is A⁻¹ \(\frac{adj A}{|A|}\) = \(\frac{1}{-1}\)\(\begin{bmatrix} 0& -1 & 2 \\[0.3em] 2 & -9 &23 \\[0.3em] 1 &-5 & 13 \end{bmatrix}\) = \(\begin{bmatrix} 0& 1 & -2 \\[0.3em] -2 & 9 &-23 \\[0.3em] -1 &5 & -13 \end{bmatrix}\)

Now, given system of equations are 

2x − 3y + 5z = 11 

3x + 2y − 4z = −5 

And x + y − 2z = −3.

The matrix form of given system of equations is AX = b, where A = \(\begin{bmatrix} 2& -3 & 5 \\[0.3em] 3 & 2 & -4 \\[0.3em] 1 & 1 & -2 \end{bmatrix}\), X = \(\begin{bmatrix} x \\[0.3em] y \\[0.3em] z \end{bmatrix}\)and b = \(\begin{bmatrix} 11 \\[0.3em] -5 \\[0.3em] -3 \end{bmatrix}.\)

Since, AX= b ⇒ X = A⁻¹ b = \(\begin{bmatrix} 0& 1 & -2 \\[0.3em] -2 & 9 &-23 \\[0.3em] -1 &5 & -13 \end{bmatrix}\)\(\begin{bmatrix} 11 \\[0.3em] -5 \\[0.3em] -3 \end{bmatrix}\) 

 \(\begin{bmatrix} 0& -5 +6 \\[0.3em] -22 & -45 +69 \\[0.3em] -11&-25+69 \end{bmatrix}\) = \(\begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\)

Hence, X = \(\begin{bmatrix} x \\[0.3em] y \\[0.3em] z \end{bmatrix}\) = \(\begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\)

Hence, the solution of given system of equations is x =1, y = 2 and z = 3.