Let A=$\begin{bmatrix}
1 & 3 & 2 \\
2 & 5 &t \\
4&7-t&-6 \\ \end{bmatrix}$, then the values of t for which inverse of A does not exist
- -2, 1
- 44622
- 2, -3
- 3, -1
The Correct Option is C
Solution and Explanation
The inverse of a matrix does not exist when its determinant is zero.
So,
\(\begin{vmatrix} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{vmatrix} = 0\)
Expanding along the first row:
\((1)\begin{vmatrix}5 & t \\ 7-t & -6\end{vmatrix} - 3\begin{vmatrix}2 & t \\ 4 & -6\end{vmatrix} + 2\begin{vmatrix}2 & 5 \\ 4 & 7-t\end{vmatrix} = 0\)
Now compute minors:
\(1[(5)(-6) - t(7-t)] - 3[(2)(-6) - 4t] + 2[(2)(7-t) - 20] = 0\)
Simplifying:
\((-30 - 7t + t^2) - 3(-12 - 4t) + 2(14 - 2t - 20) = 0\)
\(-30 - 7t + t^2 + 36 + 12t - 12 - 4t = 0\)
\(t^2 + t - 6 = 0\)
Factorising:
\((t+3)(t-2)=0\)
Hence,
\(t = 2 \text{ or } -3\)
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Concepts Used:
Invertible matrices
A matrix for which matrix inversion operation exists, given that it satisfies the requisite conditions is known as an invertible matrix. Any given square matrix A of order n × n is called invertible if and only if there exists, another n × n square matrix B such that, AB = BA = In, where In is an identity matrix of order n × n.
For example,
It can be observed that the determinant of the following matrices is non-zero.
